Electric Current In Conductors

With three resistors if equal value, there can be the following combinations of resistors:

A. All resistors in series,

With, all resistors in series the combined resistance will be,

B. All resistors in parallel:

With all resistors in parallel, the combination of resistors will provide,

C. Two resistors in parallel and one resistor in series

When two resistors are in parallel, their resistance combined will be,

And when this resistance when combined with the resistor in series we have,

D. Two resistors in series and one resistor in parallel

When two resistors are in series, their resistance combined will be,

And when this resistance when combined with the resistor in parallel we have

Therefore, by combining the resistances we will have a resistances of 10 , 15 , 45 and 20 .

Electric current flows in the direction of moving charges and hence the beam will constitute an electric current in the east to west direction since the charge is positive.

Since there is a movement of charged particles, a current will be developed and the current will be in the direction of positive ions i.e. from left to right.

In a TV tube, when the electrons move from rear to the front, they are moving from cathode to the anode, i.e. from negative electrode to positive electrode and therefore the current will be in the direction of anode to cathode i.e. from front to rear.

Since an electron, during a drift, travels in a very random and discontinuous path, we have to take an average of the distances over a long interval of time, and since limiting ∆t ⇒ 0 corresponds to a very short time interval we don’t define drift velocity as such.

The fallacy in the argument of the friend is that in the previous chapters the focus was on stationary charges and not moving charges and hence with no moving charges there will be no electric field inside a conductor and hence no current. However, in case of moving charges or current there will be an electric field due to the applied potential difference.

No, since the free electrons move throughout the entire circuit the electrons, those entering the conductor from the circuit replenish which leave the conductor, simultaneously. Hence, the charge neutrality of the conductor is maintained.

Copper has a resistivity of 1.72 x 10^-8 Ωm whereas aluminum has a resistivity of 2.82 x10^-8 Ωm. Since aluminum has a higher resistivity, more power is lost in heat than used to run the fan than copper and hence copper wound motor fans consume less fans than aluminum wound motor fans.

In the expression

both V and i are time dependent quantities and hence we cannot say that U is proportional to i. However, in the expression

only i is the time dependent quantity and hence we can say that U is proportional to i².

The work done on the resistor by the battery is dissipated as heat, which is the thermal energy, developed by the resistor and hence both the phrases represent the same physical quantity.

In case of real battery, the battery also has some internal resistance and hence the work done is equal to the thermal energy developed due to both resistances in the circuit and the internal resistance of the battery. In case of a capacitor, the work done by the battery is stored as electrical energy in the capacitor and not thermal energy.

A non-ideal battery has internal energy and hence the work done is equal to the thermal energy developed due to both resistances in the circuit and the internal resistance of the battery.

If the battery is ideal, then there will be no internal resistance and the work done will solely be the thermal energy developed in the resistor.

Yes, the statement is correct. The electric current flowing inside the resistor increases the potential energy of the resistor, which increases the temperature of the resistor to be greater than the surroundings. This temperature difference is the cause of the heat.

When a current is going through a wire, charges are flowing through the wire, which constitute electric current.

A Potentiometer will be preferred to measure the emf of a battery as it uses the null pointer method, hence draws very little to no current from the circuit, and hence gives an accurate measure of the emf. In a voltmeter the equivalent resistance of the circuit changes and hence the potential difference to be measured changes. To minimize this change the voltmeter resistance needs to be very high.

When a current pass through a circuit, the free electrons in the valence band of the conductor jump to the conduction band and drift throughout the conductor and as no extra electrons are provided to the conductor, the conductor does not get charged.

No, the potential difference across a battery cannot be greater than the emf as the emf is the maximum potential difference across the terminals of the battery. The potential across the battery, however, drops when connected to a circuit due to the drop across the internal resistance of the battery.

If the number of collisions of the free electrons with the lattice is decreased then the drift velocity of the electrons will increase.

The current is given by

Where

I is the current

n is the number of electrons

e is the charge of an electron

A is the area of cross section of a conductor

is the drift velocity

From the above formula current i is directly proportional to drift velocity. So when the drift velocity is increased then the current will increase.

If the number of collisions of the free electrons is decreased then the current will increase. Option A is correct.

Given :

Resistance of resistor A, R_A< Resistance of resistor B, R_B

Formula used : Resistance is given by the formula

Where

R is the resistance

ρ is the resistivity

l is the length

A is the area of cross section

Resistance is

Resistance is

Only the relation between resistance values of and is given. R is depends on ρ, l and With the information given we cannot conclude the relation between resistivity of the two resistors.

If information about ρ, l and A is given then we can say the relation between ρ_A and ρ_B. So option D is correct.

The resistivity of a conductor is given by

Where

is the resistivity of the conductor

is the conductivity of the conductor

The product of resistivity and conductivity is

The product of ρ and σ is unity. So it does not depends on anything. So option D is correct.

Resistance is depends on temperature. If temperature is increase, resistivity will increase. Increase in resistivity will leads to decrease in conductivity. The relation is given below.

Where →resistivity

σ →Conductivity

The product of resistivity and conductivity is not depend on temperature. Because resistivity and conductivity of a metallic resistor are nullifying the change in the temperature.

If the temperature of a metallic resistor is increased, the product of its resistivity and

Conductivity may increase or decrease. Option D is correct.

Battery is connected to an electric circuit. We don’t know what type of circuit it is. May be the battery is charging or discharging. Generally the flow of electrons is opposite to the direction of flow of positive charge. While discharging of a battery the positive charge will flows from negative terminal to positive terminal. And in charging the positive charge will flows from positive terminal to negative terminal.

Conclusion : If a battery is connected to an electric circuit the charge(positive charge) may go from positive terminal to negative terminal(in charging) or go from negative terminal to positive terminal(in discharging). So option B is correct.

Resistance R is connected to a ideal battery. Internal resistance of an ideal battery is zero. So it provides constant potential difference between two terminals.

Power dissipated by the resistor is given by

Power,

Where

V is the voltage or potential difference

R is the Resistance of the resistor

Power P is inversely proportional to Resistance R. If the value of R is decreased, the value of power P will be increase.

A resistor R is connected to an ideal battery. If the value R is decreased, the power dissipated in the resistor will increase. So option A is correct.

Kinetic energy of electrons= K₁

Kinetic energy of metal ions = K₂

Electrons are free to move and metal ions are bounded at their positions and can’t move freely as electrons. Due to thermal energy metal ions are just vibrate due to collision with electrons. When a current is passes through a resistor, because of the free movement the velocity of the electrons is greater than the metal ions.

Velocity of the electrons is grater than the metal ions. So kinetic energy of electrons is greater than the kinetic energy of metal ions.

Thus, option C is the correct option.

Two resistors are connected in series. In series, the current flows through both the resistors is same.

Formula used:

Energy is given by the formula

Energy dissipated,

Where

I is the current

R is the resistance

T is the time period

Energy developed in the resistor R is

Energy developed in the resistor 2R is

The ratio of energy developed in R and 2R is =

When two resistors R and 2R are connected in series then the ratio of thermal energy developed is 1:2.

Conclusion : Two resistors R and 2R are connected in series, the thermal energy developed in R and 2R are in the ratio 1:2 respectively. So option A is correct.

Two resistors are connected in parallel. In parallel connection voltage is same. The voltage across the two resistors is same

Where E is the energy dissipated

V is the voltage

R is the resistance

T is the time

Energy developed in the resistor R is

Energy developed in the resistor 2R is

The ratio of energy developed in R and 2R is

When two resistors R and 2R are connected in parallel the ratio of energy developed is 2:1. Option B is correct.

Formula used :

Resistance of a wire is given by

Resistance,

Where

ρ→ resistivity of the material

l→ length of the wire

A → cross sectional area of the wire

l is proportional to R.

When a uniform wire of resistance 50Ω is cut into 5 equal parts then resistance of each part is

10 Ω.

Now they are connected in parallel.

Equalent resistance is calculated as

Ω

A uniform wire of resistance 50 Ω is cut into 5 equal parts. If the parts are connected in parallel then equivalent resistance is Ω. So option A is correct.

Kirchhoff’s junction law : The sum of all the currents directed towards a node is equal to sum of all the currents leaving the same node. It follows from the conservation of charge where the charge neither be created nor be destroyed, but it just transfers from one point to another. The net quantity of charge is equal to positive charge minus negative charge.

Kirchhoff’s loop law : The algebraic sum of potential differences along a closed path in a circuit is zero. It follows from the Conservative nature of electric field. Electro static force is a conservative force and the work done by it in any closed path is zero.

Kirchhoff’s junction law follows from conservation of charge and Kirchhoff’s loop law follows from the conservative nature of electric field. So both statements A and B are correct. Option A is correct.

Let e₁ and e₂ be the emf of battery 1 and battery 2 respectively. And r₁ and r₂ be the internal resistance of battery 1 and battery 2 respectively.

Emf is nothing but the voltage across the terminals. The two batteries are connected in series.

The equivalent emf is e = e₁+e₂

The equivalent internal resistance r= r₁+r₂

: In series connection the equivalent emf and equivalent internal resistance are becomes larger. So statement A is correct and B is wrong. Option B is correct.

Let and be the emf of battery 1 and battery 2 respectively. And and be the internal resistance of battery 1 and battery 2 respectively.

The two batteries are connected in parallel.

Equivalent emf

The equivalent internal resistance

if two batteries are connected in parallel then the equivalent emf is larger than either of the two emfs and the equivalent internal resistance is smaller than either of the two internal resistances. Statement B is correct ad A is wrong. So option C is correct.

We always use ammeter in series to calculate the current drawn by the element from voltage source. If the net resistance of an ammeter is high, because of the series connection it will add up the net resistance of an ammeter. Then we can’t get the current values accurately. If net resistance of an ammeter is low, it only shows a very small changes in current to be measured those can be negligible.

The net resistance of the ammeter should be small. Option D is correct.

Voltmeter is always connected in parallel to measure the voltage or potential difference across the elements. When the voltmeter is connected across some element in a circuit, it will change the overall resistance in the circuit. It will effect the current values also.

To minimize the error voltmeter should have the large net resistance. Then large resistance in parallel with small resistor will have only a very slight change.

The net resistance of a voltmeter should be large to ensure that it does not appreciably change the potential difference to be measured. Option D is correct.

Formula used :

Charge of a capacitor is given by

Charge of a capacitor,

Where

ε is the emf of a battery

C is the capacitance

R is the resistance of a resistor which is in series

T is the time period

charge developed on the capacitor in first interval of 10ms is

charge developed on the capacitor in first interval of 20ms is

Charge developed on the capacitor in the interval 10ms to 20ms is

Compare with

(Here )

The time taking for 10μC to developed on the plates of capacitor is

The time taking for 10μC to developed on the plates of capacitor

Divide (1) by (2)

Take log on both sides

Q₁ is the charge developed on the capacitor in a time interval of 10 ms and Q₂ is the charge developed on the capacitor in the next time interval of 10ms. If a10μC charge be deposited in a time interval t₁ and the next 10 μC charge is deposited in the next time interval t₂. Then Q₁ > Q₂, t₁ < t_2. So option B is correct.

Electric field is in the direction form right to left.

Let velocity of an electron at stop A is

velocity of an electron at stop A is

Potentials are increased in the opposite direction of electric field.

So

Potential energy of the electron at point A is

Potential energy of the electron at point B is

Because

Kinetic energy of an electron at point A is

Kinetic energy of an electron at point B is

Applying conservation of mechanical energy , we get

(Mechanical energy =potential energy+kinetic energy)

(Because

Speed of electron is more at stop B than stop A. Option A is correct.

capacitor is connected to a battery at t = 0. We have to consider two points. Point A is on the connecting wires and B is in between the plates. The current flows through the battery up to the capacitor is fully charged. So there is a current through point A as long as the charging is not complete.

Capacitor has no dielectric. Without a medium, how can a current flows through the plates of a capacitor. So there is no current flow through the point B.

Conclusion : A capacitor with no dielectric is connected to a battery at t = 0, no current will pass through capacitor plates and there is only current pass through connecting wire which is used to connect the capacitor with battery up to capacitor is fully charged. So option B and C are correct.

When there is no current passing through a conductor then there is no charge will flow through conductor. So net charge will be zero. Because of no net charge all the electrons will be in a random motion. Due to random motion there is no net charge transfer. So the average velocity of a free electron over a large time period will be zero. If we see the average velocities of a free electrons it will also zero at an instant because of absence of net charge.

Conclusion : When no current is passed through a conductor. The average velocity of a free electron over a large period of time is zero and the average of the velocities of all the free electrons at an instant is zero. Option C and D are correct.

Resistor is heated due to the current flowing through it. Thermal energy is increase.

Formula used :

Where i→ current

A→ area of the cross section

n→ electrons per unit area.

→ Drift speed

R→ resistance

ρ→ resistivity

l→ length

A→ area of cross section

If thermal energy is increase the resistance will increase. Resistance is directly proportional to resistivity, so resistivity also increase. Increase in resistance leads to decrease in current. Current is directly proportional to drift velocity. So drift speed. So only the number of electrons remains same.

If the resistor connected to a battery is heated due to current flowing through it the drift speed, resistance, resistivity all are change except number of free electrons. Option D is correct.

Formula used : Resistivity is given by

Where is the resistivity

σ is the conductivity

When temperature of a conductor is increases, its resistivity will increase and conductivity will decrease.

Ration of resistivity to conductivity is

Conclusion : When the temperature of conductor increases, the ratio of resistivity to conductivity will increase. Option A is correct.

Formula used : current density is given by

current density ,

Where

i→ current

A→ area of the cross section

n→ electrons per unit area.

→ Drift speed

Current density is inversely proportional to area of the Ross section. So area of the cross section is depends on current density.

Drift speed is also inversely proportional to area of the cross section. So area of the cross section is depends drift speed.

It does not depends on free electron density and the charge crossing in a given time interval.

A current passes through a wire of non uniform cross-section. Then it does not depends on The charge crossing in a given time interval and free electron density. Option A and D are correct.

Ammeter is always connected in series with a circuit that the current to be measured. If ammeter has a large resistance, the net resistance will be high. It will effect the total measures of circuit. Then we can’t get the accurate values of the current drawn from the voltage source. If the ammeter have a small resistance, it will not show an appreciable change in net resistance.

Voltmeter is always connected in parallel with the element to measure the voltage across the element. Voltmeter should have a large resistance. If voltmeter have small resistance, definitely draw the current from the source. It is not supposed to draw any current from the source. It has to measure the potential difference across the element only.

An ammeter should have small resistance and a voltmeter should have large resistance. Option A, D are correct.

Given : C = 500 μF

R=10

Capacitor is connected to a battery thought a resistor. Initially capacitor starts charge. If it is get fully charged then it starts discharge through the element connected to it. Typically for the Charging or discharging the time constant is in the order of mille seconds. Generally for 99% of the charging of a capacitor 4 to 5 time constants are sufficient. Given that the charge stored on the capacitor is larger in the first 5s. So it larger than 5s, 50s, 500s and 600s.

A capacitor of capacitance 500 μF is connected to a battery through a 10kΩ resistor. The charge stored on the capacitor is large in the first 5s. All the options given are correct.

Given:

Both the capacitors are connected to a same battery forma long time. So capacitors fully charged. The two capacitors connected to an equal resistors separately at t=0. Now the capacitors starts discharge at the same time. So at t=0 the current in both discharging circuits are equal but not zero. Capacitors are in the ratio 1:2. So C₁ loses 50% of its initial charge sooner than C₂ loss 50% of its initial charge.

If the charged capacitors connected to the same resistance separately then the currents in the two discharging circuits at t = 0 are equal but not zero and C₁ loses 50% of its initial charge sooner than C₂ loss 50% of its initial charge. Option B and D are correct.

a) IT^-1, I, IT b)53A

Given,

Charge as a function of time is Q(t) = At² + Bt + C.

The principle of homogeneity states that each term on the either side of an equation has the same dimensions.

a) Each terms on the Right Hand Side of the equation has the same unit, and hence the dimension of that of the term on the Left Hand Side.

So, each term on RHS is having same dimensions as of the quantity Charge, Q.

We know that, Charge Q is

Where I is current with dimension ‘I’ and t is time in seconds with dimension ‘T’.

Hence the dimension of Q or Q(t)is ‘IT’.

By inspection, we can see that the term C in RHS is devoid of any other quantities and hence C also has the dimension ‘IT’ (Ans.)

We know that dimension of the term At² is also ‘IT’, and t represents time (Dimension T).

So,

Or

Similarly,

Or,

So dimensions of A, B, and C are IT^-1, I, IT respectively.

b) The expression for the charge at time t can be rewritten by assigning values to the constants as.

We know that instantaneous current, I can be expressed as

By substituting the given expression in the above equation, we get,

Or,

For t=5s, I becomes

Hence the current at t=5s is 53A

3.2×10^-3 A

Given,

The number of electrons emitted=2.0 × 10¹⁶

Time, t, in seconds in which 2.0 × 1016 electrons are emitted=1s

Formula Used:

The current flowing from an electron gun or through a circuit, I, due to the movement of charges, q, through it can be expressed as,

Where q is the charge flowing and t is time in seconds.

Also, for n number of electrons, q is

Where e is the charge of 1 electron= 1.6× 10^-19C

Hence, in the given problem, the total charge flowing from the gun is, q

And the corresponding current, I is

So the corresponding current is 3.2×10^-3 A.

6.0×10^-4C

Given,

The electric current in the tube= 2.0 μA

Time for which charge transfer is to be calculated= 5 min=300s

Formula used

The amount of charge, q, transferred in t seconds, with a current I is

Solution,

The discharge tube carries a current of 2.0 μA. So the charge transferred across the cross-section in 300s, by the above relation, is

So the charge transferred across the cross-section is 6.0×10^-4C

300C

Given,

The expression of current through the wire is

i = i₀ + αt

where i₀=10A, time for which current passes, t= 10s, and a=4 As^–1

Formula Used:

For a given current i, the charge q is expressed as

Solution,

For the given expression of current, charge q is,

Or,

On integration,

By substituting the given values,

Hence, the charge crossed through a section of the wire in 10 seconds is 300C.

0.074mm/s

Given,

Current in the wire, I= 1A

Cross section of the wire, A= 1mm²=10^-6m²

Density of Copper, d=9000 kg m^–3

Formula Used

The current due to ‘n’ freely bounded electrons per unit volume with a drift speed ‘V_d’ can be expressed as,

Where ‘A’ is the cross-sectional area of the material through which electrons are passing; and ‘e’ is the charge of the electron, which is

1.6×10^-19C.

In the given problem, n is not directly given. But we know that 63.5 grams of Copper have Avogadro number (6.022×10²³) of atoms. So ‘m’ Kilograms have,

Also, in terms of density, d, the mass m can be replaced in the above expression as,

And for Unit volume, the number of atoms are,

So, the number of free electrons/ atoms are,

From eqn.1, the expression for Drift Velocity, V_d

Substituting the known values, it becomes,

Hence the drift speed of free electrons is 0.074mm/s

Given,

Length of wire, l = 1m

Radius of wire, r=0.1mm=0.1×10^-3m

Resistance of the wire, R =100 Ω

Formula used,

The resistivity, ρ, of a wire with cross-sectional area A and length l is expressed as

Where R is the resistance offered by the wire.

Solution,

Firs, we find the area of cross-section of the wire as,

By substituting the value of r, Area becomes

Now substitute all the given values in the expression for resistivity. So, ρ is

So, the resistivity of the material is π×10^-6 Ωm

400 Ω

Given,

Initial Resistance, R₁ of the wire= 100 Ω

Initial length of wire=l₁, Final length of wire=l₂= 2 l₁

Formula used

The expression for the resistance, R, of a wire is

Where ρ is the resistivity, A is the area of cross-section and l is the length of the wire.

Hence, by knowing the final length and area we can calculate the final resistance by comparing it with that of the initial case. Also, the information that the volume of the wire would not change on recast, should be used.

Solution,

We know that the volume remains same after the recast.

If we represent the volume as a function of area and length, the above information can be expressed as,

Where subscripts ‘1’ and ‘2’ denotes the initial and final cases.

It is given that, l₂= 2 l₁, so the final area A₂ can be represented as,

We use this to compare the two resistance,

The initial resistance, R₁ is

Similarly, the final resistance, R₂ is

Since the material is the same in both cases, ρ₁= ρ₂. By dividing eqn.3 by eqn.2, we get,

By substituting the relation between length and area at the initial and final cases, the above expression can be re-written as,

Or,

Or,

the resistance of the wire is 400 Ω

8.9 hours

Given,

Number of free electron per unit volume, n=10²⁹

Area of cross-section, A=1 mm²=10^-6 m²

Length of wire, l= 4m

Current through the wire, I=2A

Formula used

We know that the expression for drift velocity, V_d of ‘n’ free electrons through a wire of cross-sectional area A, for a current ‘i' is

Where e is the charge of 1 electron= 1.6×10^-19 C

V_d can be expressed in terms of the travel length, l and travel time, t as,

Combining the above two expressions, we can write as,

Or,

Solution,

Substituting the given values in eqn.1, we get the time taken by the free electrons to cross the length as,

So the average time taken by an electron to cross the length of the wire is approximately 8.9 hours.

0.6km

Given,

Resistance of the wire, R=1000Ω

Resistivity of copper, = 1.7 × 10^–8 Ω m

Area of cross-section, A=0.01 mm²=0.01×10^-6 m²

Formula used,

The expression for the resistance, R, of a wire is

Where ρ is the resistivity, A is the area of cross-section and ‘l’ is the length of the wire.

Or the length of the wire for given area and Area of a cross section is,

Solution,

Substituting the values in eqn.2, we get the required length as,

Hence the length of the copper wire for the given resistance is 0.6km.

We have to find the resistance associated with the truncated cone as shown in the figure.

For that purpose, we take an elemental area of this uniformly increasing cone and find the resistance of that element. By integrating that value from radius a to b, we would be able to find the resistance associated with the truncated cone.

The expression for the resistance, R, of a cylinder with length ‘l’ and area of cross-section A is,

We assume that the cone is made up of an infinite number of cylinders with length ‘dx’.

So for that element, resistance dR is

Hence the total resistance R is

Now the resistance for an element with a radius ‘y’ at x distance from left with length ‘dx’ should be found out.

The mean area of the element is,

Hence eqn.1 becomes

From the figure, we can write a relation connecting x and y as,

Or,

Differentiating the above expression, we get,

On re-arranging

Substituting this value in eqn.1 and combining with eqn.3, we get,

On integration,

Or,

Or,

Or,

Hence the resistance of the truncated cone is

a) 1.25×10¹⁷ b) 6.37×10⁵ A/m²

Given,

Radius of the wire, r= 0.1mm=0.1×10^-3m

Resistance of the wire, R= 1000Ω

The voltage across the wire, V= 20V

Formula used,

The wire is having a resistance, R and hence the current, I through it can be found from the equation,

Where V denotes the potential difference across the resistance.

We know that the current, I through a resistor can be calculated from the charge, q flowing through it per seconds, or

Where t is the time in seconds for which the current is flown.

But the charge, q can be written in terms of number of electrons, n flowing by the relation,

Where e is the charge of 1 electron= 1.6×10^-19 C

Also, the current density, J through a material with constant area of cross section A, can be represented as

Solution,

a) From the above equations, eqn.1 can be re-written with the help of eqn.2 and eqn.3 as,

Or

For time t= 1s, and by substituting the given values in the above relation, we can calculate the number of electrons passed in 1s through the copper wire with resistance 1000Ω as,

Hence the number of electrons transferred per second between the supply and the wire at one end is 1.25×10¹⁷

b)

To find the current density, we have to calculate the area of cross-section, A of the wire with radius r. Hence

By substituting the value for r, we get the area as,

Also, the current through the wire, from eqn.1, is

Now, by substituting the known values in the expression for current density, eqn.4, we get J as

Hence the current density in the wire is around 6.37×10⁵ A/m²

8.5mVm^-1

Given,

Area of cross-section of wire, A=2.0 mm²=2×10^-6 m²

Current flowing through the wire, I=1A

Resistivity of copper, ρ= 1.7 × 10^–8 Ω m

Formula used,

From electrostatics, we know that Electric field, E is

Or it can be written as

Where l is the distance over which potential difference V has an effect

From Ohm’s law and the concept of current density, the relation connecting resistance, R and resistivity, ρ can be written as,

Where ‘l’ is the length of the conductor (or wire) and ‘A’ is the area of cross-section of the conductor.

Solution,

We know that potential difference, V is

Using eqn.2, the above expression can be modified as,

Now, in order to find the electric field by eqn.1, we can replace V in eqn.1 with the above expression as,

On simplification, it becomes,

We have all the values for the above expression. So, on substitution, Electric field will be,

Hence the electric field in a copper wire is 8.5mVm^-1

25Vm^-1

Given,

Length, l of the wire= 2m

Resistance, R of the wire=5 Ω

Current I, passing through the wire= 10A

Formula used,

From electrostatics, we know that Electric field, E is

Or it can be written as

Where l is the distance over which potential difference V has an effect

Solution,

We know that the potential difference, V in the wire with current I passing and with a resistance R is

By substituting the given values in the above expression, we get V as

Substituting this value of potential difference in eqn.1, we can find the electric field,

So,

Or

Hence the electric field existing inside the wire is 25Vm^-1

84.5°C

Given,

Resistance of iron wire, R_{Fe, i} at 20°C= 3.9 Ω

Resistance of Copper wire, R_{Cu, i} at 20°C = 4.1 Ω

Initial temperature of both the wires, T_i= 20°C

Temperature coefficient of resistivity for iron, α_Fe=5.0 × 10^–3 K^–1

Temperature coefficient of resistivity for copper, α_Cu=4.0 × 10^–3 K^–1

Formula used,

For most of the conducting materials, the relation connecting the resistance with the change in temperature can be represented as,

Where R_f is the final resistance after the change in temperature, R_i is the initial temperature, α is the Temperature coefficient of resistivity and ∆T is the change in temperature from the initial temperature of the material.

Solution:

We have the expression connecting the change in temperature and the resistance of the material.

First, let us assume that the final temperature is T_f at which the resistance of both the wires will be same.

So, change in temperature ∆T is

Now the final resistance of iron wire, R_{Fe, f} at temperature T_f can be written based on eqn.1 as,

Or by substituting the known values, we can write it as,

Similarly, the final resistance of iron wire, R_{Cu, f} at temperature T_f can be written based on eqn.1 as,

Or by substituting the known values, we can write it as,

At the final temperature, it is given that the resistance of both the wires are same. So we can equate eqn.3 and eqn.4.

So,

Or,

By solving,

Or,

Or,

We know that ∆T is the difference between final and initial temperature, and hence T_f from eqn.2 is

Or,

By substituting the known values, we get T_f as

Hence the temperature in which the resistances are equal is 84.5°C

0.4V

Given,

Ammeter reading of current, i₁ in the 1^st case=1.75A

Ammeter reading of current, i₂ in the 2^nd case=2.75A

Voltmeter reading for current i₁, V₁=14.4V

Voltmeter reading for current i₂, V₂=22.4V

Formula used

The ammeter shows the accurate reading while the voltmeter deviate from showing the corresponding potential difference value due to the presence of Zero error, V_e.

Hence the original potential difference can be calculated as, V_o

Where V is the shown voltage in the defected voltmeter.

We also know that the potential difference across the conductor is, V

Where R is the resistance of the conductor

Solution:

For the 1^st and 2^nd case, the relation between absolute current and voltage can be written, from eqn.2, as

And

By dividing these expressions, we get

The terms V_o1 and V_o2 in above equation can be replaced by eqn.1 as

By substituting the given values in the above expression, we get

Or,

Or,

Or,

Or,

Hence the zero error associated with the measurement is 0.4V.

1.52V, 0.07Ω

Given,

Voltage reading, V₁ with switch is open=1.52V

Voltage reading, V₂ with switch is closed=1.45V

Current through the ammeter, i= 1A

Formula used

Kirchhoff’s loop rule says that the algebraic sum of the voltage in a loop is always zero. Or

For a cell with an internal resistance r and emf ϵ, the voltage drop, V across it, if a current ‘i’ is passed through the circuit can be written using the loop rule as,

Solution:

a) When switch is open, the current would circulate through loop ABCA but not through the loop ACDEA. Since the internal resistance is very small compared to the resistance of Voltmeter, the voltage drop occurs completely across the voltmeter. This voltage drop will be measured in the meter and it will be almost equal to the emf of the cell.

Hence the e.m.f of cell= volt meter reading.

Or,

b) When the switch is closed, a current ‘i’ will flow through the loop ACDEA. Now the volt meter will show the potential drop across the cell and the internal resistance combined. So, using eqn.1 , we can find the internal resistance,

Where V=Volt meter reading= 1.45V, i= Ammeter reading=1A, and ϵ= 1.52V as we

By re-arranging, we can find the expression for internal resistance as,

By substituting the given values,

Hence the internal resistance of the cell is 0.07Ω

29Ω

Given,

The emf of the cell, E=6V

The value of internal resistance, r= 1 Ω

The volt meter reading across the setup, V=5.8V

The value of external resistance= R Ω

Formula used:

Figure shows the overview of the setup.

The voltage across the cell will be equal to that across the external resistance. So,

Solution:

The current flowing through the internal and external resistance are the same.

By substituting the given values in the eqn.1, we get i as

Or,

Or,

This current also circulate through the external resistance R. Hence from the eqn.1, we can write its expression as,

By substituting the values of ‘i’ and V, we get

Or,

Hence the resistance of the external resistor is 29Ω.

0.6Ω

Given,

The potential difference across the setup, V= 7.2V

The emf of the cell, E=6V

The current flowing through the circuit, i= 2A

Formula used:

When the battery is charging the potential difference, V across the cell can be written as,

Where ‘E’ is the emf of the cell and ‘r’ is internal resistance of the cell.

Solution:

As the battery is getting charged, the internal resistance can be calculated from the eqn.1, as

And by substituting the given values,

Hence the internal resistance of the cell is 0.6Ω

a)0.3A, b)3A

Given,

The emf of the battery, E=6V

The internal resistance of the battery, r₁ when discharged=10Ω

The internal resistance of the battery, r₂ when charged=1Ω

Potential difference provided by the charger, E_c=9V

Formula used:

When the accumulator battery is connected to a charger, the current through the internal resistance depends on the net emf available across the resistor.

Hence Net emf across the resistor in the case of charging will be the difference between the provided potential difference and the potential difference rating of the battery.

Solution:

a) When the battery is being charged, the net emf, E_net across the resistance, r₁ will be,

Or,

Hence from the relation, , we can find the current, i by substituting r₁ as 10 Ω,

Or,

b) When the battery is completely charged, the internal resistance r₂ will be 1Ω. The net emf across the resistance will be the same, and is 3V.

Hence from the relation, we can find the current, i through the resistance by substituting r₂ as 1Ω,

Or,

Hence the current through the internal resistance while charging and after completely charged are 0.3A and 3A.

a) 0.57, b)1, c)1.75

Formula used:

Kirchhoff’s loop rule state that the algebraic sum of all the voltages in a loop will be zero. Or,

Solution:

a)

Given resistances are: 1 Ω,1 Ω,0.1Ω

Applying Loop rule, we can write,

Or,

And i₁ will be,

In the next figure,

Let i be the current that passes through the middle branch and hence i₂-i will pass through the upper branch, as shown in figure.

By applying loop rule in AFEDA, we can write as,

Or,

Similarly, applying the loop rule in ADCBA, we can write as,

Or,

Replacing ‘i’ in eqn.1 using eqn.2, we get,

Or,

Or,

Now i₁/i₂ is,

b)

Given resistances are: 1 Ω,1 Ω,1Ω

Applying Loop rule, we can write,

Or,

And i₁ will be,

In the next figure,

Let i be the current that passes through the middle branch and hence i₂-i will pass through the upper branch, as shown in figure.

By applying loop rule in AFEDA, we can write as,

Or,

Similarly, applying the loop rule in ADCBA, we can write as,

Or,

Replacing ‘i’ in eqn.1 using eqn.2, we get,

Or,

Or,

Now i₁/i₂ is,

c)

Given resistances are: 1 Ω,1 Ω,10Ω

Applying Loop rule, we can write,

Or,

And i₁ will be,

In the next figure,

Let i be the current that passes through the middle branch and hence i₂-i will pass through the upper branch, as shown in figure.

By applying loop rule in AFEDA, we can write as,

Or,

Similarly, applying the loop rule in ADCBA, we can write as,

Or,

Replacing ‘i’ in eqn.1 using eqn.2, we get,

Or,

Or,

Now i₁/i₂ is,

a) ; b) n₁r=n₂R

Solution:

a)

When n₁ cells each with emf ‘E’ are connected in series, the total emf, E_net in one branch is,

Since, n₂ of such the branches are connected in parallel, the Total emf in every branches are the same, E_net.

The resistance of n₁ cells each with resistance ‘r’ in series is,

The total resistance for such n₂ number of branches, connected in parallel, is

It is given that the whole setup is connected to an external resistance R. Hence the total net resistance will be,

Hence the current through the external resistor is,

Or,

b)

We know that the relation connecting n₁, n₂, R and r is

To get the minimum current through the resistor R, the denominator in the above expression should be minimum.

In order to minimize the term ‘n₁r+n₂R’, we re-write it as,

SO, this term should be minimized. But, the above value is minimum only when the term in the bracket is zero.

So,

Or,

Hence i is maximum when n₁r=n₂R

10mA

Given,

Emf of the battery, E= 100V

Resistance of series resistor, r= 10kΩ=10000kΩ

Resistance of external resistor, R=0-100Ω

Formula used:

A constant-current source is a power source that supply constant current to an external load, even if there is a change in load resistance.

When the battery is connected to the external resistance that vary from 0 to 100 Ω, the effective resistance will change across the potential difference provided by the battery.

Solution:

Let’s find out the current when R=0Ω or when there is no external resistance is connected,

We know that, current i for a series resistor connection is

By substituting the known values, we get i as,

Now let’s take R as 2 Ω (or a low value like 1 Ω or so)

The value of current, i is

Where R_tot is the effective resistance across the battery. From the figure, R_tot can be calculated as,

Hence by putting R=2 Ω, we get i as,

Similarly, by putting R=100Ω , the highest possible, we get the current i as,

So, as the principle predicted, the value of the current, 1mA, does not change much, or it stays consistent till two significant digits.

1.6A, 4A

Given,

Current through A₁ ammeter, i-i₁= 2.4A

Formula used:

Depending on the resistance offered by each paths of a circuit, the current split inversely. The voltage across the two branches will be same as they are connected parallel.

Solution:

The current going through Ammeter A₃ will split into two. To find the current in two branches, let’s equate the voltage in each branches.

Hence,

By substituting the value of i-i₁= 2.4A, we get i₁ that goes through ammeter A₂ as,

Hence the total current i, passing through A₃

Hence the current through A₂ and A₃ are 1.6A and 4A.

0.15A, 0.83A

Given,

The resistance of the rheostat, R= 30Ω

The emf of the battery, E=5.5V

Formula used:

The rheostat can vary the resistance from 0Ω to maximum, and will be added as a series resistance to the given setup.

Solution:

The 10 Ω and 20 Ω connected parallel to each other. This can be reduced to a single resistance as they both connected to same potential difference. So the effective resistance, R_eff between 10 Ω and 20 Ω will be,

The rheostat resistance will be added in series to the above resistance.

The minimum current will be marked when the total resistance is maximum, which happens when rheostat resistance R=30Ω.

So the current will be,

Where R_tot is

Hence the minimum current, i_min is

Similarly, maximum current, i_max can be obtained when Rheostat resistance R is minimum, R=0Ω.

So, R_tot is

Hence the current is,

Hence the current in the ammeter vary from 0.15A to 0.83A

a)1A, b)0.67A, c)0.33A

Given,

Emf of the battery, E=60V

Resistance of each bulb, r=180Ω

Formula used:

The potential difference across each bulb will be equal as they are connected in parallel across the same cell.

Also, we know that, for parallel connection of resistors, r with equal resistance, the effective resistance, R_eff will be

Where n is the number of resistors.

a)

When all the switches are closed (switched on), the current will split equally across each of the resistors. And the total current will be the ratio between the potential difference and the effective resistance.

The effective resistance will be,

Or,

Hence the current,i will be

b)

In this case two resistors (or bulbs) are connected in parallel. So the effective resistance will be,

Or,

Hence the current, i will be

c)

In this case one resistor (or bulb) only connected to the battery. So the effective resistance will be,

Hence the current, i will be

The current in the setup due to the connection of 3 bulbs, 2 bulbs and 1 bulb is respectively 1.0A, 0.67A and 0.33A

and

Given, resistances , ,

Maximum resistance occurs when the three resistances are connected in series.

hence,

Minimum resistance is when they are parallel to each other.

and

Given, Voltage of the battery,

Power operated P=5W, 10W or 15W

We know,

Now,

The parallel combination of resistances is always less than individual resistances. Therefore, has to be parallel combination while and are the resistances.

4mA, 8mA, 1340V.

Here, current in resistor is .

Let current through be and that from be such that

now, the potential across and has to be same as they are connected in parallel.

Current flowing through is 4mA and that from is 8mA.

To calculate potential difference, we need to find the equivalent resistance.

The current is

Therefore, potential between A and B is

Given, current in resistor,

Current in a parallel setup,

Let the resistor be R

Voltage in the first case, —————(1)

Let R’ be the equivalent resistance.

Voltage in the second case, —————(2)

now, (1) should be equal to (2).

r/3

Given, resistance of each resistor is r.

The equivalent circuit is given as follows:

Therefore, equivalent resistance is

, and

Resistance of wire is .

Resistance of each arm is

The following figure shows the setup of resistor.

All other arms are collectively in series and parallel to arm AB.

The equivalent resistance of all other arms are:

Hence, the resistance across AB is

BAll other arms are collectively in series and parallel to arm AC. AC has two series resistors.

The equivalent resistance of all other arms are:

Hence, the resistance across AB is

CAll other arms are collectively in series and parallel to arm AD. AD has three series resistors.

The equivalent resistance of all other arms are:

Hence, the resistance across AB is

0.1A and 0.3 A

Given, resistor , and .

A If the switch is open, both resistors are in series.

Total resistance,

Current,

B If the switch is closed, is short circuited.

Total resistance,

Current,

0.2A

Here, the resistor is short circuited. Let the current be .

Using KVL in the loop,

Hence, the current through resistors is 0.2A

1A, 0.4A, 0.6A

Given, , and .

Let the potential at the joint be

Current through is

Current through is

Current through is

Now, Kirchhoff’s junction rule,

Therefore,

1A,0A,1A,0A

Given, resistance of resistor,

emf of each cell,

A Equivalent circuits are

For current through a is

As for b, the two emf cancel each other, thus total potential is zero. Hence, the current is zero.

B Equivalent circuits are

For current through a is

Again for b, the two emf cancel each other, thus total potential is zero. Hence, the current is zero.

There are two current loops,

KVL in loop 1 gives

—————(1)

And in loop 2,

—————(2)

Solving (1) and (2)

Now,

B On rotating the figure we find that this circuit is similar to A

hence,

2V,1A,0A,1A

Here, we will apply KVL

KVL in loop 1 gives

—————(1)

KVL in loop 2 gives

—————(2)

Solving (1) and (2) gives

Current through topmost branch is 1A.

Current through middle branch is 0A.

Current through bottommost branch is 1A.

Therefore, potential difference is

0A

Here, there are two loops,

KVL on loop 1 gives

—————(1)

KVL on loop 2 gives

—————(2)

Solving (1) and (2) gives,

and

thus current through is

This question can be solved by critical analysis of the circuit. Three loops are present. In each loop there are 2 cells of equal emf opposing each other. Thus, the total effective emf is zero. Thus, there will be no current from either arms.

Any value of R

This is an example of balanced Wheatstone bridge.

In such a setup, the current through the middle branch is zero irrespective of the resistance. Hence, any value R would suffice for current to be zero.

(a)

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

The given circuit can be rewritten as

As we wish to find the resistance between points a and b, we have proceeded to add a voltage source of emf ϵ between the points a and b.

Let the net current passing through the upper branch be 2I. We can see that the upper branch is symmetric i.e. its upper and lower portion are the identical. Thus, the current should divide equally when branching out. So, current through ec and ed is I.

Now,

Hence,

As there is no potential difference across dc, there is no current passing through dc.

Hence, we can rewrite the circuit without the resistor across dc.

This circuit is equivalent to:

We can see that uc and cv are in series.

Also, nd and dm are is series.

Hence, we can rewrite the circuit as follows:

Clearly, uv and nm and gh are in parallel.

(b)

Note that because of symmetry,

Hence, we can rewrite the circuit as cg and gd are is series and eg and gf are in series.

The blue resistors are in series. Net resistance =

Now, the blue resistors are in parallel.

Now, the equivalent resistance of the upper branch and the lower branch is:

The middle branch:

Now, the circuit looks like this:

All the resistance are now in parallel:

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Let the current through the ammeter be . We can see that the upper and lower part are completely identical. So , there is no reason they should have different currents. Hence, the current equally distributes into each branch due to symmetry.

Due to this symmetry, d and h; e and i are the same potential. Hence, no current passes through de and ei.

Hence, we can redraw the circuit by removing the resistors across dh and ei.

We can see that the resistors across cd, de and ef are in series. Hence,

Similarly,

We can rewrite the circuit as follows:

We can easily see that cf and gj are in parallel.

Now, the circuit looks like this:

By Ohm’s law,

(a)

Concept/Formula used:

Kirchhoff’s loop rule:

The sum of potential differences around a loop is zero.

Let the current flowing through the circuit be .

Applying Kirchhoff’s loop rule through the whole loop abcdea,

(b)

Concept/Formula used:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

The potential difference across R₁ is:

(c)

The potential difference across R₂ is:

(d)

Let us consider the second circuit.

Applying Kirchhoff’s loop rule through the whole loop abcdea,

The potential difference across R₁ is:

The potential difference across R₂ is:

As we need to find the equivalent resistance across a and f. We connect these two a source of emf ϵ.

Now, we will exploit the symmetry in the cube to find the current in each branch. Let the net current be 6I. Then, it distributes equally in all 3 branches due to symmetry. Hence, ab, ah and ad have current of 2I. Now, it further divides into two branches of current I. All this is in accordance with the junction rule.

Now, following the junction rule, we can find out the rest of the currents.

Applying Kirchhoff’s loop rule on loop adgfia,

Now, we need the find an r_eq such that

Using (1), we get

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

(a)

The resistors across dc and cb are in series.

Now, the two resistors across db are in parallel.

The resistor across db and ad are in series.

The circuit can be redrawn as follows:

The resistances a’b’ and ab are in parallel.

(b)

The point b is connecting the ends of three resistors; the other ends of the resistors are attached to a fourth one with one of its end being point a.

Hence, we can redraw the circuit as follows:

We can see that the resistors across bc are in parallel.

We can redraw the circuit as follows:

Note that the resistors across bc and ca are in series.

(c) We can redraw the given circuit as follows:

Note that there is symmetry in the circuit; the upper and the lower parts are identical. Hence, the potential at d should be the same as potential at c. Consequently, there is no potential difference across dc. We can neglect the resistors across dc.

We can redraw this circuit as follows:

We can easily see that uc and cv are in series.

Similarly,

Now, resistors across uv and nm are in parallel.

(d)

Note that one end of all four resistors is connected to one point a. The other ends are connected to a circle with no resistance. Hence, all points in the circle are same i.e. point b. Thus, all the other ends are connected to the same point b. This simply describes resistors in parallel.

(e)

The circuit can be drawn as follows:

As we wish to find the resistance between points a and b, we have proceeded to add a voltage source of emf ϵ between the points a and b.

Let the net current coming out of the battery be 2I. We can see that the circuit is symmetric i.e. its upper and lower portion are the identical. Thus, the current should divide equally when branching out. So, current through ac and ad is I.

Now,

Hence,

As there is no potential difference across dc, there is no current passing through dc.

Hence, we can rewrite the circuit without the resistor across dc.

The circuit can be redrawn as:

We can see that uc and cv are in series.

Also, nd and dm are is series.

Hence, we can rewrite the circuit as follows:

Now, uv and nm are in parallel.

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

(a)

Let the equivalent resistance between A and B be r_eq Ω .

This means that we can rewrite the circuit as:

Where r_eq has replaced the following infinite combination:

We can redraw the infinite circuit as

Note that de and cb are in parallel. The equivalent resistance is given by:

Now, the 1 Ω resistor and r’_eq are in series.

Rearranging and dropping the subscript, we get a quadratic equation:

The roots of this equation are 2 and -1. As resistance can’t the negative the equivalent resistance between A and B is 2 Ω.

(b)

Let the net current be 2I. This current passes through the 1 Ω resistor. Than splits up equally due to symmetry as there are two 2Ω resistors.

Now,

Hence, the current passing through the nearest 2 Ω resistor is 1.5A.

Given:

Emf,

External resistor,

Internal resistance,

Resistance of Ammeter,

Resistance of Voltmeter,

(a)

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

We are trying to find the equivalent resistance of the circuit first.

We will treat the measuring devices as ordinary resistors. Now, voltmeter and R= 50Ω are in parallel. The equivalent resistance is given by:

We can redraw the circuit as follows:

Now, we can see that all the devices are in series.

Hence,

Now, the current that passes through R_eq also passes through the three series components : internal resistor, r’ and ammeter.

Hence,

Using Ohm’s law,

Now, the potential difference across R is the same as potential difference across r’.

Hence, the ammeter reading is 0.1A and the voltmeter reading is 4V.

(b)

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Let us find the equivalent resistance, R_eq of the circuit.

We will treat the ammeter and the voltmeter as ordinary resistors to find the equivalent resistance.

Note that the ammeter and the external resistor are in series.

Hence,

We can redraw the circuit as follows:

Note that the voltmeter and r’ are in parallel.

The circuit can be redrawn as follows:

Now, the r’’ and the internal resistance are in series.

By Ohm’s law, the current coming out of the battery is

Now, potential difference across r’’ = 41.27 Ω is the same as across r = 52 Ω .

Voltmeter Reading:

Now, the current passing through the ammeter is the one passing through r’.

The ammeter reading is 0.08A and the voltmeter reading is about 4.3 V.

Concepts/Formula used:
Resistors in Series:

Resistors in parallel:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

(a)

The given circuit is:

We can easily see that the voltmeter and R₁ are in parallel. The equivalent resistance, r’ is given by:

We can redraw the circuit as follows:

Note that r’ and R₂ are in series. The equivalent resistance is given by:

The current coming out of the battery also passes through r’. By Ohm’s law,

Now, the voltmeter reads the potential different across R₁ .

Hence, the voltmeter reads 24V.

(b)

The given circuit is now:

The equivalent resistance is :

The current is the same throughout the circuit as all components are in series and is given by:

Now, the potential difference across R₁ is given by:

Hence, the potential difference across the 100Ω resistor before connecting the voltmeter was 28V.

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Kirchhoff’s junction rule:

The sum of currents entering a junction is equal to the sum of currents leaving it.

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

Let us call the 24Ω resistance be R₁ and 50Ω R₂ . Let the resistance of voltmeter be R_V .

It is given that the voltage across the voltmeter, V = 18V.

The given circuit can be drawn and labelled as follows:

Applying Kirchhoff’s loop rule,

Using Ohm’s law,

The potential difference across voltmeter and R₂ is the same as they are in parallel.

Using Ohm’s law again,

Using Kirchhoff’s junction rule at X,

Finally, using Ohm’s law for the voltmeter, we get

Hence, the voltmeter has resistance 130Ω.

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Note that 1A = 100mA.

Note that the current at full deflection is .

The voltmeter can be represented as follows:

It is given that the resistances are in series.

Potential difference is maximum when there is full scale deflection and is given by Ohm’s law:

Thus, maximum potential difference that can be measured is 6V.

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

The maximum current that can be measured is .

The current through the coil when there is full scale deflection is .

The resistance of the coil is

Let the resistance of the shunt be and current when there is full-scale deflection is .

By Kirchhoff’s junction rule,

Using Ohm’s law, we have

and

As r_s and r are in parallel, the potential difference across them is the same.

Hence, the shunt resistance is 0.125Ω .

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Ammeter:

It consists of a galvanometer coil in parallel with a stunt resistance.

Kirchhoff’s junction rule:

The sum of currents entering a junction is equal to the sum of currents leaving it.

The given voltmeter looks like this:

The maximum potential difference that can be measured is ,and let the current through the voltmeter for maximum deflection be .

Note that the coil (r) and the other resistor (R) are in series.

(Note that 1kΩ = 1000Ω )

Now, using Ohm’s law,

The ammeter we want looks like the following diagram:

The maximum current that can be measured is .

From our previous calculations we know that the current through the coil for maximum deflection is .

Note that the shunt resistance (r_s) and the coil (r) are parallel in an ammeter.

By Kirchhoff’s junction rule,

Using Ohm’s law, we have

and

As r_s and r are in parallel, the potential difference across them is the same.

Hence, the shunt resistance is 0.251Ω.

Concepts/Formula used:

Wheatstone bridge:

The condition for no deflection through the galvanometer is

Here, and

Let the potentiometer wire be at l cm from A when there is no deflection through galvanometer.

Let the resistance per cm of the potentiometer wire be be ρ .

Hence, and .

Using the equation when there is no deflection for a Wheatstone bridge/potentiometer, we have

Hence, the free end of the galvanometer must be 16cm from point A.

Concepts/Formula used:

Wheatstone bridge:

The condition for no deflection through the galvanometer is

Here and .

When no deflection occurs,

Let the resistance per cm of the potentiometer wire be be ρ .

Hence, and .

Using the equation when there is no deflection for a Wheatstone bridge/potentiometer, we have

(a)

As ,

Let us account for the potential differences when moving from A to C.

(b)

Let the area of cross section of wire be A and resistivity be ρ.

Then,

where V is the voltage across a wire segment of length l.

As, all the quantities except for length are the same for all sections of wire AB,

Hence,

Hence, D is 66.67cm away from A.

(c) As , there is no potential difference across CD and hence, the current through it is zero.

(d)

(a)

As ,

Let us account for the potential differences when moving from A to C.

(b)

Let the area of cross section of wire be A and resistivity be ρ.

Then,

where V is the voltage across a wire segment of length l.

As, all the quantities except for length are the same for all sections of wire AB,

Hence,

This is not possible. As the length of the wire is only 100cm. There is no such point D.

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Let the area of cross section of the wire (resistor) be A and resistivity be ρ.

Then,

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

(a)

We consider the circuit when there is no deflection:

Applying Kirchhoff’s rule on loop ABCF,

Using ,

Let the area of cross section of wire be A and resistivity be ρ.

Then,

where V is the voltage across a wire segment of length l.

As, all the quantities except for length are the same for all sections of wire AB,

Hence,

(b)

Let the current coming out of the main battery be I₁ and the current through the galvanometer be I₂.

Let the area of cross section of wire be A and resistivity be ρ.

Then,

where V is the voltage across a wire segment of length l.

As, all the quantities except for length are the same for all sections of wire AB,

Hence,

Using

Now, R_AD and R_DB are in series.

Applying Kirchhoff’s rule on loop ABCFA,

……………(1)

Applying Kirchhoff’s rule on loop ADEFA,

………………(2)

From (i) and (ii)

Concepts/Formulas used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Capacitance:

If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as

At steady state, no current passes through 6μF . Hence, we can ignore the capacitor to find the equivalent resistance.

Now, 10Ω and 20Ω are in series.

The current is given by Ohm’s law:

Now, this is the same current that passes through 10Ω resistor.

The potential across it is given by:

Now, as the capacitor is in parallel with the 10Ω resistor the,

Now,

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Kirchhoff’s junction rule:

The sum of currents entering a junction is equal to the sum of currents leaving it.

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

Energy stored by a capacitor:

If the potential difference between the two conductors of the capacitor is V and its capacitance is C, its energy is given by:

(a)

Using Kirchhoff’s law on loop FEABF,

........(1)

Using Kirchhoff’s law on loop ACDBA,

……………(2)

Solving (1) and (2), we get

Hence,

(b)

At steady state, no current passes through the capacitor; hence, the results are the same as in (a).

Now, the potential as the capacitor is in parallel with the 20Ω resistor,

Hence, the energy stored by the capacitor is 32μJ.

Concepts/Formula used:

Ohm’s Law:

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

Kirchhoff’s junction rule:

The sum of currents entering a junction is equal to the sum of currents leaving it.

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

Capacitance:

If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as

Applying Kirchhoff’s loop rule on ABCDEA,

Applying Kirchhoff’s loop rule on ABHGFA,

Now,

As the 3Ω resistor and the 3μF capacitor are in parallel,

Now, we know that

Also,

Now, we know that

Now,

Now, we know that

Now,

Now, we know that

Capacitors in parallel:

If capacitors C₁, C₂, C₃ , … are in parallel, then the equivalent capacitance is given by:

Capacitors in series:

If capacitors C₁, C₂, C₃ , … are in sereis, then the equivalent capacitance is given by:

Note that the charge is same on each capacitor in series in steady-state.

The circuit can be redrawn as follows:

We can see that C₁ and C₂ are in parallel,

Also, we can see that C₃ and C₂₄ are in parallel,

The circuit can be redrawn as follows:

Now, the two capacitors and are in series.

Note that there is no current in equilibrium. Hence,

As and are in series,

Now,

Also,

Energy stored in a capacitor:

The energy stored in a capacitor with capacitance C , charge is given by:

where V is the potential difference across the capacitor.

Power supplied by the battery:

If a battery of emf ϵ gives a current I, then the power supplied by the battery is given by:

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:

(a)

By Kirchhoff’s loop rule,

As ϵ is constant,

At t = 0, there is no charge on the capacitor. Hence . So, V_r is maximum at t = 0.

(b) Initially, current flows through the circuit treating capacitor as short-circuit. But as charge accumulates on the capacitor the current reduces with time. It is maximum at t =0.

Using Kirchhoff’s loop rule,

I is max at t = 0 and

(c)

Using Kirchhoff’s loop rule,

As R and ϵ are constant

Current is minimum at equilibrium when capacitor acts as an open switch and current is zero. Hence,

(d)

The energy stored by the capacitor is given by:

As C is constant,

We know from (c) that the maximum potential difference across the capacitor is ϵ. Hence,

(e)

Power delivered by the battery is given by:

As ϵ is a constant,

Using the result from (b), we get

(f)

The resistor converts energy into heat:

As R is constant,

Now, using the result from (b),

Concepts/Formulas Used:

Time constant for capacitor:

Where R is the resistance through which the capacitor is being charged/discharged and C is the capacitance.

Capacitance of a parallel plate capacitor:

A capacitor consists of two conducting plates of area of cross section A each separated by a distance d. The capacitance if there is only vacuum between the plates is given by:

where ϵ₀ is the permittivity of free space.

Given,

Area of the plate,

(

Now, distance of separation:

Resistance ,

We know that,

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed

The charge is at any t>0 is given by:

Note that is known as time constant.

Given,

Capacitance,

EMF of battery,

Resistance,

We also know that at ,

We know that,

Taking natural logarithm on both sides, we get

Now,

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed

The charge is at any t>0 is given by:

Note that is known as time constant.

Given,

Capacitance,

EMF of battery,

Resistance,

Now,

Also,

We know that,

At t = 2.0ms,

Hence, the charge on the capacitor at t = 2.0ms is 76μC.

Charge on Capacitor during Discharging (RC Circuit):

A capacitor of capacitance with charge C is being discharged through a resistor of resistance R. A switch S is also connected in series with the capacitor. The switch is initially open. At t=0, the switch is closed. The charge on the capacitor at any time t>0 is given by:

where and Q_i is the initial charge on the capacitor.

Note that the capacitor begins charging at t = 0.

This is simply a discharging circuit.

Given,

Capacitance,

Resistance,

Initial Charge,

Now,

( and )

We know that

(a) At t = 0,

(b) At t = 30μs

(c) At t = 120μs

(d) At t = 1.0ms

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current and is the time constant.

Given,

Capacitance,

EMF of battery,

Resistance,

(a)

Just after the connections are made, there is no potential difference across the capacitor and it acts as a short circuit; hence, the current can simply be calculated form Ohm’s law:

(b)

We know that,

Now, at t=τ,

Using the result from (a), we get,

Formulas/Concepts Used:

Capacitance:

If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as

Capacitance of a Capacitor in presence of a dielectric:
The capacitance of the capacitor is initially C₀ and then a dielectric medium of dielectric constant K is inserted between the plates. The new capacitance is

Also for parallel plate capacitors,

Where ϵ0 is the permittivity of free space, A is the area of plate and l is the distance between the plates.

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

The charge is given by:

Note that is known as time constant.

Given,

Area of the plate,

(

Now, distance of separation:

Emf of battery,

Resistance ,

Time,

Now,

Time constant,

We want to find the charge on the capacitor at t = 10ns

Now,

Energy stored in a capacitor:

The energy stored in a capacitor with capacitance C , charge is given by:

where V is the potential difference across the capacitor.

Capacitance of a Capacitor in presence of a dielectric:
The capacitance of the capacitor is initially C₀ and then a dielectric medium of dielectric constant K is inserted between the plates. The new capacitance is

Also for parallel plate capacitors,

Where ϵ₀ is the permittivity of free space, A is the area of plate and l is the distance between the plates.

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

The charge is given by:

Note that is known as time constant.

Given,

Area of the plate,

(

Now, distance of separation:

Dielectric constant,

Emf of battery,

Resistance ,

Time,

Now,

Time constant,

We want to find the charge on the capacitor at t = 8.9μs

Now, energy of the capacitor is given by:

Hence, the capacitor stores of energy after 8.9μs.

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

The charge is given by:

Note that is known as time constant.

Given,

Resistance of the resistor,

Capacitance,

Emf of the battery,

Now, time constant ,

(a)

The current is given by:

At t =0, we have y-intercept I₀.

Now,

Hence, we need to plot

At t = 10min = 600s,

(b)

Now,

Note that the current is in Ampere and time in seconds. The graph will represent exponential decay.

We need to plot

Note that t is in seconds and Q in coulombs.

At t = 10 min = 600s,

The charge will keep on increasing and will almost touch the asymptope at Q = 0.0024C.

Current when capacitor is discharging:

A capacitor of capacitance C is being discharged through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I0 is the initial current.

Note that is known as time constant

Current when capacitor is charging:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

Note that is known as time constant.

In both the cases wish to find time t>0 such that

I(t) is given by the same formula in both the cases.

Hence, 0.69 time constants will elapse.

Concepts/Formulas used:

Charge on Capacitor during Discharging (RC Circuit):

A capacitor of capacitance with charge C is being discharged through a resistor of resistance R. A switch S is also connected in series with the capacitor. The switch is initially open. At t=0, the switch is closed. The charge on the capacitor at any time t>0 is given by:

where and Q_i is the initial charge on the capacitor.

Note that the capacitor begins charging at t = 0.

Capacitance:

If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as

Let a capacitor of capacitance C be discharged through a resistor of resistance R. Switch S in attached is series and in initially open. It is closed at t = 0.

Charge is maximum when the capacitor is fully charged. Hence,

Now, we have at any time t>0,

We want to find t such that

Hence, we need 6.9 time constants.

Concepts/Formulas used:

Charge on Capacitor during Charging (RC Circuit):

A capacitor of capacitance with charge C is being charged with a battery of emf ϵ through a resistor of resistance R. A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The charge on the capacitor at any time t>0 is given by:

where

Note that the capacitor begins charging at t = 0.

Energy stored in a capacitor:

The energy stored in a capacitor with capacitance C , charge is given by:

where V is the potential difference across the capacitor.

Let a capacitor of capacitance C with no initial charge be attached to a battery of emf ϵ through a resistor of resistance R. Switch S in attached is series and in initially open. It is closed at t = 0.

Equilibrium is when no current flows and the potential across the battery is the same as the potential across the capacitor. The energy stored at equilibrium is:

Suppose the capacitor begins charging at t = 0.

Now, at any time t > 0, the energy stored is

Substituting the value for Q(t),

We want to find t when

Now, or

We reject the former as the capacitor begins charging at t = 0.

Hence, 1.23 time constants elapse .

Concepts/Formula used:

Current when capacitor is charging:

A capacitor of capacitance C is being charged using a batteru of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

Note that is known as time constant

Power supplied by the battery:

If a battery of emf ϵ gives a current I, then the power supplied by the battery is given by:

Now, at anytime t > 0,

Now, power is maximum when the current is maximum i.e. when t = 0

We wish to find time t such that

The time in 0.69 times the time constant.

Formula/Consepts Used:

Energy stored by capacitor:

For a capacitor of capacitance C , with charge Q, and potential difference V across it, the energy stored is given by:

Charging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R, a switch, and battery of emf ϵ . It is uncharged at first. The switch is closed at t = 0, then at time any time t the charge stored on the capacitor is given by

Energy stored in the capacitor is

For maxima,

Taking the natural logarithm on both sides,

(Note that and

Now, the maximum rate is

Concepts/Formula used:

Current when capacitor is charging:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I₀ is the initial current.

Note that is known as time constant

Power supplied by the battery:

If a battery of emf ϵ gives a current I, then the power supplied by the battery is given by:

Energy dissipated by a resistor :

A resistor of resistance R with current I through it, dissipates energy U given by:

in time Δt.

Its power is given by:

The capacitor is being charged

Given,

Capacitance,

Resistance,

Emf of the battery,

Time,

Now, time constant,

(a)

The initial current is :

Now,

At t = 12.0μs

(b)

The power supplied by the battery is:

At t = 12.0μs,

(c)

The power dissipated as heat:

At t = 12.0μs,

(d)

By conservation of energy,

Energy supplied by battery = Energy stored by capacitor + Energy dissipated as heat.

Dividing by time, gives us

Power supplied by battery = Power dissipated as heat + rate at which energy is stored in the capacitor.

Hence, using the previous results, we have,

Rate at which energy is stored in the capacitor:

Concepts/Formulas Used:

Energy dissipated by a resistor :

A resistor of resistance R with current I through it, dissipates energy U given by:

in time Δt.

Its power is given by:

Current when capacitor is discharging:

A capacitor of capacitance C is being charged through a resistance R , the current through the circuit is given by:

Where I₀ is the initial current.

Energy stored by capacitor:

For a capacitor of capacitance C , with charge Q, and potential difference V across it, the energy stored is given by:

Discharging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R and a switch. Before the switch is closed, it has charge Qi . If the switch is closed at t = 0, then at any time t, the charge on the capacitor is given by:

where

The initial energy of the capacitor,

As the capacitor is discharged, it looses Charge ,and the potential difference across it also decreases.

Note that

Now, at t= τ,

The energy lost is dedicated as heat and is equal to:

Now let us find the energy dissipated by another method:

Substituting ,

Note that and

Both ways give us the same result!

Concepts/Formulas Used:

Energy dissipated by a resistor :

A resistor of resistance R with current I through it, dissipates energy U given by:

in time Δt.

Its power is given by:

Current when capacitor is charging:

A capacitor of capacitance C is being charged by a battery of emf V through a resistance R is series , the current through the circuit is given by:

Where

Suppose a capacitor of capacitance C is being charged by a battery of emf V through a resistance R.

Now, the power of the resistor is given by:

Substituting ,

Substitute

This is the same as the energy stored in the capacitor when it is fully charged! (Note that when the capacitor is fully charged, the potential difference across it is V.)

Time constant for capacitor:

Where R is the resistance through which the capacitor is being charged/discharged and C is the capacitance.

Capacitance of a Capacitor in presence of a dielectric:
The capacitance of the capacitor is initially C₀ and then a dielectric medium of dielectric constant K is inserted between the plates. The new capacitance is

Also for parallel plate capacitors,

Where ϵ₀ is the permittivity of free space, A is the area of plate and l is the distance between the plates.

Resistance and Resistivity:

For a material of length l and uniform cross-section A and reistivity ρ, the resistance is given by:

Note the area of cross section of the material is the same as the area of the capacitor plates and the length of the material is the same as the distance of separation between the plates.

Now,

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R, a switch, and battery of emf ϵ . It is uncharged at first. The switch is closed at t = 0, then at time any time t the charge stored on the capacitor is given by

Capacitors in parallel:

If capacitors C₁, C₂, C₃ , … are in parallel, then the equivalent capacitance is given by:

If the charges on the capacitors are Q1, Q2, Q3, .. are in parallel, then the charge on the capacitor with equivalent capacitance is given by:

We can replace the two capacitors by another capacitor of capacitance C. As the capacitors are in parallel.

Now,

We know that

Here,

Also, and

Hence,

Let the charge on both the capacitors be Q. As both have the same capacitance and potential (, both must have the same charge. Note that they both are in parallel.

Hence,

When the switch is closed and the circuit is in steady state, no current passes through the capacitor.

Applying Kirchhoff’s loop rule on loop bedcb,

Substitute I₁ = 0 as there is no current through the capacitor at steady state.

Now, as they are in parallel

Note that V_af is the potential difference across the capacitor.

When the switch is just opened, the potential difference across it is the same for a moment as the charge and capacitance is the same. However, the charge and potential decrease slowly as the capacitor starts discharging.

Discharging:

We know that when discharging,

Applying Kirchhoff’s loop rule on fabef,

At , and

Here, R = 10Ω and C = 25μF

Now,

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R, a switch, and battery of emf ϵ . It is uncharged at first. The switch is closed at t = 0, then at time any time t the charge stored on the capacitor is given by

Discharging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R and a switch. Before the switch is closed, it has charge Qi . If the switch is closed at t = 0, then at any time t, the charge on the capacitor is given by:

Given,

Emf = 6V

Capacitance,

Resistance,

Time for charging = time for discharging =

When charging,

When discharging,

Now,

Hence,

Concepts/Formulas used:

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero.

Capacitance:

If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as

Capacitors in series:

If capacitors C₁, C2, C3 , … are in series, then the equivalent capacitance is given by:

We can replace C₁ and C₂ by C_eq. As C₁ ans C₂ are in series,

Let us drop the subscript and call C_eq just C.

Let the potential across the capacitor C be at time t be V_c. Let the charge at time t be q.

Note that as C₁ and C₂ are in series,

Applying Kirchhoff’s loop rule ,

We know that

Where B is a constant

Let

Substitute q = 0 at t = 0,

Substituting the value of A back,

where