Current Electricity

The diagram is given as:

Given: Electromotive force (E) of battery = 12V

Internal Resistance r of Battery = 0.4 Ω

Let the maximum current drawn from the battery be I.

Therefore, as per Ohm’s Law,

E = Ir

Hence maximum current drawn from battery is 30Ampere.

The diagram is given as:

Given: Electromotive force (E) of battery = 10 volt.

Internal Resistance r of the battery = 3 Ω

Current in the Circuit (I) = 0.5 A

Let the resistance of the required resistor is R Ω

Let the Terminal voltage of the register be V.

As per Ω ’s law,

⇒

⇒

⇒R + r = 20 Ω

∵ r = 3Ω

Resistance of Resistor is 17 Ω.

Also, as per Ohm’s law,

⇒V = IR

⟹ V = 0.5A × 17Ω =

⇒ V = 8.5V

∴ Terminal voltage of the register is 8.5V.

The diagram is given as:

When three resistors are connected in series then the total resistance is equal to sum of individual resistance.

Total resistance is equal to = 1 + 2 + 3 = 6 Ω

Let the current flowing in the circuit be I.

Electromotive force EMF, (E) = 12V

Total resistance in the circuit R = 6 Ω

According to Ohm ’s law, V = IR

Potential drop across 2 Ω resistor = 2 × 2 = 4Volt, as V = I × R by Ohm ’s Law

Potential drop across 1 Ω resistor = 2 × 1 = 2Volt

Potential drop across 3 Ω resistor = 2 × 3 = 6Volt

Since three resistors are connected in parallel, hence total resistance(R) of the combination is given by

Hence, Total Resistance is 1.05 Ω

Given, EMF of the battery E = 20Volt

Current flowing through 2 Ω resistor is given by

Current flowing through 4 Ω resistor is given by

Current flowing through 5 Ω resistor is given by

Current flowing through 2 Ω resistor is 10Ampere; 4 Ω resistor is 5Ampere and through 5 Ω resistor is 4Ampere respectively.

Given: Temperature coefficient of filament,

α = 1.70 × 10^-4 per °C

Let T₁ be the temperature of element, R₁ = 100Ω (Given: T₁ = 27°C)

Let T₂ be the temperature of element, R₂ = 117Ω

To find T₂ = ?

The formula is: R₂ = R₁[1 + α(T₂-T₁)]

⇒ R₂ - R₁ = R₁α(T₂ - T₁)

⇒ R₂ - R₁ = R₁α (T₂ - T₁)

⇒

⇒ T₂-27°C = 1000°C

⇒ T₂ = 1000°C + 27°C

Hence Temperature of element, T₂ is 1027 °C.

Note: The temperature coefficient of resistance (α) gives the change in resistance per degree change in temperature. For pure metals, it is positive means- The resistance increases with temperature.

Given, the length of wire l = 15 metre

Area of cross section of wire a = 6.0 × 10^-7 metre square

Resistance of material of wire, R = 5 Ω

Resistivity of material of wire = ρ

Hence Resistivity of material of wire is 2 × 10^-7 metre

Resistance of silver wire at 27.5 °C R₁ = 2.1 Ω

Resistance of silver wire at 100 ° C R₂ = 2.7 Ω

Let temperature coefficient of silver wire is given by α.

The formula for α is: R₂ = R₁[1 + α(T₂-T₁)]

⇒ R₂ - R₁ = R₁α(T₂ - T₁)

⇒ R₂ - R₁ = R₁α (T₂ - T₁)

⇒

⇒ α = 0.0039°C

Temperature Coefficient of silver wire is 0.0039 per °C.

Given Supply Voltage = 230 Volt

And it draws current I₁ = 3.2 Ampere

Resistance is given by R₁ = V/I

R₁ = 230V/3.2A

R₁ = 71.87 Ω

Steady value of current I₂ = 2.8Ampere

Resistance R α is given by R₂ = 230V/2.8A

R₂ = 82.14 Ω

Temperature coefficient of nichrome wire α = 1.70 × 10^-4 per °C

Initial temperature of nichrome T₁ = 27 ° C

Steady Temperature attained by nichrome = T₂

The formula is: R₂ = R₁[1 + α(T₂-T₁)]

⇒ R₂ - R₁ = R₁α(T₂ - T₁)

⇒ R₂ - R₁ = R₁α (T₂ - T₁)

⇒

⇒ T₂-27°C = 840.5°C

⇒ T₂ = 840.5°C + 27°C = 867.5° C

Hence Steady Temperature attained by nichrome = 867.5 ° C.

The current through the labelled diagram shows the current flowing the respective branches:

For the closed circuit ABDA, potential is zero i.e.,

10I₂ + 5I₄ − 5I₃ = 0

2I₂ + I₄ −I₃ = 0

I₃ = 2I₂ + I₄ ... (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I₂ − I₄) − 10(I₃ + I₄) − 5I₄ = 0

5I₂ + 5I₄ − 10I₃ − 10I₄ − 5I₄ = 0

5I₂ − 10I₃ − 20I₄ = 0

I₂ = 2I₃ + 4I₄ ... (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I₂ + 10I₁ − 5I₄

3I₂ + 2I₁ − I₄ = 2 ... (3)

From equations (1) and (2), we obtain

I₃ = 2(2I₃ + 4I₄) + I₄

I₃ = 4I₃ + 8I₄ + I₄

− 3I₃ = 9I₄

− 3I₄ = + I₃ ... (4)

Putting equation (4) in equation (1), we obtain

I₃ = 2I₂ + I₄

− 4I₄ = 2I₂

I₂ = − 2I₄ ... (5)

It is evident from the given figure that,

I₁ = I₃ + I₂ ... (6)

Putting equation (6) in equation (3), we obtain

3I₂ + 2(I₃ + I₂) − I₄ = 2

5I₂ + 2I₃ − I₄ = 2 ... (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I₄) + 2(− 3 I₄) − I₄ = 2

− 10I₄ − 6I₄ − I₄ = 2

17I₄ = − 2

I₄ = -2/17Ampere

Equation (4) reduces to

I₃ = − 3(I₄)

I₃ = -3 × -2/17

I₃ = 6/17Ampere

I₂ = -2(I₄)

I₂ = -2 × -2/17

I₂ = 4/17Ampere

I₂ - I₄ = 6/17Ampere

I₃ + I₄ = 6/17Ampere

I₁ = I₂ + I₃ = 4/17 + 6/17 = 10/17 Ampere

Therefore, current in branch AB = 4/17Ampere

In branch BC = 6/17Ampere

In branch CD = -4/17Ampere

In branch AD = 6/17Ampere

In branch BD = -2/17Ampere

Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.

The diagram is shown below:

Balance Point from end A, l₁ = 39.5 centimeter

Resistance of resistor Y = 12.5 Ω

Balance condition is given by,

⇒ X = 8.2 ohm

When X and Y gets interchanged then l1 and 100-l1 gets interchanged.

Now the balance point of bridge will be 100-l1 from A

100-l1 = 100-39.5 = 60.5cm

Hence the balance point is 60.5cm from A.

When the galvanometer and cell are interchanged at balance point of bridge then no current would flow through galvanometer because of no deflection.

Electromotive force EMF of battery = 8Volt

Internal resistance of battery r = 0.5 Ω

Supply Voltage V = 120Volt

Resistance of resistor R = 15.5 Ω

Effective voltage in circuit = V1

Since Resistance R is connected in series Hence We can write V1 = V-E

V₁ = 120V-8V = 112 V

Current flowing in the circuit:

⇒

⇒

⇒

I = 7Ampere

By Ohm ’s Law, Voltage across resistor R is given by V = IR

V = 7A × 15.5Ω

⇒V = 108.5Volt

Supply Voltage = Terminal Voltage of battery + Voltage drop across Resistor R

Therefore Terminal Voltage of battery = 120 V-108.5 V = 11.5Volt

Series resistor in charging circuit reduces the current drawn from external supply and current will be too high in its absence.

Given EMF of the cell E1 = 1.25Volt

Balance point of the potentiometer l1 is given 35cm

Since the cell is replaced by another cell of EMF E2

New balance point of potentiometer l2 is 63cm

Since balance condition is given by

E2 = 1.25V × (63m/35m)

E2 = 2.25Volt.

Number density of free electrons in a copper conductor, n = 8.5 × 10²⁸ per meter cube

Length of Copper Wire l = 3 meter

Area of cross section of wire A = 2 × 10^-6 meter square

Current drawn by wire is given 3 Ampere

Since I = nAeV_d

Here V_d is the drift velocity and “e” is the electron charge 1.6 × 10^-19 Coulomb

And V_d Drift velocity is

Therefore

I = nAel/t

Or t = nAel/I

⇒ t = 2.7 × 10⁴ second

Surface Charge density of earth d = 10^-9 Coulomb per meter square

Current over the entire globe = 1800 Ampere

Radius of earth r = 6.37 × 10⁶ meter

Surface area of earth A = 4 × π × radius × radius

A = 4 × π × 6.37 × 10⁶ × 6.37 × 10⁶

A = 5.09 × 10¹⁴ m²

Charge on the earth surface q = d × A

q = 10^-9 × 5.09 × 10¹⁴ m²

q = 5.09 × 10⁵ Coulomb

Let the time taken to neutralize earth surface = t

Current

t = 5.09 × 10⁵C/1800 A

t = 282.78 seconds.

Number of secondary cells n = 6

Emf of each secondary cell E = 2Volt

Internal resistance of each cell r = 0.015 Ω

Resistance of resistor R = 8.5 Ω

Let current drawn from supply = I

⇒ I = 1.39Ampere

Hence current drawn from supply is 1.39Ampere

Emf of secondary cell E = 1.9Volt

Internal resistance of cell r = 380 Ω

⇒ I = 1.9V/380Ω

⇒ I = 0.005Ampere

Hence maximum current drawn from cell is 0.005Ampere and since a large current is required to start the motor of a car and hence cell cannot be used to start the motor.

Resistivity of aluminium, p₁ = 2.63 × 10⁻⁸ Ω m

Relative density of aluminium, d₁ = 2.7

Let l1 be the length of aluminium wire and m1 be its mass.

Resistance of the aluminium wire = R₁

Area of cross-section of the aluminium wire = A₁

Resistivity of copper, p₂ = 1.72 × 10⁻⁸ Ω metre

Relative density of copper, d₂ = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

We can write,

R₁ = (p₁ × l₁)/A₁

R₂ = (p₂ × l₂)/A₂

Since R₁ = R₂

(p₁ × l₁)/A₁ = (p₂ × l₂)/A₂

And l₁ = l₂

Therefore p₁/A₁ = p₂/A₂

A₁/A₂ = (2.63 × 10-8)/(1.72 × 10^–8)

⇒ A₁/A₂ = 2.63/1.72

Mass of the aluminium wire,

m₁ = Volume × Density

= A₁ × l₁ × d₁ = A₁ l₁ d₁ .....1

Mass of the copper wire, m₂ = Volume × Density

= A₂ × l₂ × d₂ = A₂ l₂ d₂ .......2

Dividing equation (1) by equation (2), we obtain

m₁/m₂ = (A₁ l₁ d₁)/(A₂ l₂ d₂)

Since l₁ = l₂

m₁/m₂ = A₁ d₁/A₂ d₂

Since A₁/A₂ = 2.63/1.72 calculated above

⇒ m₁/m₂ = (2.63 × 2.7)/(1.72 × 8.9)

⇒ m1/m2 = 0.46

Since m1 is smaller than m2 Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.

From the given information shown in the table, the ratio of voltage and current is a constant value and the value is equal to 19.7 Hence manganin is a Ω ic conductor.

According to Ohm’s law the ratio of voltage and current is resistance of conductor so Resistance of manganin is 19.7 Ω .

(a) When a steady current flows in a metallic conductor of non-uniform cross section, the current flowing through the conductor is constant.

Since Current density, electric field and drift speed are inversely proportional to area of cross section therefore they are not constant.

(b) No, Ω ’s law is not applicable for all conducting elements. Example of non Ω ic conductor is Vacuum diode semiconductor.

(c) According to Ω ’s law V = I × R

If Voltage V is low, then Resistance R must be very low so that high current can be drawn from source.

(d) A high Tension supply must have a very large internal resistance so that it can limit the current from exceeding the safety limit.

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.

Alloys are the homogeneous mixture of two or more metals and due to which alloys have better conducting properties as compared to pure metals. Pure metals have large number of free electrons and simple atomic arrangement but in alloys these free electrons number as well as their atomic arrangement changes. Hence alloys have greater resistivity than metals.

(b) Alloys usually have lower temperature coefficient of resistance than pure metals.

The resistance change factor per °C of temp change is called the temperature coefficient. Coefficient approaching zero can be obtained by alloying certain metals. So we can say that alloys have lower temperature coefficient of resistance than pure metals.

(c) The resistivity of alloy, manganin, is nearly independent of increase of temperature.

The resistivity of alloys is high and atoms are not arranged in proper order and with the increase in temperature resistivity increases due to collision and random motion of atoms.

(d) The resistivity of a typical insulator is greater than that of a metal by a factor of order of 10²².

In case of insulator, resistivity decreases with increase of temperature.

Total number of resistors = n

Resistance of each resistor = R

(i)

When n resistors are connected in series, effective resistance R1is the maximum,

given by the product nR.

Hence, maximum resistance of the combination, R1 = nR

(ii)

When n resistors are connected in parallel, the effective resistance (R₂) is the minimum, given by the ratio: R/n

Hence, minimum resistance of the combination, R₂ = R/n

(iii) The ratio of the maximum to the minimum resistance is,

⇒

(i)

The resistance of the given resistors is,

R1 = 1 Ω , R2 = 2 Ω , R3 = 3 Ω

Connect 2 Ω and 1 Ω resistor in parallel and 3 Ω resistor in series to them

Equivalent resistance R’ is {(2 × 1)/(2 + 1)} + 3

Hence R’ is equal to 11/3 Ω

(ii)

Now connect 2 Ω and 3 Ω resistor in parallel and one Ω resistor in series to it

Equivalent Resistance R” = {(2 × 3)/(2 + 3)} + 1

R” = 11/5 Ω

(iii)

Now connect all three resistor 1 Ω 2 Ω and 3 Ω in series

Hence total resistance R”’ is 1 + 2 + 3 = 6 Ω

(iv)

Now connect all three resistors in parallel;

Equivalent Resistance is R”” = (1 × 2 × 3)/(1 × 2 + 2 × 3 + 3 × 1)

R”” = 6/11 Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors

of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1 + 1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Now the given circuit diagram can be explained as

Now draw 4 loops each having 2 resistors of 2 Ω and 4 Ω connected in parallel.

Hence Equivalent Resistance R_e of each loop is

⇒ Re = 4/3 Ω

Since all 4 loops will be connected in series and each loop will have 4/3 Ω resistance

Hence, equivalent resistance of the given circuit is 4 × 4/3 = 16/3 Ω .

(b)

From the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R + R = 5R

The resistance of each resistor connected in the given circuit, R = 1 Ω

Equivalent resistance of the given circuit = R’

The network is infinite. Hence, equivalent resistance is given by the relation,

R’ = 2 + (R’/R’ + 1)

or R’² -2R’ -2 = 0

On solving R’ = (2+√4+8)/2

(Negative value of R’ cannot be accepted)

R’ = 1 + √3 Ω

R’ = 1 + 1.73 = 2.73 Ω

Internal resistance of the circuit, r = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, V = 12 V

According to Ω ’s Law, current drawn from the source is given by the ratio, = 12/3.23 = 3.72 A

(a) Constant emf of the given standard cell, E₁ = 1.02 V

Balance point on the wire, l₁ = 67.3 cm

A cell of unknown emf, ε, replaced the standard cell. Therefore, new balance point on the wire,

l = 82.3 cm

The relation connecting emf and balance point is,

E₁/l₁ = E/l

E = l × E₁/l₁

E = 82.3 cm × 1.02 V/67.3 cm

E = 1.247 Volt.

The value of unknown emf is 1.247 V.

(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

(c) The balance point is not affected by the presence of high resistance.

(d) The point is not affected by the internal resistance of the driver cell.

(e) The method would not work if the driver cell of the potentiometer has an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then we cannot obtain balance point on the wire.

(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a high percentage of error. The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

Balance point for open circuit, l₁ = 78.3 cm
Resistance of the external resistor, R = 9.5 Ω

Balance point for this resistor, l₂ = 64.8 cm

The relation connecting internal resistance r and balance point is,

Therefore, the value of the unknown resistance, X, is 1.685 Ω.