Chapter 12 Electricity

The Current will be given by: I = V/R

R is the total resistance of the circuits

Since V and R are equal in all the three circuits. So, Current will be same in all the three circuits and in series circuit current is same and only potential difference varies.

The current recorded in the ammeter will be same in all the three circuits.

In figure (i), the value of net resistance R=2 Ω

In figure (ii), the value of net resistance R= 4 Ω

In figure (iii), the value of net resistance R = 1Ω

Heat is given by the formula: H = (V²× Time) /R

Since net resistance is minimum in figure (iii) and heat dissipated is inversely proportional to the resistance, so Heat dissipated will be maximum in case (iii)

Electrical Resistivity is given by the relation:

ρ is inversely proportional to the number density of free of electrons i.e. nature of the material.

Current, I= 1 A

Time, t = 16 seconds

No. of electrons, N = I×T/e

= 10²⁰ electrons

Circuit shown in figure (ii) is properly connected. In other figures, the connections are not correct. For e×ample: in figure (i), voltmeter is connected in series and not in parallel. Similarly in figure (iii), ammeter is in parallel which should be in series. In figure (iv), the negative terminal of battery has been connected to positive terminal of ammeter whereas it should be connected to negative terminal and not positive terminal of ammeter.

When each resistor is of 1/5 Ω and when all the five resistors are connected in series, then the net resistor will be:

R= R₁+R₂+R₃+R₄+R₅

R = 1/5 +1/5+1/5+1/5+1/5

R= 1 Ω

When each resistor of 1/5 Ω is connected in parallel:

1/R= 1/R₁+1/R₂+1/R₃+1/R₄+1/R₅

1/R = 5+5+5+5+5

1/R = 25 Ω

R = 1/25 Ω

The maximum potential in above connection will be when all are connected in series circuit. The negative terminal of cell should be connected to positive terminal and it should be continued which is correct in figure (i)

Work done is given by the relation :

W= charge × potential difference

W= Q × V

As we know that Q= I × t

Therefore putting in above equation

W= I × t × V

V= Work done/ Current × time

Case 1: when length is l and Area of cross section is A

Resistivity will be

Case 2: when length is 2l and resistance is R and let area is A’

Resistivity will be

As given in the question both conductors are of same material and has same area of cross-section, therefore, resistivity will be equal

And therefore, A’= 2A

Since the slope of the above graph is inversely proportional to the resistance. As in the graph, slope of R₁ is maximum, so resistance will be minimum. Similarly Slope of R₃ is minimum, so it will have maximum resistance. So, the correct order is:

R3 > R2 > R1

Case 1: power is given by

P₁= I²R

Case 2: when current is increased by 100%

It means that current becomes doubled i.e. 2I

Power will be P₂ = (2I)²R

i.e. Power = 4 I²R

Change in dissipated power = 4 I²R- I²R =3 I²R

Increase in percent of power dissipated = 3P₁/P₁ ×100% = 300%

The resistivity depends on the nature of the material and temperature. It doesn’t vary with the shape of the resistor.

Since bulb having high power rating will led to production of more heat as well as more light. So, Brightness of bulb B will be more than that of A

Since resistors are connected in series. So, net resistance R=

R_{1_}+ R₂

R = 2+4 = 6 Ω

Current will be calculated using ohm’s law

I= V/R

= 6/6 = 1A

Heat dissipated will be given by: H= I²Rt

for 4 Ω resistor in 5 s

H = (1)² × 4 × 5

H = 20 J

Power = 1 kW

In Watt, P= 1000 W

Current is given by I = P/V

I= 1000/220

I = 4.5 A

So, A fuse wire of rating slightly greater than 4.5A must be used i.e. 5A.

In series circuit, current flowing through the resistors are same and has difference potential difference.

The S.I. unit of electric power is: Watt

1 Watt is given by : 1 Watt = 1 Volt 1 Ampere

1 Watt = volt Ampere

In the above circuit, the ammeter has been connected in parallel and the voltmeter is connected in series which is not correct. The ammeter should be connected in series and voltmeter is parallel. The correct circuit diagram will be as follows:

Resistance = 2 Ω

Power = 18W

P = I²R

I = (P/R)^1/2

I = (18/2)^1/2

I = 3A

The resistance of an ammeter should be low because ammeter is connected in series in the circuit and if resistance will be high, no current will flow through the circuit.

Circuit diagram consisting of a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination is as follows:

Net resistance connected in parallel R = 2 Ω

Since resistance of 2 Ω and effective resistance of parallel combination of two 4 Ω resistances in series , same current will flow through them and therefore potential difference will be same across the 2 Ω resistor as across the parallel combination of 4 Ω resistors.

An electric fuse is wire which is made of a material of high resistance and low melting point. Using an electric fuse prevents the flow of unduly high electric current and protects the appliances from the damage. Due to Joule heating, the fuse melts to break the electric circuit.

Resistance of a uniform metallic conductor is directly proportional to the length (l) and inversely proportional to the area of cross-section.

i.e. R

R

Or R =

ρ is constant of proportionality and is known as electrical resistivity of the material

When length of wire is doubled, the current decreases because Resistance of a uniform metallic conductor is directly proportional to the length (l) and according to Ohm’s law, Resistance is inversely proportional to the current and therefore when the length of wire is increased, there is decrease in the value of current.

The commercial unit of electrical energy is Kilowatt hour (kWh). It can be converted into joules as follows:

1 kWh= 1000 watt seconds

1 kWh = 3.6 10⁶ watt second

1 kWh =3.6 10⁶ Joule

When 5Ω and resistance of lamp say R is connected in series:

Effective resistance R’ = 5+R

Applying ohm’s law:

V= IR

R = V/I = 10/1 = 10 Ω

Therefore, resistance of lamp, R= 10-5 Ω = 5 Ω

When a 10 Ω resistance is connected in parallel with the series connection

Effective resistance 1/R’’= 1/10 +1/10

R’’ = 5 Ω

Current through the circuit I = V/R = 10/5 = 2A

So, current flowing through each circuit will be of 1 V

Potential difference across the lamp will be same because voltage is same across the circuit.

If different appliances in a domestic are connected in series then if one of the appliance is switched off or stops working then the other appliances will also stop. Moreover, if they are connected in series, then then there will be different voltage for different appliances and there will be loss of voltage due to add on effects of resistances. Each appliance is connected in parallel to each other so that each appliance has an equal potential difference. That’s why series arrangement is not used in domestic circuits and parallel arrangement is used in domestic wiring.

(i) There will be no effect on glow of other two bulbs and will remain same when B1 gets fused because glowing of bulb depends on power and the potential difference and resistance remains same of other two bulbs.

(ii) When there are parallel connections:

Net resistance will be 1/R = 1/R₁+1/R₂+1/R₃

Since resistance is same so, R’ = R/3

Applying ohm’s law V= IR

R =4.5Ω

Since B2 gets fused, so now only two bulbs B1 and B3 are in parallel

Therefore net resistance in parallel 1/R’ =2/R

R’ = 4.5/2 Ω

I = V/R’ = 2×4.5/4.5

I = 2A

So, current will be distributed in both the bulbs as 1 A each.

(iii) Power dissipated when all three bulbs glow together

P = V × I

P= 4.5 × 3 = 13.5 W

(a) The bulbs which are in parallel connection will glow with more brightness than with series connection bulbs because in parallel connection, bulbs are connected with the same source. So, power in each bulb will be:

P = V² /2R whereas in series Power through each bulb will be P= V² /9R Therefore, bulbs in parallel connection will glow more.

(b) If one bulb in series circuit is fused then all the bulbs in the series circuit will get fused but in cause of parallel circuit, if one bulb fused the other two fused will continue to glow because in case of series circuit, same amount of current is distributed whereas in case of parallel circuit, the current is divided among electrical gadgets and is therefore, each bulb has different value of resistance and requires different current to operate.

According to Ohm’s law:

The electrical current flowing through a conductor is directly proportional to the potential difference across its ends providing the temperature remains same.

R

R= V/I

R is resistance of conductor

V is potential difference

I is current

It can be verified experimentally by following:

1. Set up an electrical circuit consisting of nichrome wire of length ×Y say 0.5m , an ammeter, a voltmeter and four cells of 1.5 V each .

2. Firstly, use one cell as source in the circuit. Note the current in the ammeter and voltmeter reading across the nichrome ×Y in the circuit.

3. Connect two cells in the circuit and note down the reading from the given circuit for the current and potential difference.

4. Repeat the same experiment for three cells and then four cells and note down the reading from Ammeter and Voltmeter.

5. Calculate the value of ratio to each pair of potential difference V and current I.

It is observed that the value of V/I is approximately same for each case and V-I graph obtained is a straight line which verifies Ohm’s law.

No, it doesn’t hold good under all conditions because the value of current is different for different components. Certain components offer an easy path for flow of resistance while others resists the flow.

Resistance of a uniform metallic conductor is directly proportional to the length (l) and inversely proportional to the area of cross-section.

i.e. R

R

Or R =

ρ is constant of proportionality and is known as electrical resistivity of the material

S.I. unit of electrical resistivity of a material is: Ω m.

An experiment to study the factors on which the resistance of conducting wire depends can be demonstrated as follows:

1. Complete an electrical circuit which consists of a cell, an ammeter, a nichrome wire of length l.

2. Note down the reading from Ammeter.

3. Replace the nichrome wire with twice the length of earlier nichrome wire having same thickness.

4. Note down the ammeter reading in the given circuit.

5. Replace the nichrome wire with thicker nichrome wire having a large cross-section area. Note down the current through circuit.

6. Now, replace the nichrome wire with copper wire having same length and same area of cross-section.

7. Note down the value of current and observe the difference in the current in all the cases.

Following observations were noted:

1. The ammeter reading decreases to one-half when the length of wire is doubled.

2. The ammeter reading is increased when thicker wire of same material and same length is used.

3. Change in ammeter reading is observed when copper wire is used.

So, Resistance of conducting wire depends on various factors:

1. Length

2. Area of cross-section

3. Nature of material

The same current flows through every part of the circuit containing three resistances in series connected to a battery can be prove with the help of following experiment:

1. Connect three resistors of different values in series connection.

2. Connect the resistors with an ammeter, a plug key and a battery.

3. Let the resistors be of 1Ω, 2 Ω and 3 Ω and battery of 6V.

4. Plug the key and note down the ammeter reading.

It will be observed that the same amount of current value flows through the circuit, independent of its position in the given circuit.

The same potential difference exists across three resistors connected in a parallel arrangement to a battery can be concluded as follows:

1. Make a parallel combination of three resistors ×Y of three resistors R₁, R₂ and R₃. Connect the resistors with a battery, an ammeter and a plug key.

2. Connect a voltmeter in parallel with the resistors.

3. Plug the key and note the ammeter reading and also note the voltmeter reading.

4. The potential difference across each resistor is V.

5. Now, take out the plug key from the circuit. Remove the ammeter and voltmeter from the given circuit.

6. Insert the ammeter in series with the resistor R₁ and note down the ammeter reading I₁.

7. Similarly, note the I₂ and I₃ across the resistors R₂ and R₃.

The total current through the circuit will be given by:

I = I₁+I₂+I₃

Applying Ohm’s law to the parallel combination of resistors

I = V/ R_P

For each resistor, ohm’s law will be given by:

I₁ = V/ R₁ ;I₂ = V/ R₂ ;I₃ = V/ R₃

Now, V/ R_P = V/ R₁+ V/ R₂+ V/ R₃

1/ R_P = 1/ R₁+ 1/ R₂+ 1/ R₃

Hence, the same potential difference exists across three resistors connected in a parallel arrangement to a battery.

If the electrical circuit is purely resistive, that is a configuration of resistors only connected to battery, the source energy continually gets dissipated in the form of heat. This is known as Joule’s heating effect of electric current. Joule’s law of heating is given by the relation:

H= I²Rt. The law implies that : heat produced in a resistor is

(i) Directly proportional to the square of current for a given resistance

(ii) Directly proportional to the resistance of given circuit.

(iii) Directly proportional to the time for which the resistance flows through the resistor.

It can be demonstrated experimentally by following:

When an electric field is applied across the ends of a conductor, the free electrons starts moving towards the electric field. These electrons suffers collision with the atoms which have lost electrons. Due to the collision, energy is transferred to the atoms and they vibrate as they gain energy due to which heat is developed in the conductor. The more is the current, the more will be heat. This is Joule’s heating effect.

An electric fuse is wire which is made of a material of high resistance and low melting point. Using an electric fuse prevents the flow of unduly high electric current and protects the appliances from the damage. Due to Joule heating, the fuse melts to break the electric circuit.

The four applications in daily life are:

1. In electrical heaters.

2. In electrical iron

3. In electrical fuse wire

4. In electrical bulb

(i) since resistances are connected in parallel

Therefore, effective resistance 1/R’ = 1/8+1/8 = 1/4

R’ = 4Ω

(ii) Current can be calculated with the help of Ohm’s law:

I = V/R = 8/4 = 2 A

(iii)Potential difference is given by, V= IR

V= 4 × 2 =8 Watt

(iv)Reading in ammeter A₁, I = V/R = 8/4 = 2 A

Reading in ammeter A₂ = 8/4

= 2 A

Difference in ammeter readings = 2- 2= 0