CBSE - Electricity

A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.

The unit of electric current is ampere (A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.

One electron possesses a charge of 1.6 ×10^-19 C, i.e., 1.6 ×10^-19 C of charge is contained in 1 electron.

∴ 1 C of charge is contained in  1 / 1.6 x 10^-19

= 6.25 x 10¹⁸ = 6 x 10¹⁸

Therefore, 6 x 10¹⁸ electrons constitute one coulomb of charge.

Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.

If 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.

The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.
Now we know that,
Potential difference = Work Done/Charge

∴ Work done = Potential difference × charge
Where, Charge = 1 C and Potential difference = 6 V

∴ Work done = 6 × 1
= 6 Joule.

The resistance of a conductor depends upon the following factors:
→ Length of the conductor
→ Cross-sectional area of the conductor
→ Material of the conductor
→ Temperature of the conductor

The current will flow more easily through thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross – section. If thicker the wire, less is resistance and hence more easily the current flows.

According to Ohm’s law
V = IR
⇒ I=V/R …                   (1)
Now Potential difference is decreased to half
∴ New potential difference Vʹ=V/2
Resistance remains constant
So the new current Iʹ = Vʹ/R
= (V/2)/R
= (1/2) (V/R)
= (1/2) I = I/2

Therefore, the amount of current flowing through the electrical component is reduced by half.

The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

Use the data in Table 12.2 to answer the following �
Table 12.2 Electrical resistivity of some substances at 20�C
 

	Material	Resistivity (Ω m)
Conductors	Silver	1.60 � 10−8
	Copper	1.62 � 10−8
	Aluminium	2.63 � 10−8
	Tungsten	5.20 � 10−8
	Nickel	6.84 � 10−8
	Iron	10.0 � 10−8
	Chromium	12.9 � 10−8
	Mercury	94.0 � 10−8
	Manganese	1.84 � 10−6
	Constantan (alloy of Cu and Ni)	49 � 10−6
Alloys	Manganin (alloy of Cu, Mn and Ni)	44 � 10−6
	Nichrome (alloy of Ni, Cr, Mn and Fe)	100 � 10−6
	Glass	1010 − 1014
Insulators	Hard rubber	1013 − 1016
	Ebonite	1015 − 1017
	Diamond	1012 − 1013
	Paper (dry)	1012

(a) Resistivity of iron = 10.0 x 10^-8 Ω
Resistivity of mercury = 94.0 x 10^-8 Ω
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Three cells of potential 2 V, each connected in series therefore the potential difference of the battery will be 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V and a plug key which is closed means the current is flowing in the circuit.

An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.

The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,
V = IR,
Where,
Potential difference, V = 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5 + 8 + 12 = 25Ω
I = V/R = 6/25 = 0.24 A
Potential difference across 12 Ω resistor = V1
Current flowing through the 12 Ω resistor, I = 0.24 A
Therefore, using Ohm’s law, we obtain
V1 = IR = 0.24 x 12 = 2.88 V
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.

(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.



Therefore, equivalent resistance ≈ 1 Ω

(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:

Let R be the equivalent resistance.



Therefore, equivalent resistance = 0.999 Ω

Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Potential difference of the source, V = 220 V
These are connected in parallel, as shown in the following figure.


Let R be the equivalent resistance of the circuit.

According to Ohm's law,

V = IR

Where,

Current flowing through the circuit = I

7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let R' be the resistance of the electric iron. According to Ohm's law,

There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.
The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors.


Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by


This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω
Hence the total resistance of the circuit is 4 Ω.

(b) The following circuit diagram shows the connection of the three resistors.


All the resistors are connected in series. Therefore, their equivalent resistance will be given as


Therefore, the total resistance of the circuit is 1 Ω.

There are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively.

(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by



Therefore, 2 Ω is the lowest total resistance.

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very law so it does not glow.

The amount of heat (H) produced is given by the Joule's law of heating as

Where,

Voltage, V = 50 V

Time, t = 1 h = 1 � 60 � 60 s

Amount of current,




Therefore, the heat generated is .4.8 � 10⁶  J

The amount of heat (H) produced is given by the joule’s law of heating asH= Vlt
Where,
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V

H= 100 x 5 x 30 = 1.5 x 10⁴ J.

Therefore, the amount of heat developed in the electric iron is 1.5 x 10⁴J.

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Power (P) is given by the expression,P = VI
Where,
Voltage,V = 220 V
Current, I = 5 A
P= 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 x 60 x 60 = 7200 s
∴ P = 1100 x 7200 = 7.92 x 10⁶ J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 x 10⁶ J

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is -
(a) 1/25
(b) 1/5
(c) 5
(d) 25

Ans (d) 25

Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R

Ans (b) IR²

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Ans (d) 25 W

(d)Energy consumed by an appliance is given by the expression,





Where,

Power rating, P = 100 W

Voltage, V = 220 V

Resistance, R =

The resistance of the bulb remains constant if the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as



Therefore, the power consumed will be 25 W

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Ans (c) 1:4

(c) The Joule heating is given by, H = i²Rt

Let, R be the resistance of the two wires.

The equivalent resistance of the series connection is RS= R + R = 2R

If V is the applied potential difference, then it is the voltage across the equivalent resistance.



The heat dissipated in time t is,



The equivalent resistance of the parallel connection is RP=

V is the applied potential difference across this RP.



The heat dissipated in time t is,



So, the ratio of heat produced is,



Note: H α R also H α i² and H α t. In this question, t is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H α R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i²Rt to find the ratio.

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

A copper wire has diameter 0.5 mm and resistivity of 1.6 � 10⁻⁸Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer

Area of cross-section of the wire, A =π (d/2) ²
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that



Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.

The plot between voltage and current is called IV characteristic. The voltage is plotted on x-axis and current is plotted on y-axis. The values of the current for different values of the voltage are shown in the given table.

The IV characteristic of the given resistor is plotted in the following figure.

The slope of the line gives the value of resistance (R) as,

Slope = 1/R = BC/AC = 2/6.8

R= 6.8/2 = 3.4 Ω

Therefore, the resistance of the resistor is 3.4 Ω.

Resistance (R) of a resistor is given by Ohm�s law as,V= IR
R= V/I
Where,
Potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10^-3 A



Therefore, the resistance of the resistor is 4.8 kΩ

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

V= IR

I= V/R

Where,

R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R.

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Potential difference, V= 9 V

I= 9/13.4 = 0.671 A
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.

For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm�s law asV= IR
R= V/I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R,given as



Therefore, four resistors of 176 Ω are required to draw the given amount of current.

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2 = 3 Ω is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel



Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be



The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
(b) Two resistor in series



Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.

The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be



Therefore, the total resistance is 4 Ω.

Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because R=V²/P



∴ Number of electric bulbs connected in parallel are 110.

Supply voltage, V= 220 V
Resistance of one coil, R= 24 Ω



(i) Coils are used separately

According to Ohm�s law,
V= I₁R₁
Where,

I1 is the current flowing through the coil

I1 = V/R₁ = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, R₂ = 24 Ω + 24 Ω = 48 Ω

According to Ohm�s law,V = I₂R₂
Where,

I₂ is the current flowing through the series circuit

I₂ = V/R₂ = 220/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel

Total resistance, R₃ is given as =



According to Ohm�s law,

V= I₃R₃
Where,
I₃ is the current flowing through the circuit I3 = V/R₃ = 220/1₂ = 18.33 A
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I= 6/3 = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,
P= (I)2R = (2)2 x 2 = 8 W

(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
P= V2/R = 42/2 = 8 W
Therefore, the power used by 2 Ω resistor is 8 W.

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,Power = Voltage x Current
Current =  Power/Voltage = 60/220 A
Hence, current drawn from the line = 100/220 + 60/220 = 0.727 A

Energy consumed by an electrical appliance is given by the expression,H= Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×10⁵ J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600
= 7.2×10⁵ J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

Rate of heat produced by a device is given by the expression for power as, P= I²R
Where,
Resistance of the electric heater, R= 8 Ω
Current drawn, I = 15 A

P= (15)² x 8 = 1800 J/s

Therefore, heat is produced by the heater at the rate of 1800 J/s.

(a) The melting point and of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.
(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.
(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa. 

R α 1/A
(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of figure12.1. The current recorded in the ammeter will be

(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases
Ans. (d) The same in all the cases
Explanation: None of the conditions change in any of the circuits. Hence, current would be same in all circuits.

2. In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) same in all the cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)
Ans. (c) Maximum in case (ii)
Explanation: In this case, two resistors are in series. Hence, their sum will be equal to their arithmetic sum. In figure (iii) the total resistance will be less than individual resistances because they are connected in parallel. A higher resistance produces more heat so option (c) is correct.

3. Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of the material
Ans. (d) Nature of the material

4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly
(a) 1020
(b) 1016
(c) 1018
(d) 1023

ans A
Explanation: We have current I=1 A and time t=16s
We know;
I = Q / t or 1 A = Q/ 16 s

Q= 16C

Charge contained in 1 electron 1.6 x 10-19 C
So, 16 C charge is contained in following number of electrons;

5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Ans. (b) (ii)

6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω
Ans. (d) 1Ω

7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω ?

(a) 1/5 Ω
(b) 1/25 Ω
(c) 1/10 Ω
(d) 25 Ω
Ans. (b) 1/25 Ω
Explanation: When resistors are connected in parallel then we get the minimum resistance out of the combination.
When given resistors are connected in parallel, the resistance of combination can be calculated as follows:

8. The proper representation of series combination of cells (Figure12.4) obtaining maximum potential is

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Ans. (a) (i)
Explanation: In this combination, positive terminal of next cell is adjacent to negative terminal of previous cell.

9. Which of the following represents voltage?

Ans a

10. A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Ans. (c) 2A

Explanation: Both the conductors are of same material hence their resistivity is same.


11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1 R2 and R3 respectively (Figure. 12.5). Which of the following is true?

(a) R1= R2= R3
(b) R1> R2 > R3
(c) R1> R2> R3
(d) R1 > R2 > R3

ANS C
Explanation: Current is inversely proportional to resistance. So, higher resistance would allow less current to pass; which is shown by resistance R3.

12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

(a) 100 %
(b) 200 %
(c) 300 %
(d) 400 %
Ans. (c) 300 %
Explanation: The heat generated by a resistor is directly proportional to square of current. Hence, when current becomes double, dissipation of heat will multiply by 22 =4. This means there will be an increase of 300%.

13. The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed
Ans. (c) the shape of the resistor is changed

14. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B
Ans. (c) Brightness of bulb B will be more than that of A
Explanation: Since bulbs are connected in parallel so resistance of combination would be less than arithmetic sum of resistance of all the bulbs. So, there will be no negative effect on flow of current. As a result, bulbs would glow according to their wattage.

15. In an electrical circuit two resistors of 2 O and 4 O respectively are connected in series to a 6 V battery. The heat dissipated by the 4 O resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Ans. (c) 20 J
Explanation: Total resistance of combination =2 Ω + 4 Ω = 6Ω
Current through the circuit can be calculated as follows:
I = V / R
= 1A
Heat dissipation by 4Ω can be calculated as follows:
H = I² RT = (IA)² x 4Ω x 5s = 20 J

16. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A
Ans. (d) 5 A
Explanation: Power=1 kW= 1000W
Current flowing through the kettle can be calculated as follows:
P = V x I
Or 1000W = 220 V x I

Or I = 1000w / 220V = 4.24A
So, required rating of fuse wire=5A

17. Two resistors of resistance 2 Ω  and 4 Ω  when connected to a battery will have
(a) same current flowing through them when connected in parallel
(b) same current flowing through them when connected in series
(c) same potential difference across them when connected in series
(d) different potential difference across them when connected in parallel
Ans. (b) same current flowing through them when connected in series.
Explanation: In series combination, current does not change because each resistor receives a common current. In other words, current does not get divided into branches.

18. Unit of electric power may also be expressed as
(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second
Ans. (b) kilowatt hour

A child has drawn the electric circuit to study Ohm�s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

ANS
Correct diagram is as follows:

Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors?


Current can be calculated as follows:
P =I²R

or, 18W = I²× 2Ω

I² = 18W / 2Ω = 9

or I = 3A

Resistance of ammeter should ideally be zero so that ammeter should not affect the flow of current in circuit. Hence, resistance of an ammeter should be very low because zero resistance is not possible in real life.

Total resistance for parallel combination of 4Ω resistors can be calculated as follows:

1/ R = 1/4 + 1/4 =1/2 or R = 2Ω

Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 2Ω resistance will be same as potential difference across the other two resistors which are connected in parallel.

Fuse wire has low resistance than current rating of the main wiring. So, whenever there is a surge in electric current, the fuse wire melts and breaks the circuit. This prevents any damage to the electrical appliances. Thus, a fuse wire protects electrical appliances.

The inherent property of a conductor because of which it resists the flow of electric current is called resistivity. Resistivity for a particular material is unique. Resistance varies directly as length of the conductor. Current varies inversely as resistance. So, when length of the wire is doubled, its resistance becomes double. When resistance becomes double, current becomes half. This explains why the reading of ammeter decreases to half when the length of the wire is doubled.

The commercial unit of electrical energy is kilowatt hour.
 1 kWh = 11000 W h
 1000 W h= (1000 Joule/second) ×hour 
 
 = 1000 x J / S x 60 x 60 s 
 = 3600 x 1000 J 
 = 3.6 x 10⁶ J

Total resistance of circuit can be calculated as follows: 

R= V / I 

= 10 / 1

= 10  Ω

Since lamp and conductor are in series so resistance of lamp =10 Ω − 5Ω = 5Ω .The new resistance in parallel to earlier combination has same value, i.e. 10Ω as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 5Ω conductor. Now, resistance remains the same but current has become half. Using Ohm formula, potential difference across the lamp can be calculated as follows: V = IR = 0.5A x 5Ω = 2.5V

In parallel arrangement, total resistance of the circuit is less than arithmetic sum of resistors in the wiring. This helps in reducing the load on the wiring. In parallel arrangement, each appliance can have its independent switch. Fault in one section is not going to affect any other section. These advantages make parallel arrangement ideal of domestic wiring

B1, B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
Ans. In parallel circuit, potential difference does not get divided. Hence, glow of other bulbs will not be affected when bulb B1 gets fused.

(ii) What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
Ans. Ammeter A shows a reading of 3A. This means each of the A1, A2, and A3 show 1A reading.
When the bulb B2 gets fused, no current flows through this bulb. So, all the current is equally divided between remaining two bulbs. So, ammeters A1 and A2 will show 3/2 = 1.5 A current each. Ammeter A3 will show zero current.

(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
Ans. For finding power, we need to first calculate the resistance in the circuit

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
Ans. Voltage gets divided in series combination, So bulbs in series combination will glow with less brightness. Voltage does not get divided in parallel combination, so bulbs in parallel combination will glow with more brightness.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
Ans. In case of series combination, fault in even a single component will break the circuit.
So, when one of the bulbs gets fused; another bulb would continue to glow.

Ohm’s Law: At constant temperature, potential difference across a conductor is directly proportional to electric current passing through it. If V is the potential difference and I is electric current, then as per Ohm’s law;

VαI

or V/ I = R

Where R is the constant of proportionality and it is called resistance.

Verifying Ohm’s Law:

• Take a nichrome wire, some electric cells, key, voltmeter and ammeter.
• Connect the component to make a circuit; as shown in given figure.
• Start with 1 cell in the circuit. Switch on the key and take the reading of ammeter.
• Increase the number of cells to 2. Switch on the key and take the reading of
ammeter.
• Increase the number of cells in similar increments and take the reading of
ammeter in each case.
Tabulate your data and plot a V-I graph.

The graph shows that potential difference varies directly as electric current. Ohm’s law holds good under normal conditions. But, it may not hold good under exceptional conditions.

The inherent property of a conductor because of which it resists the flow of electric
current is called resistivity. Resistivity for a particular material is unique.
The SI unit of resistivity is Ωm (Ohm metre).

Experiment to study the factors on which resistance of conducting wire depends:
• Take an ammeter, electric cell, plug key, nichrome wire and wires of different materials.
• Make the circuit as shown in figure.
• Start the experiment with nichrome wire. Attach it in the circuit and take ammeter reading.
• Change the length of nichrome wire and take ammeter reading.
• Change the thickness of nichrome wire and take ammeter reading.
• After above steps, use copper wire for the experiment. Attach a copper wire in the circuit and take ammeter reading.
• Change the length of copper wire and take ammeter reading.
• Change the thickness of copper wire and take ammeter reading.
• Repeat above steps with wires of different materials. 

Observations:
• It is seen that resistance depends on material of conductor.
• Resistance depends on length of conductor.
• Resistance depends on area of cross-section. 

• Take three resistances R1, R2 and R3.
• Connect them in series combination to complete a circuit; as shown in figure.
• First of all, connect the ammeter between battery and R1 and take its reading.
• After that, connect the ammeter between R1 and R2 and take its reading.
• Then, connect the ammeter between R2 and R3 and take its reading.

Observation:
It is observed that ammeter shows same reading in all situation. This shows that same current flows through every part of the circuit containing three resistances in series.

• Take three resistors R1, R2, and R3 and connect them in parallel to make a circuit; as shown in figure.
• Use voltmeter to take reading of potential difference of three resistors in parallel combination.
• Now, remove the resistor R1 and take the reading of potential difference of remaining resistors’ combination.
• Then, remove the resistor R2 and take the reading of potential difference of remaining resistor.

Observation: Voltmeter reading was same in each case. This shows that the same potential difference exists across three resistors connected in a parallel arrangement. 
 

The Joule’s Law of Heating states that the heat produced in a resistor is:
 (a) Directly proportional to the square of current for a given resistor.
 (b) Directly proportional to resistance for a given current, and
 (c) Directly proportional to the time for which the current flows through the resistor.
 This can be expressed by following equation:
 
 H =I²Rt
 
 Here; I is electric current, R is resistance, t is time and H is heating effect.
 
Experiment to Demonstrate Joule’s Law of Heating:
• In this experiment, we will show the effect of current on heating.
• Take a water heating immersion rod and connect to a socket which is connected to regulator. It is important to recall that a regulator controls the amount of current flowing through a device.
• Keep the pointer of regulator on minimum and count the time taken by immersion rod to heat a certain amount of water.
• Increase the pointer of regulator to next level. Count the time taken by immersion rod to heat the same amount of water.
• Repeat above step for higher levels on regulator to count the time.

Observation: It is seen that with increased amount of electric current, less time is required to heat the same amount of water. This shows Joule’s Law of Heating. 
Application: Electric toaster, oven, electric kettle and electric heater etc. work on the basis of heating effect of current. 
 

Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination
Ans. These resistors are in parallel, so effective resistance can be calculated as follows:

1/ R = 1/8 + 1/8
=1/4 Ω

R= 4Ω

(b) Current flowing through 4 Ω resistor
Ans. For this, we first need to find effective resistance in the circuit,
= 4Ω +4 Ω =8 Ω
v / I = R or

I = 1 A

(c) Potential difference across 4 Ω resistance
Ans. Potential difference is getting divided into two resistance of 4Ω each. Hence,
potential difference across 4Ω = = 8 V/ 2 = 4V

(d) Power dissipated in 4 Ω resistor
Ans. P= I2Rt = (IA)2 x 4Ω =4W

(e) Difference in ammeter readings, if any.
Ans. Since resistors are connected in series, so there will be no difference in ammeter reading.