12

ELECTRICITY

Revolt of 1857
 ELECTRICITY 2

1. A cylindrical conductor of length ‘l’. and uniform area of (d) Neither of the two circuits
cross-section ‘A’ has resistance ‘R’. The area of cross- [CBSE, Term 1, Set 1, 2016]
section of another conductor of same material and same 10. A student found that when a resistance of 3 Ω was joined
resistance but of length ‘2l’ is with 3 V battery as per figure shown below, the current
A 3A flowing through it was 1 A. He then joined another
(a) (b) resistance of 6 Ω in parallel with 3 Ω resistance. The
2 2 reading in the ammeter will now be:
(c) 2A (d) 3A

[CBSE OD, Set 1, 2020]

2. The maximum resistance which can be made using four

1
resistors each of resistance Ω is:
2 (a) 9 A (b) 1.5 A
(a) 2Ω (b) 1Ω
(c) 1 A (d) 6 A
(c) 2.5 Ω (d) 8Ω
[CBSE, Term 1, Set 1, 2016]
[CBSE OD, Set 1, 2020]
11. Two resistances are connected in series as shown in the
3. At the time of short circuit, the electric current in the diagram. The potential difference across 12 Ω resistor
circuit : will be:
(a) vary continuously (b) does not change
(c) reduces substantially (d) increases heavily
[CBSE Delhi, Set 1, 2020]
4. Two bulbs of 100 W and 40 W are connected in series.
The current through the 100 W bulb is 1 A. The current
through the 40 W bulb will be :
(a) 0.4 A (b) 0.6 A
(c) 0.8 A (d) 1 A
[CBSE Delhi, Set 1, 2020]
5. Why are the heating elements of electric toasters and (a) 6 V (b) 2.4 V
electric irons made of an alloy rather than a pure metal? (c) 2.8 V (d) 12 V
[CBSE OD, Set 1, 2019]
6. Should the resistance of a voltmeter be low or high? Give [CBSE, Term 1, Set 1, 2015]
reason.
12. The current flowing through a resistor connected in a
[CBSE OD, Set 2, 2019]
circuit and the potential difference developed across its
7. Name and define the SI unit of current. ends are as shown in the diagram by milli-ammeter and
[CBSE Delhi, Set 2, 2019] voltmeter readings respectively:
8. Write the function of voltmeter in an electric circuit. (a) What are the least counts of these meters?
[CBSE Delhi, Set 3, 2019] (b) What is the resistance of the resistor?
9. To find the equivalent resistance of two resistors
connected in series, the connection of ammeter is correct
in the circuit:

[CBSE OD, Set 1, 2019]

13. While studying the dependence of potential difference
(a) Circuit A (V) across a resistor on the current (I) passing through it,
(b) Circuit B in order to determine the resistance of the resistor, a
(c) Both the circuits student took 5 readings for different values of current and
 ELECTRICITY 3

plotted a graph between V and I. He got a straight line

graph passing through the origin. What does the straight
line signify? Write the method of determining resistance
of the resistor using this graph.
[CBSE Delhi, Set 1, 2019]
14. The values of current (I) flowing through a given resistor
of resistance (R), for the corresponding values of
potential difference (V) across the resistor are given
below:

Plot a graph between current (I) and potential difference

(V) and determine the resistance (R) of the resistor.
[CBSE, 2018]
15. Find the least count of a miliammeter in which there are
20 division between 400 mA and 500 mA marks. (i) Select the circuit diagrams which are correct.
[CBSE, Term I, Set I, 2016]
(ii) Give reason for the circuit diagrams which are not
16. In an experiment, to study the dependence of potential
correct.
difference (V) on the electric current (I) across a
[CBSE Delhi, Term 2, Set 1, 2015]
conductor (resistor), if the circuit is on for long time,
18. A V-I graph for a nichrome wire is given below. What do
then–select two correct options from the following:
you infer from this graph?
(i) Zero error of an ammeter will be changed.
Draw a labelled circuit diagram to obtain such a graph.
(ii) Zero error of a voltmeter will be changed.
[CBSE Delhi, Set 1, 2020]
(iii) Value of a resistance will be changed.
(iv) Resistor will be heated.
[CBSE, Term 1, Set 1, 2015]
17. To study the dependence of potential difference (V) on
current I across Resistor (R), four circuit diagrams are
prepared.

19. (a) Write the mathematical expression for Joule’s law of

heating.
(b) Compute the heat generated while transferring 96000
coulomb of charge in two hours through a potential
difference of 40 V.
[CBSE Delhi, Set 1, 2020]
20. (a) State the relation correlating the electric current
flowing in a conductor and the voltage applied across it.
Also draw a graph to show this relationship.
(b) Find the resistance of a conductor if the electric
current flowing through it is 0.35 A when the potential
difference across it is 1.4 V.
[CBSE Delhi, Set 2, 2020]
21. (a) For the combination of resistors shown in the
following figure, find the equivalent resistance between
M & N.
 ELECTRICITY 4

(b) State Joule’s law of heating.

(c) Why we need a 5 A fuse for an electric iron which
consumes 1 kW power at 220 V?
(d) Why is it impracticable to connect an electric bulb
and an electric heater in series?
[CBSE OD, Set 1, 2020]
22. (a) Define power and state its SI unit.
(b) A torch bulb is rated 5 V and 500 mA.
Calculate its
(i) Power
(ii) Resistances
(iii) Energy consumed when it is lighted for 2½ hours.
[CBSE OD, Set 2, 2020]
23. (a) An electric bulb is rated at 200 V; 100 W. What is its
resistance?
(b) Calculate the energy consumed by 3 such bulbs if
they glow continuously for 10 hours for complete month
of November.
(c) Calculate the total cost if the rate is 6.50 per unit.
[CBSE OD, Set 3, 2020]
24. (a) What is meant by the statement, “The resistance of a
conductor is one Ohm”?
(b) Define electric power. Write an expression relating
electric power, potential difference and resistance.
(c) How many 132 W resistors in parallel are required to
carry 5 A on a 220 V line?
[CBSE OD, Set 3, 2020]
25. (a) How will you infer with the help of an experiment
that the same current flows through every part of a circuit
containing three resistors in series connected to a battery?
(b) Consider the given circuit and find the current
flowing in the circuit and potential difference across the
15 Ω resistor when the circuit is closed.

[CBSE OD, Set 1, 2019]

 ELECTRICITY 5

Solutions:
1. (c) 2A
2. (a) 2Ω
3. (d) Increases heavily
4. (d) 1A
5. The resistivity of an alloy is generally higher than that
of its constituent metals. 1
 R
Alloys do not oxidize (burn) readily at higher
Slope of V  I graph
temperatures. Therefore, conductors of electric heating
14. We know that, V = IR
devices, such as toasters and electric irons, are made up
of an alloy rather than pure metal.
6. The resistance of a voltmeter should be high, because
voltmeter is connected parallel to the component of a
circuit and it also takes negligible current from the circuit
in order to measure the potential difference accurately.
7. ‘Ampere’ is the SI unit of current.
1 Ampere current can be defined as a unit charge flowing
per second in the circuit.
From the graph, V = 0.5
1Coulomb
 1Amp  I = 0.1
1Sec ond ⇒ 0.5 = 0.1 × R
8. Voltmeter measures the potential difference across 0.5
 R
two points in a circuit. It is always connected in parallel 0.1
in the circuit..  R  5
9. (c) Both the circuits
15.
10. (b) 1.5 A
Range
11. (b) 2.4 V Least count 
12. (a) 10 mA and 0.1 V No.of division
(b) V = 2.4 volt, I = 250 mA = 0.25 A 500  400
 
From Ohm’s law, 20
100
V 2.4    5mA
R   9.6 20
I 0.25 16. (iii) Value of resistance will be changed.
13. The straight line in the graph signifies that potential (iv) Resistor will be heated.
difference and current are directly proportional to each 17. (i) Circuit diagrams (I) and (III) are correct.
other. (ii) Circuit diagram (II) is incorrect because ammeter is
The method of determining resistance of resistor using always connected in series with the resistor and voltmeter
the graph is by Ohm’s law, V = IR and by calculating the is always connected in parallel to the resistor.
Circuit diagram (IV) is incorrect because the negative (–
slope from the points mentioned on the graph
ve) terminal of the ammeter is connected to the positive
(+ve) terminal of the battery.
18. From the graph, we infer that as the current in the
nichrome wire increases, the potential difference across it
increases linearly. Thus, the graph follows the Ohm’s
law.
 ELECTRICITY 6

Mathematically; V = IR
where, V = Voltage applied in volts
I = Current flowing in circuit in amperes.
R = Resistance of conductor (proportionality constant)
Graphically; slope of will determine resistance of
conductor.

19. (a) According to the Joule’s law of heating, heat

produced in a resistor is directly proportional to the :
(i) square of current (I) for a given resistance.

(ii) resistance (R) for a given current.

(iii) the time (t) for which the current. flows through the
resistor.

Mathematical form of Joule’s law of heating is: (b) Given, current flowing through conductor (I) = 0.35 A
Potential difference across conductor
H = I2Rt (V) = 1.4 V
(b) Given, charge (q) = 96000 C We know, according to Ohm’s law,
V = IR
Time (t) = 2 hrs =120 min = 7200 s ⇒ 1.4 V = (0.35 A) × R
Potential difference (V) = 40 volt  1.4V 
 R   4
We know that,  0.35 A 
Heat (H) = V It, where I is current...(1) Therefore, resistance of conductor is 4 ohms.
21. (a) (R3 × R4/R3 + R4) + R1 + R2
Also, (b) Joule’s law of heating implies that heat produced in a
resistor is (i) directly proportional to the square of
I = q/t, where q is charge and t is time in seconds. ...(ii) current for a given resistance, (ii) directly proportional to
resistance for a given current and (iii) directly
∴By putting I = q/t in eqn. (1) we get,
proportional to the time for which the current flows
H = (V × q/t) × t = Vqt/t = Vq through the resistor.
H = I2Rt
⇒ H = 40 × 96000 joule where
I = Current
⇒ H = 3840000 joule
R = Resistance
20. (a) The Ohm’s law states that the current carrying in a T = Time taken
conductor is directly proportional to the voltage applied (c) For an electric iron which consumes 1 kW electric
across the ends of the conductor, keeping the resistance power when operated at 220 V, a current of (1000/220)
constant. A, i.e., 4.54 A will flow in the circuit. In this case, a 5 A
fuse must be used.
(d) In a series circuit the current is constant throughout
the electric circuit. Thus, it is obviously impracticable to
connect an electric bulb and an electric heater in series,
because they need currents of widely different values to
operate properly.
22. (a) Power is defined as the rate of doing work, it is the
work done in unit time.
The SI unit of power is Watt (W) which is joules per
According to ohm’s law; V ⇒ I second (J/s).
 ELECTRICITY 7

(b) Given: Potential difference = V = 5 V

500 24. (a) Ohm’s law states that current flowing through a
Current = I = 500 mA =  0.5 A
1000 A conductor is directly proportional to the potential
difference maintained across the two ends of a conductor
(i) Power = V × I = 5 × 0.5 at constant temperature and pressure.
Let current flowing through a conductor is I, V is the
5
 5  2.5W potential difference maintained across the two ends of a
10 conductor.
(ii) Resistance = R = ? As per Ohm’s law, V ∝ 1
Thus, V = IR
By Ohm’s law: V = IR Here R is a proportionality constant called resistance.
As per the question, the resistance of a conductor is 1
V
R Ohm.
I Let V = 1 volt and I = 1 A, then R = 1
5 Ohm Hence, the resistance is said to be 1 Ohm if 1
R ampere of current flows through a circuit due to the
0.5
potential difference of 1 volt.
50
R  10 ohms (b) Electric power is the rate at which work is done of
5 energy is transformed in an electrical circuit.
The formula for electric power is given by :
1
(iii) Energy consumed when it is lighted for 2 hours P = VI
2 Where,
Energy = Power × Time P is the power
V is the electric current
t = 2.5 × 60 × 60 sec. Power can also be written as,
So, Energy= 2.5 × 2.5 × 60 × 60 P  I 2R
= 22500 Joules/sec. V2
 P
23. (a) Here, potential difference across the bulb,
R
The above two expression are given by using Ohm’s law,
V = 200 V
Power of the bulb, P = 100 W where voltage, current and resistance are related by the
following relation.
V 2  200V 
2
V2 Where,
As P  ,R  
R P 100W R is the resistance in the circuit.
V is the potential difference in the circuit
4 104
   400 I is the electric current
100 (c) For x number of resistors of resistance 132 W
(b) Electric energy consumed by 1 bulb in 10 hours for Supply voltage, V = 220 V
30 days, i.e. Current, I = 5 A
W = Pt Equivalent resistance of the combination = R, given as
= 100 W × 10 × 30 1  1 
= 30000 Wh  x 
Electric energy consumed by 3 bulbs
R  132 
= 30000 × 3 132
R
= 90000 Wh x
= 90 kWh From Ohm’s law:
= 90 unit
120
(c) Cost of 90 units of electric energy 132 x 
= 90 × 6.50 = Rs. 585. 5
 ELECTRICITY 8

⇒ 132x = 44
⇒ 132 = 44x
⇒x=3
So, 3 resistors are having resistance 132 Ω
25. (a) Let three resistors R1, R2 and R3 are connected in
series which are also connected with a battery, an
ammeter and a key as shown in figure.

When key is closed, the current starts flowing through the

circuit. Take the reading of ammeter. Now change the
position of ammeter to anywhere in between the resistors
and take its reading. We will observe that in both the
cases reading of ammeter will be same showing same
current flows through every part of the circuit above.

(b) Given,

R1 = 5 Ω, R2 = 10 Ω, R3= 15 Ω, V = 30 V

Total resistance, R = R1 + R2 + R3

[ 5 Ω, 10 Ω and 15 Ω are connected in series]

= 5 + 10 + 15

= 30 Ω

Potential difference, V = 30 V

Current in the circuit, I =?

From Ohm’s law.

V 30V
I   1A
R 30
∴ Current flowing in the circuit = 1 A

Potential difference across 15Ω resistors = IR3

= 1A × 15Ω

= 15 V