Worksheet - Electricity
Electricity - Solved Worksheet
- Step-by-Step
Approach: Break down the problem into smaller steps
and solve each step methodically.
- Understand
the Question: Ensure a clear understanding of what the
problem is asking before attempting a solution.
- Identify
Given Information: Identify and list down the known values,
equations, or formulas relevant to the problem.
- Choose
the Right Formula: Select the appropriate formula or
equation that applies to the given problem.
- Substitute
and Solve: Plug in the given values into the formula
and solve for the unknown.
- Check
Units and Dimensions: Ensure consistency in units and dimensions
throughout the calculations.
(a) ρ
(b) ρ/2
Answer: a. The resistivity is a constant. It depends on the material of the conductor. However It depends on temperature.
2. A battery of 10 volt carries 20,000 C of charge through a resistance of 20 Ω. The work done in 10 seconds is-
(a) 2 × 10² joule
(b) 2 × 10⁵ joule
(c) 2 × 10⁴ joule
(d) 2 × 10² joule
Answer: b
Given: Q = 20,000 C, V = 10 V, R = 20 Ω
We know,
W = V x Q
= 10 x 20000
= 2 x 10⁵
3. A boy records that 4000 joule of work is required to transfer 10 coulomb of charge between two points of a resistor of 50 Ω. The current passing through it is-
(a) 2 A
(b) 4 A
(c) 8 A
(d) 16 A
Answer: c
Given: Q = 10 C, W = 4000 J, R = 50 Ω
We know V = W/Q
= 4000/10
= 400 V
We know I = V/R (ohm’s law)
= 400/50
= 8 A
4. To get 2 Ω resistance using only 6 Ω resistors, the number of them required is -
(a) 2
(b) 3
(c) 4
(d) 6
Answer: b
We know, in Parallel combination –
1/ Rp = 1/R + 1/R + 1/R
Rp = R/3
= 6/3
= 2 Ω
5. The least resistance obtained by using 2 Ω, 4 Ω, 1 Ω and 100 Ω is-
(a) < 100 Ω
(b) < 4 Ω (
c) < 1 Ω
(d) > 2 Ω
Answer: c
It is known that the least resistance will be obtained when the resistances would be connected in parallel and the equivalent resistance would be even smaller than the least resistance of the circuit.
6. 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
Answer: c
We know
R = ρ x l/A – (1)
R = ρ x 2l/A’ – (2)
ρ x l/A = ρ x 2l/A’
A’ = 2 A
7. 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
Answer: d
Maximum Resistance is when resistors are in series combination. I.e.
R = 1/5 + 1/5 + 1/5 + 1/5 + 1/5
= 1 Ω
8. 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
Answer: a
1 A = 1 C/1 second, and 1 C = Charge of 6.25 x 10¹⁸ Electrons = (1/ (1.6 x 10⁻¹⁹)
I = Q/t or Q = I x t
= 1 x 16
= 16 C
16 C = 16 x 6.25 x 10¹⁸
= 10²⁰
9. A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply. The rating of fuse to be used is
(a) 2.5 A
(b) 5.0 A
(c) 7.5 A
(d) 10 A
Answer: d
We know, P = V x I,
Total Current required is = 1500/200 + 500/200
= 10 A
10. 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 %
Answer: c
We know, P = I² x R
Now I is increased by 100%,
I’ = I + 100 x I/100 = 2I
P’ = I′² x R
= (2I)² x R
= 4 I² x R
= 4P
Change in Power dissipation = P’ – P
= 4P – P
= 3 P
% increase in Power dissipation = (P’ – P)/P x 100
= 3P/P x 100
= 300%
11. Electric power is inversely proportional to
(a) resistance
(b) voltage
(c) current
(d) temperature
Answer: a We know, P = V x I
= V² / R , P α 1/R keeping V constant
12. An electric bulb is connected to a 220V generator. The current is 0.50 A. What is the power of the bulb?
(a) 440 W
(b) 110 W
(c) 55 W
(d) 0.0023 W
Answer: b
Given: V = 220V, I = 0.50 A,
We know P = V x I
= 220 x 0.5
= 110 W
13. Two appliances of rating 200 watt-250 volts and 100 watt-250 volts are joined in series to a 250 volts supply. Total power consumed in the circuit is
(a) 46 watt
(b) 67 watt
(c) 10 watt
(d) 30 watt
Answer: b We know, P = V x I
= V² / R, 0r R = V² /P
R1 = 250 x 250 /200
= 625/2 Ω
R2 = 250 x 250 /100
= 625 Ω
Total Resistance Rs = R1 + R2
= 625/2 + 625
= 1875/2 Ω
Total Power consumed = V2 / Rs
= 250 x 250 x 2/1875
= 66.67 W
= 67 W approx.
14. Two electric bulbs have resistances in the ratio 1:2. If they are joined in series, the energy consumed in them is in the ratio.
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:1
Answer: b We know, P = V x I = I² x R
Energy consumed E = P x t = I² x R x t
In series combination, the current is same.
E1 = I² x R x t,
E2 = I² x 2R x t
E1/E2 = 1/2
15. Two heater wires of equal length are first connected in series and then in parallel with a battery. The ratio of heat produced in the two cases is:
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
Answer: c We know, H = I² x R x t
= V² x t /R, (V = I x R)
For series combination, H1 = V² x t/2R
For Parallel combination, H2 = V² x t/R/2
H1/H2 = 4/1
16. 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
Answer: d We know, Prated = Vrated x Irated
= V²rated /R
R = V²rated / Prated
= 220 x 220/100
= 22 x 22 ohms
Now, when operated on 110V,
P₁₁₀ = 110 x 110/22 x 22 = 25 W
17. Two heater wires of same length and material but of different thickness are connected in series across a power supply. Thus power dissipated :
(a) Will be same in both
(b) Will be more in thinner wire
(c) Will be more in thicker wire
(d) Cannot be predicted
Answer: b
H α I² x R x t
Heat generated directly proportional to the Resistance of the conductor.
R α 1/Area of Cross-section
The thinner wire will have more resistance; hence heat dissipated will be more in thinner wire.
18. The resistance of hot filament of the bulb is about 10 times the cold resistance. What will be the resistance of 100 W-220 V lamps, when not in use?
(a) 48 Ω
(b) 400 Ω
(c) 484 Ω
(d) 48.4 Ω
Answer: d
We know, Prated = Vrated x Irated = V²rated / Rhot
Rhot = V²rated / Prated
= 220 x 220/100
= 22 x 22 ohms
= 484 ohms
Rold = Rhot /10
= 484/10
= 48.4 ohms
20. A 24 V potential difference is applied across a parallel combination of four 6 Ω resistor. The current in each resistor-
(a) 1 A
(b) 4 A
(c) 16 A
(d) 36 A
Answer: c
We know, Equivalent Resistance in case of Equal resistances in Parallel combination = R/n
Where n is the number of resistors.
R = 6 Ω and n = 4
Rp = 6/4 = 1.5 Ω
From ohm’s law, I = V/ Rp
= 24/1.5 = 16 A
21. A wire is drawn such that its radius changes from r to 2r. The new resistance is –
(a) 1 times
(b) 4 times
(c) 8 times
(d) 1/16 times
Given: Old Radius = r units New Radius = 2r units
Volume of wire remains constant after stretching also.
Volume of old resistance = Volume of new resistance
A x l = A’ x l’
Π x r² x l = Π x (2r)² x l’
l’ = ¼ x l
We know, R = ρ x l/A = ρ x l/π r²
Rold = ρ x l/π r² and Rnew = ρ x l’/π (2r)² = ρ x ¼ x l /π 4r² = 1/16 x ρ x l/π r²
Rnew = 1/16 Rold
22. A fuse wire is inserted in a –
(a) Live Wire
(b) Neutral Wire
(c) Earth Wire
(d) Any of the above Wire
Answer: a
23. The type of current supplied by a Cell or Battery is –
(a) AC
(b) DC
(c) Both (A) and (B)
(d) None of these
Answer: b
24. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(a) I R²2
(c) VI
(d) V²/R
Answer: b
25. Two devices are connected between two points say A and B in parallel. The physical quantity that will remain the same between the two points is
(a) current
(b) voltage
(c) resistance
(d) None of these
Answer: b
Reviewed by Syed Hafiz Choudhary
on
December 30, 2023
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