HSLC EXAM QUESTIONS – ELECTRICITY
MCQ & Very Short Questions (1 Mark Each)
Answers are provided after each question.
1. The unit of specific resistance of a material in S.I. system is
(a) Ω (b) Ωm (c) Ωm⁻¹ (d) Ωm²
✅ Answer: (b) Ωm
2. A piece of wire of resistance 2 Ω is cut into four equal parts. These are connected in parallel. If R is the equivalent resistance, then
(a) 16 (b) 1/16 (c) 1/8 (d) 8
✅ Answer: (b) 1/16 Ω
3. Two wires have lengths, resistances and resistivities in the ratio 1 : 2. What is the ratio of diameters?
(a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 1 : √2
✅ Answer: (d) 1 : √2
4. Which of the following does NOT represent electrical power?
(i) IR² (ii) I²R (iii) V²/R (iv) VI
Options:
(a) IR² only (b) I²R (c) V²/R (d) VI
✅ Answer: (a) IR² (This is incorrect dimensionally.)
5. The SI unit of electric charge is
(a) coulomb (b) volt (c) watt (d) joule
✅ Answer: (a) coulomb
6. Which of the following is used to measure potential difference of a circuit?
(a) Ammeter (b) Voltmeter (c) Galvanometer (d) Multimeter
✅ Answer: (b) Voltmeter
7. The device used for producing electric current is called
(a) motor (b) ammeter (c) galvanometer (d) generator
✅ Answer: (d) generator
8. What will be the work done in moving 2 C charge through a potential difference of 12 V?
(a) 12 J (b) 24 J (c) 6 J (d) 4 J
Work = QV = 2 × 12 = 24 J
✅ Answer: (b) 24 J
9. If three resistances in parallel are in relation R₁ > R₂ > R₃, the relation between currents is
(a) I₁ > I₂ > I₃
(b) I₃ > I₂ > I₁
(c) I₂ > I₃ > I₁
(d) I₁ = I₂ = I₃
Current ∝ 1/R
Smallest resistance → highest current
R₃ lowest → I₃ highest
R₁ highest → I₁ lowest
✅ Answer: (b) I₃ > I₂ > I₁
10. How many electrons contribute in 1 second in causing 1 microampere current?
Current = 1 μA = 10⁻⁶ A
Charge = 10⁻⁶ C/sec
n = Q / e = 10⁻⁶ / (1.6 ×10⁻¹⁹) ≈ 6.25×10¹²
✅ Answer: (c) 6.25 × 10¹²
11. A 220 V, 100 W bulb is used in 110 V. The power consumed is
P ∝ V²
P = 100 × (110/220)² = 25 W
✅ Answer: (b) 25 W
12. The practical unit of electrical energy 1 kWh means
Options:
(i) 36 × 10⁶ J
(ii) 3.6 × 10⁶ J
(iii) 3.6 × 10³ J
(iv) 36 × 10⁴ J
1 kWh = 3.6 × 10⁶ J
✅ Answer: (ii)
13. In an experiment, the variation of current with applied voltage is shown. Which conclusions are true?
(i) Current increases linearly with voltage.
(ii) Resistance is different for different values of V.
(iii) Slope of graph same for every V and I.
Straight line graph → constant resistance
(i) True
(ii) False
(iii) True
Correct options: (i) and (iii)
✅ Answer: (b) (i) and (iii)
14. 1 kWh = how many joules?
(a) 3.6×10⁶ (b) 3.6×10⁵ (c) 3.6×10⁷ (d) 3.6×10⁸
Correct: 3.6×10⁶
✅ Answer: (a)
15. 1 eV = ? J
Correct value = 1.6 × 10⁻¹⁹ J
Options:
(a) 1.602×10⁻¹⁸ (b) 1.602×10⁻¹⁹
(c) 1.601×10⁻¹⁸ (d) 1.601×10⁻¹⁶
✅ Answer: (b) 1.602 × 10⁻¹⁹ J
16. A current of 0.5 A flows for 10 minutes. Total charge?
Q = It = 0.5 × 600 = 300 C
Options:
(a) 330 C (b) 310 C (c) 320 C (d) 300 C
✅ Answer: (d) 300 C
17. The device used for producing electric current is called
(a) motor (b) ammeter (c) galvanometer (d) generator
✅ Answer: (d) generator
18. An electric bulb is connected to a 200 V generator. The current is 0.4 A. What is the power of the bulb?
(a) 100 W (b) 90 W (c) 80 W (d) 110 W
Power = VI = 200 × 0.4 = 80 W
✅ Answer: (c) 80 W
19. A student studied Ohm’s law (V = IR). Table is given:
| V (volt) | 0.4 | 0.8 | 1.2 | Y |
|---|---|---|---|---|
| I (amp) | 0.1 | 0.2 | X | 0.5 |
Find X and Y.
Using V = IR
-
For X:
1.2 = R × X
R = 0.4 / 0.1 = 4 Ω
So X = 1.2 / 4 = 0.3 A -
For Y:
V = IR = 0.5 × 4 = 2 V
✅ Answer: X = 0.3 A, Y = 2 V
20. Define the unit of current.
Answer:
The SI unit of current is ampere (A).
One ampere is defined as the flow of 1 coulomb of charge per second.
21. State the S.I. unit of electric charge.
Answer:
The SI unit of electric charge is coulomb (C).
B. Short Type Questions (2/3 Marks)
1. Four resistors R₁ = 7 Ω, R₂ = 10 Ω, R₃ = 30 Ω, R₄ = 5 Ω and a 12 V battery are connected. Calculate:
(i) Total resistance
From the diagram:
- R₂, R₃, R₄ are in parallel.
- Their combination is in series with R₁.
Parallel part:
\frac{1}{R_p}=\frac{1}{10}+\frac{1}{30}+\frac{1}{5}
\frac{1}{R_p} = 0.1 + 0.0333 + 0.2 = 0.3333
R_p = 3\ \Omega
Total resistance:
R_{total} = R_1 + R_p = 7 + 3 = 10\ \Omega
(ii) Current in the circuit
I = \frac{V}{R} = \frac{12}{10} = 1.2\ \text{A}
✅ Answer:
- Total resistance = 10 Ω
- Current = 1.2 A
2. Show how to connect three resistors of 6 Ω to get:
(i) 9 Ω
To get 9 Ω:
- Connect two 6 Ω resistors in parallel →
R_p = \frac{6 \times 6}{6 + 6} = 3\ \Omega
Total R = 3 + 6 = 9 Ω
(ii) 2 Ω
To get 2 Ω:
- Connect all three 6 Ω resistors in parallel
R = \frac{1}{\frac{1}{6}+\frac{1}{6}+\frac{1}{6}} = 2\ \Omega
3. Draw a diagram with 4 cells (3 V each) and four resistors 5 Ω, 10 Ω, 15 Ω, 20 Ω in series. Ammeter is in series; voltmeter across 20 Ω. Calculate readings.
Total voltage:
4 cells × 3 V = 12 V
Total resistance:
R_{total} = 5+10+15+20 = 50\ \Omega
Current:
I = \frac{12}{50} = 0.24\ \text{A}
Voltmeter reading across 20 Ω:
V = IR = 0.24 \times 20 = 4.8\ \text{V}
✅ Answer:
- Ammeter reading = 0.24 A
- Voltmeter reading = 4.8 V
4. How is the life of a filament bulb prolonged?
Answer:
The life of a filament bulb is prolonged by:
- Using inert gases like argon inside the bulb
- Keeping the filament thicker
- Avoiding sudden switching on/off
- Using low voltage operation
- Preventing overheating
5. Electric heaters are rated 11 W for 220 V. Max supply current is 5 A.
(i) How many heaters can be connected in parallel?
Power of each heater: 11 W
Voltage: 220 V
Current per heater:
I = \frac{P}{V} = \frac{11}{220} = 0.05\ \text{A}
Maximum current supply = 5 A
Number of heaters:
n = \frac{5}{0.05} = 100
(ii) Total energy consumed in 20 minutes
Total power = 100 × 11 W = 1100 W = 1.1 kW
Time = 20 min = 20/60 = 1/3 hour
Energy:
E = Pt = 1.1 \times \frac{1}{3} = 0.366\ \text{kWh}
✅ Answer:
- Number of heaters = 100
- Energy consumed = 0.366 kWh
6. A 12 V battery gives 2.5 mA current. Find resistance.
I = 2.5\text{ mA} = 0.0025\ \text{A}
R = \frac{V}{I} = \frac{12}{0.0025} = 4800\ \Omega
✅ Answer: 4800 Ω
7. What is the commercial unit of electrical energy? Which uses more energy—250 W TV for 1 hour or 1200 W iron for 10 minutes?
Commercial unit:
kWh (kilowatt hour)
Energy used by TV:
E = 0.25 \times 1 = 0.25\ \text{kWh}
Energy used by Iron:
Time = 10 min = 1/6 hr
E = 1.2 \times \frac{1}{6} = 0.2\ \text{kWh}
TV uses 0.25 kWh
Iron uses 0.2 kWh
More energy used:
250 W TV for 1 hour uses more energy.
8. A current of 0.5 A flows for 10 minutes. Find charge.
t = 10\ \text{min} = 600\ \text{s}
Q = It = 0.5 \times 600 = 300\ \text{C}
✅ Answer: 300 C
9. How much energy is given to each coulomb of charge passing through a 6 V battery?
Energy per coulomb = Voltage = 6 J
✅ Answer: 6 joule
10. A wire 1 m long has resistance 26.0 Ω at 20°C. Diameter = 0.03 cm. Find resistivity.
Diameter = 0.03 cm = 0.0003 m
Radius = 0.00015 m
Cross-sectional area:
A = \pi r^2 = \pi (0.00015)^2 = 7.07 \times 10^{-8}\ \text{m}^2
Resistivity:
\rho = \frac{RA}{L} = 26 \times (7.07 \times 10^{-8}) = 1.84 \times 10^{-6}\ \Omega m
✅ Answer:
11. 100 J of heat is produced each second in a 4 Ω resistor. Find potential difference.
Heat per second = Power = 100 W
P = \frac{V^2}{R}
100 = \frac{V^2}{4}
V^2 = 400 \Rightarrow V = 20\ \text{V}
✅ Answer: 20 V
12. A current of 0.6 A flows for 5 min. Find charge.
t = 5\text{ min} = 300\text{ s}
Q = It = 0.6 \times 300 = 180\ \text{C}
✅ Answer: 180 C
13. How electrical charges flow inside a wire? Explain.
Answer:
Electrical charges flow due to the motion of free electrons present in the metal.
When a potential difference is applied:
- the electrons drift from negative terminal to positive terminal,
- creating an electric current.
14. Write SI unit of resistivity. A motor takes 5 A from a 220 V source. Find its power and energy consumed in 2 hours.
SI unit:
Ω·m (ohm metre)
Power:
P = VI = 220 \times 5 = 1100\ \text{W} = 1.1\ \text{kW}
Energy consumed:
E = Pt = 1.1 \times 2 = 2.2\ \text{kWh}
✅ Answer:
- Power = 1.1 kW
- Energy = 2.2 kWh
15. An 8 Ω heater draws 15 A for 2 hours. Calculate rate of heat developed.
Rate of heat = Power
P = I^2 R = 15^2 \times 8 = 225 \times 8 = 1800\ \text{W}
Or
P = VI = (15 \times 8 \times 15) = 1800 W
Energy in 2 hours:
E = P \times t = 1800 \times 2 = 3600\ \text{Wh} = 3.6\ \text{kWh}
But the question only asks rate of heat:
✅ Answer: 1800 W (1.8 kW)
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