Chapter 4 — Electricity: Multiple Choice Questions
Instructions: Each question carries 1 mark. Choose the best answer.
1. The SI unit of electric charge is ______.
Answer
(a) Ampere (b) Volt (c) Coulomb (d) Ohm
✅ Answer: (c) Coulomb
ExplanationThe SI unit of electric charge is the Coulomb (C). One Coulomb is the charge transported by a current of 1 Ampere in 1 second (Q = It). The charge of one electron is 1.6 × 10⁻¹⁹ C.
2. Ohm's Law can be stated as ______.
Answer
(a) V = I/R (b) V = IR (c) I = VR (d) R = VI
✅ Answer: (b) V = IR
ExplanationOhm's Law: V = IR, where V = potential difference (Volts), I = current (Amperes), R = resistance (Ohms). Rearrangements: I = V/R and R = V/I.
3. In a series circuit, the current ______.
Answer
(a) is different in each component (b) is the same through all components (c) is zero at the last component (d) increases with each component
✅ Answer: (b) is the same through all components
ExplanationIn a series circuit, the same current flows through all components. There is only one path for current to flow. The voltage, however, divides across each component.
4. In a parallel circuit, the voltage ______.
Answer
(a) is different across each component (b) divides among components (c) is the same across all components (d) is zero
✅ Answer: (c) is the same across all components
ExplanationIn a parallel circuit, all components are connected between the same two points, so they all have the same potential difference (voltage). The current, however, divides — more current flows through lower resistance branches.
5. The total resistance in a series combination of R₁ and R₂ is ______.
Answer
(a) R₁ × R₂ (b) R₁ + R₂ (c) R₁R₂/(R₁ + R₂) (d) 1/R₁ + 1/R₂
✅ Answer: (b) R₁ + R₂
ExplanationFor series combination: R_total = R₁ + R₂ + R₃ + ... Total resistance is always greater than any individual resistance. For parallel combination: 1/R_total = 1/R₁ + 1/R₂ (total is always less than smallest individual resistance).
6. The SI unit of electric power is ______.
Answer
(a) Joule (b) Ampere (c) Watt (d) Volt
✅ Answer: (c) Watt
ExplanationThe SI unit of power is the Watt (W). 1 W = 1 J/s = 1 V·A. Power = VI = I²R = V²/R. A 100 W bulb consumes 100 J of electrical energy per second.
7. The commercial unit of electrical energy is ______.
Answer
(a) Watt (b) Joule (c) Kilowatt-hour (kWh) (d) Volt-ampere
✅ Answer: (c) Kilowatt-hour (kWh)
ExplanationThe commercial unit (used by electricity boards for billing) is the kilowatt-hour (kWh), also called a "unit" of electricity. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. Electricity bills charge per kWh consumed.
8. A fuse wire is made of material with ______.
Answer
(a) high melting point and high resistance (b) low melting point and high resistance (c) low melting point and low resistance (d) high melting point and low resistance
✅ Answer: (b) low melting point and high resistance
ExplanationA fuse wire melts and breaks the circuit when excessive current flows, protecting appliances. For this, it must have:
- Low melting point: Melts quickly when overloaded
- High resistance: Generates enough heat even with small excess current
Fuse wires are typically made of tin-lead alloy.
9. The resistance of a conductor depends on ______.
Answer
(a) Length only (b) Cross-sectional area only (c) Length, cross-sectional area, material, and temperature (d) Voltage applied
✅ Answer: (c) Length, cross-sectional area, material, and temperature
ExplanationR = ρL/A, where:
- R = resistance (Ω)
- ρ = resistivity of material (depends on material and temperature)
- L = length (longer wire → more resistance)
- A = cross-sectional area (thicker wire → less resistance)
10. Which of the following is an example of Joule's heating effect?
Answer
(a) Charging a battery (b) Electric bulb glowing (c) Fan running (d) Speaker producing sound
✅ Answer: (b) Electric bulb glowing
ExplanationJoule's heating effect (H = I²Rt) occurs when electrical energy is converted to heat. An electric bulb filament heats up and glows due to Joule heating. Other examples: electric iron, heater, toaster, fuse wire melting.
Quick Revision — Key Electricity Concepts
| Concept | Formula | Unit |
|---|
| Ohm's Law | V = IR | V = Volt, I = Ampere, R = Ohm |
| Charge | Q = It | Coulomb (C) |
| Power | P = VI = I²R = V²/R | Watt (W) |
| Energy | E = Pt = VIt | Joule (J) or kWh |
| Series resistance | R = R₁ + R₂ | Ω (larger) |
| Parallel resistance | 1/R = 1/R₁ + 1/R₂ | Ω (smaller) |
| Joule's heating | H = I²Rt | Joule (J) |
| Resistivity | R = ρL/A | Ω·m |
Common MCQ Traps
- Series vs Parallel current/voltage: In series, same current; in parallel, same voltage
- Total resistance in parallel is LESS than the smallest individual resistance
- kWh vs Joule: kWh is commercial unit; Joule is SI unit. 1 kWh = 3.6 × 10⁶ J
- Fuse: Low melting point AND high resistance (not high, high)
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