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Electricity — Class 10 CBSE Science MCQs with Solutions
Free Class 10 CBSE Science Electricity MCQs with step-by-step solutions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — easy
SI unit of electric charge (Q) is:
A. Ampere
B. Coulomb ( ✓ Correct
C. C) Volt
D. Ohm
Solution: Electric charge is measured in Coulombs (C).
Q2 — easy
SI unit of electric current (I = Q/t) is:
A. Volt
B. Ampere (A) ✓ Correct
C. Ohm
D. Watt
Solution: Current = Charge / Time = Coulombs/sec = Amperes (A).
Q3 — easy
SI unit of electric potential difference (V = W/Q) is:
A. Ampere
B. Volt (V) ✓ Correct
C. Ohm
D. Joule
Solution: Potential Difference = Work / Charge = Volts (V).
Q4 — easy
Ohm's Law states that electric current I flowing through a conductor is directly proportional to potential difference V across its ends, expressed as:
A. V = I × R ✓ Correct
B. V = I / R
C. I = V × R
D. R = V × I
Solution: Ohm's Law: V = IR.
Q5 — easy
SI unit of electrical resistance (R) is:
A. Volt
B. Ohm (Ω) ✓ Correct
C. Ampere
D. Watt
Solution: Resistance is measured in Ohms (Ω).
Q6 — easy
Resistance R of a uniform cylindrical conductor is directly proportional to its length L and inversely proportional to its:
A. Temperature
B. Area of cross-section (A) ✓ Correct
C. Voltage
D. Current
Solution: R = ρ(L / A).
Q7 — easy
SI unit of electrical resistivity (ρ) is:
A. Ohm (Ω)
B. Ohm-metre (Ω·m) ✓ Correct
C. Volt/metre
D. Ampere/metre
Solution: Resistivity ρ = R·A/L = Ω·m.
Q8 — easy
When two resistors R₁ and R₂ are connected in SERIES, equivalent resistance Rₛ =
A. R₁ + R₂ ✓ Correct
B. (R₁ × R₂)/(R₁ + R₂)
C. 1/R₁ + 1/R₂
D. R₁ − R₂
Solution: Series equivalent Rₛ = R₁ + R₂.
Q9 — easy
When two resistors R₁ and R₂ are connected in PARALLEL, equivalent resistance Rₚ =
A. R₁ + R₂
B. (R₁ × R₂)/(R₁ + R₂) ✓ Correct
C. R₁ − R₂
D. R₁ × R₂
Solution: Parallel: 1/Rₚ = 1/R₁ + 1/R₂ → Rₚ = R₁R₂ / (R₁ + R₂).
Q10 — easy
According to Joule's Law of Heating, heat produced H in a conductor of resistance R carrying current I for time t is:
A. H = I R t
B. H = I² R t ✓ Correct
C. H = I R² t
D. H = V I² t
Solution: Joule's Heating Law: H = I²Rt.
Q11 — easy
SI unit of electric power (P = VI = I²R = V²/R) is:
A. Joule
B. Watt (W) ✓ Correct
C. Kilowatt-hour
D. Ampere
Solution: Electric power is measured in Watts (W).
Q12 — easy
An ammeter is always connected in _______ in a circuit, while a voltmeter is always connected in _______.
A. Parallel, Series
B. Series, Parallel ✓ Correct
C. Series, Series
D. Parallel, Parallel
Solution: Low-resistance ammeter in series; high-resistance voltmeter in parallel.
Q13 — easy
An electric bulb is rated 220 V, 100 W. Resistance of bulb filament is:
A. 220 Ω
B. 484 Ω ✓ Correct
C. 100 Ω
D. 440 Ω
Solution: R = V² / P = (220)² / 100 = 48400 / 100 = 484 Ω.
Q14 — medium
1 Kilowatt-hour (kWh) of electrical energy is equal to:
A. 3.6 × 10⁵ J
B. 3.6 × 10⁶ J ✓ Correct
C. 1000 J
D. 3.6 × 10⁴ J
Solution: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ Joules.
Q15 — medium
Commercial unit of electric energy is kWh. An electric iron of 1000 W used for 2 hours daily consumes how many units in 30 days?
A. 30 units
B. 60 units ✓ Correct
C. 600 units
D. 120 units
Solution: Energy = 1 kW × 2 h/day × 30 days = 60 kWh (60 units).