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Electricity Class 10 Numericals — Find Your Weakness

September 23, 2026 Pandey Sir

CBSE Class 10 Science • 2026–27

Electricity Class 10 Numericals — Step-by-Step Practice & Weakness Detector

Solve 15 carefully checked numericals, discover which Electricity skill needs more practice, then use the targeted corrective set to fix it.

Start the diagnostic

Current-syllabus scope: The official CBSE Class X Science (086) 2026–27 curriculum includes electric current, potential difference, Ohm’s law, resistance and resistivity, factors affecting resistance, series and parallel resistor combinations, heating effect of current and electric power. The practical list also includes plotting a V–I graph and determining equivalent resistance in series and parallel.

This is not another Electricity notes page. Use this page to practise calculation and diagnosis. If you first need concepts and formulas explained in detail, use the existing Electricity revision notes and formulas.

Formula Compass: Choose the Formula Before Calculating

The most common numerical mistake is not arithmetic—it is choosing the wrong relationship. Start with what the question gives you.

Charge + time?
I = Q ÷ t

V, I, R?
V = IR

Resistor combination?
Series: add R
Parallel: add reciprocals

Power / heat / energy?
P = VI
H = I²Rt
E = Pt

I = Q ÷ tcurrent = charge ÷ time

V = W ÷ Qpotential difference

V = IROhm’s law

R_s = R₁ + R₂ + …series

1/R_p = 1/R₁ + 1/R₂ + …parallel

R = ρl ÷ Aresistance and resistivity

P = VIalso P = I²R = V² ÷ R

H = I²RtJoule heating

E = Pt1 kWh = 3.6 × 10⁶ J

Quantity	Symbol	SI unit	Useful conversion
Current	I	ampere (A)	1 mA = 0.001 A
Potential difference	V	volt (V)	—
Resistance	R	ohm (Ω)	1 kΩ = 1000 Ω
Resistivity	ρ	ohm metre (Ω m)	1 mm² = 10⁻⁶ m²
Power	P	watt (W)	1 kW = 1000 W
Energy	E	joule (J)	1 kWh = 3.6 × 10⁶ J

Unit trap: Convert minutes to seconds before using I = Q ÷ t, and convert mm² to m² before using R = ρl ÷ A.

Series vs Parallel — See the Difference

Series

R_s = R₁ + R₂. Equivalent resistance is greater than either resistor.

Parallel

1/R_p = 1/R₁ + 1/R₂. Equivalent resistance is smaller than the smallest branch resistance.

V–I Graph: What the Slope Means

For a V-versus-I graph

Slope = ΔV ÷ ΔI = resistance.

At the marked point: V = 6 V

I = 1.5 A

R = V ÷ I

R = 6 ÷ 1.5 = 4 Ω

A steeper straight V–I line therefore represents a larger resistance.

Graph trap: If the axes are reversed and I is plotted vertically against V horizontally, the slope is I/V = 1/R. Always read the axis labels before interpreting slope.

Why More Resistance Means Less Current — interactive 3D

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Targeted Corrective Practice

Use the section named in your diagnostic report. These are original board-pattern practice questions, not claimed PYQs.

A. Ohm’s Law and Current

Corrective 1 — 18 V across 6 Ω: find current.

Formula: I = V ÷ R

I = 18 ÷ 6

I = 3 A

Corrective 2 — 900 C in 5 minutes: find current.

5 min = 300 s

I = Q ÷ t = 900 ÷ 300

I = 3 A

B. Series and Parallel

Corrective 3 — 4 Ω and 6 Ω in series: find equivalent resistance.

R_s = 4 + 6

R_s = 10 Ω

Corrective 4 — 12 Ω and 6 Ω in parallel: find equivalent resistance.

1/R_p = 1/12 + 1/6 = 3/12 = 1/4

R_p = 4 Ω

C. Resistance and Resistivity

Corrective 5 — If length becomes 1.2 times and area becomes one-fifth, how does resistance change?

R ∝ l/A

New factor = 1.2 ÷ (1/5) = 1.2 × 5

New resistance = 6R

Corrective 6 — A wire has ρl = 2.4 × 10⁻⁶ Ω m² and area 0.8 × 10⁻⁶ m². Find R.

R = ρl ÷ A

R = (2.4 × 10⁻⁶) ÷ (0.8 × 10⁻⁶)

R = 3 Ω

D. Power, Heat and Energy

Corrective 7 — A 6 Ω resistor is connected across 12 V. Find power.

P = V² ÷ R

P = 12² ÷ 6 = 144 ÷ 6

P = 24 W

Corrective 8 — A 0.75 kW appliance runs for 2 h. Find electrical energy used.

E = P × t

E = 0.75 × 2

E = 1.5 kWh

E. Graphs and Application

Corrective 9 — A V–I point is 9 V at 0.75 A. Find resistance.

R = V ÷ I

R = 9 ÷ 0.75

R = 12 Ω

Corrective 10 — A 1500 W appliance operates at 250 V. Find current.

P = VI, so I = P ÷ V

I = 1500 ÷ 250

I = 6 A

Board-Ready Numerical Writing Pattern

Example: A potential difference of 12 V is applied across a 6 Ω resistor. Find current.

Given: V = 12 V, R = 6 Ω

Formula: V = IR, therefore I = V ÷ R

Substitution: I = 12 ÷ 6

Answer: I = 2 A

Show the formula, substitution, working and final unit. Do not compress the whole solution into one unreadable line.

Common Numerical Mistakes

Series/parallel swap: Adding resistances directly in a parallel circuit.

Area conversion: Treating 1 mm² as 10⁻³ m². It is 10⁻⁶ m².

Minutes vs seconds: Using minutes directly in I = Q/t or H = I²Rt when SI units are required.

Graph slope: Memorising “slope = R” without checking which quantity is on which axis.

kW vs W: Calculating energy in kWh while leaving power in watts.

Missing unit: A correct numerical value without A, V, Ω, W, J or kWh is incomplete.

Continue the Electricity Cluster

Need the theory first?

Review current, potential difference, resistance, power and the main formulas before attempting the diagnostic again.

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After you improve

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Exam note: This page does not claim a chapter-wise marks guarantee or “sure-shot” questions. It is designed to practise the currently prescribed Electricity concepts and the mathematical/graphical skills emphasised in the official curriculum.