Class 10 Science Electricity Numericals: 40 Solved Problems for CBSE 2027

Electricity numericals in Class 10 Science cluster into a handful of recurring formulas — once you've seen the full landscape mapped out, "which formula do I use here" stops being the hard part. This is 40 original problems, 5 for every distinct type, each with a complete solution you can follow line by line.

and the Electricity topics covered on this page remain part of the current CBSE Class 10 Science curriculum, so every topic here remains examinable. How to use this page: Attempt each problem fully on paper before tapping "Reveal Solution" — check your working, not just your final number. Every single calculation on this page was computed and independently verified before publishing, and , so every topic here remains fully examinable.

Quick formula reference used throughout:

What it finds	Formula
Ohm's Law	V = IR
Resistance from resistivity	R = ρL/A
Series resistors	R = R₁ + R₂ + R₃ + ...
Parallel resistors	1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...
Power	P = VI = I²R = V²/R
Electrical energy	E (kWh) = Power(kW) × Time(h)
Heat produced	H = I²Rt

Every Numerical Type Covered

Ohm's Law — direct application

Resistance from resistivity

Resistors in series

Resistors in parallel

Combination circuits

Electric power

Electrical energy & cost

Heating effect of current

Type 1 · Ohm's Law

1.1

Find V

A current of 2 A flows through a resistor of 5 Ω. Find the voltage across it.

Reveal Solution

V = IR = 2 × 5

✓ V = 10 V

1.2

Find I

A resistor of 4 Ω is connected to a 12 V battery. Find the current flowing through it.

Reveal Solution

I = V/R = 12/4

✓ I = 3 A

1.3

Find R

A 24 V source drives a current of 3 A through a resistor. Find the resistance.

Reveal Solution

R = V/I = 24/3

✓ R = 8 Ω

1.4

Find V

A current of 0.5 A flows through a 20 Ω resistor. Find the voltage across it.

Reveal Solution

V = IR = 0.5 × 20

✓ V = 10 V

1.5

Find R

A household appliance draws 2 A of current from a 220 V mains supply. Find its resistance.

Reveal Solution

R = V/I = 220/2

✓ R = 110 Ω

Type 2 · Resistance from Resistivity

2.1

R = ρL/A

A copper wire of length 5 m has a cross-sectional area of 1 mm². Given the resistivity of copper is 1.7×10⁻⁸ Ω·m, find its resistance.

Reveal Solution

Convert area to m²

1 mm² = 1×10⁻⁶ m²

R = ρL/A = (1.7×10⁻⁸ × 5) / (1×10⁻⁶)

✓ R = 0.085 Ω

2.2

R = ρL/A

A nichrome wire of length 2 m has a cross-sectional area of 0.5 mm². Given resistivity 1.1×10⁻⁶ Ω·m, find its resistance.

Reveal Solution

Convert area

0.5 mm² = 0.5×10⁻⁶ m²

R = (1.1×10⁻⁶ × 2) / (0.5×10⁻⁶)

✓ R = 4.4 Ω

2.3

R = ρL/A

An iron wire of length 10 m has a cross-sectional area of 2 mm². Given resistivity 1.0×10⁻⁷ Ω·m, find its resistance.

Reveal Solution

Convert area

2 mm² = 2×10⁻⁶ m²

R = (1.0×10⁻⁷ × 10) / (2×10⁻⁶)

✓ R = 0.5 Ω

2.4

Effect of Stretching

A wire of resistance 10 Ω is stretched uniformly until its length is doubled (volume stays constant). Find its new resistance.

Reveal Solution

Volume is constant, so if length doubles, area halves

New R = ρ(2L)/(A/2) = 4 × ρL/A = 4 × original R

New R = 4 × 10

✓ New resistance = 40 Ω

⚠️ Common mistake: assuming resistance simply doubles when length doubles. Because area also changes (to keep volume constant), the actual increase is by a factor of 4, not 2.

2.5

Ratio Problem

Two wires are made of the same material and have the same length, but their radii are in the ratio 1:2. Find the ratio of their resistances.

Reveal Solution

Resistance is inversely proportional to area, and area ∝ radius²

R ∝ 1/r²

R₁:R₂ = (1/r₁²) : (1/r₂²) = r₂² : r₁² = 4 : 1

✓ Resistance ratio = 4:1 (the thinner wire has 4 times the resistance)

Type 3 · Resistors in Series

3.1

Series Circuit

Resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series to a 20 V battery. Find the total resistance and the current flowing.

Reveal Solution

R_total = 2+3+5 = 10 Ω. I = V/R = 20/10

✓ R_total = 10 Ω, I = 2 A (same through every resistor)

3.2

Series Circuit

Resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series to a 40 V battery. Find the current and the voltage drop across the 10 Ω resistor.

Reveal Solution

R_total = 4+6+10 = 20 Ω. I = 40/20 = 2 A

Voltage across 10Ω resistor = I×R = 2×10

✓ I = 2 A, voltage across 10Ω resistor = 20 V

3.3

Series Circuit

Four resistors of 1 Ω, 2 Ω, 3 Ω, and 4 Ω are connected in series to a 20 V battery. Find the current.

Reveal Solution

R_total = 1+2+3+4 = 10 Ω. I = 20/10

✓ I = 2 A

3.4

Series Circuit

Two resistors, 5 Ω and 15 Ω, are connected in series to a 40 V battery. Find the current and the voltage drop across each.

Reveal Solution

R_total = 5+15 = 20 Ω. I = 40/20 = 2 A

Voltage drops: 2×5=10V across the 5Ω, and 2×15=30V across the 15Ω

                    ✓ I = 2 A. Voltage drops: 10 V and 30 V (sum = 40 V, matching the battery ✓)

3.5

Series Circuit

Four identical 3 Ω resistors are connected in series to a 24 V battery. Find the current.

Reveal Solution

R_total = 3×4 = 12 Ω. I = 24/12

✓ I = 2 A

Type 4 · Resistors in Parallel

4.1

Parallel Circuit

Resistors of 2 Ω and 3 Ω are connected in parallel to a 12 V battery. Find the equivalent resistance and total current.

Reveal Solution

1/R = 1/2 + 1/3 = 5/6 → R = 6/5 = 1.2 Ω

Total current = V/R = 12/1.2

✓ R_equivalent = 1.2 Ω, Total current = 10 A

4.2

Parallel Circuit

Three identical 4 Ω resistors are connected in parallel to an 8 V battery. Find the equivalent resistance and the current through each resistor.

Reveal Solution

1/R = 1/4+1/4+1/4 = 3/4 → R = 4/3 Ω

Current through each (same V across each) = 8/4 = 2 A

✓ R_equivalent = 4/3 Ω ≈ 1.33 Ω, current through each resistor = 2 A

4.3

Parallel Circuit

Resistors of 6 Ω and 3 Ω are connected in parallel to a 6 V battery. Find the equivalent resistance and the current through each.

Reveal Solution

1/R = 1/6+1/3 = 1/2 → R = 2 Ω

Current through 6Ω = 6/6 = 1A. Current through 3Ω = 6/3 = 2A

✓ R_equivalent = 2 Ω. Currents: 1 A and 2 A (total = 3 A)

4.4

Parallel Circuit

Resistors of 10 Ω, 15 Ω, and 30 Ω are connected in parallel to a 30 V battery. Find the equivalent resistance.

Reveal Solution

1/R = 1/10+1/15+1/30 = 3/30+2/30+1/30 = 6/30 = 1/5

✓ R_equivalent = 5 Ω

4.5

Parallel Circuit

Resistors of 5 Ω and 20 Ω are connected in parallel to a 20 V battery. Find the current through each resistor.

Reveal Solution

Same voltage (20V) across each resistor in parallel

Current through 5Ω = 20/5 = 4A. Current through 20Ω = 20/20 = 1A

✓ Currents: 4 A and 1 A (total current from battery = 5 A)

Type 5 · Combination Circuits

5.1

Series + Parallel

A 4 Ω resistor is connected in series with a parallel combination of 6 Ω and 3 Ω resistors, all connected to an 18 V battery. Find the total resistance, total current, and voltage across the parallel section.

Reveal Solution

Step 1 — Resolve the parallel section

1/R = 1/6+1/3 = 1/2 → R_parallel = 2 Ω

R_total = 4 (series) + 2 (parallel) = 6 Ω

Total current

I = 18/6 = 3 A

Voltage across the parallel section

V = I × R_parallel = 3 × 2

✓ R_total = 6 Ω, I = 3 A, Voltage across parallel section = 6 V

5.2

Series + Parallel

A 2 Ω resistor is in series with a parallel combination of two 4 Ω resistors, connected to a 10 V battery. Find the total resistance and total current.

Reveal Solution

Parallel section

1/R = 1/4+1/4 = 1/2 → R_parallel = 2 Ω

R_total = 2+2 = 4 Ω. I = 10/4

✓ R_total = 4 Ω, I = 2.5 A

5.3

Series + Parallel

A 5 Ω resistor is in series with a parallel combination of two 10 Ω resistors, connected to a 15 V battery. Find the total resistance and total current.

Reveal Solution

Parallel section

1/R = 1/10+1/10 = 1/5 → R_parallel = 5 Ω

R_total = 5+5 = 10 Ω. I = 15/10

✓ R_total = 10 Ω, I = 1.5 A

5.4

Series + Parallel

A 3 Ω resistor is in series with a parallel combination of two 8 Ω resistors, connected to a 14 V battery. Find the total resistance and total current.

Reveal Solution

Parallel section

1/R = 1/8+1/8 = 1/4 → R_parallel = 4 Ω

R_total = 3+4 = 7 Ω. I = 14/7

✓ R_total = 7 Ω, I = 2 A

5.5

Series + Parallel

A 1 Ω resistor is in series with a parallel combination of 6 Ω and 3 Ω resistors, connected to a 9 V battery. Find the total resistance and total current.

Reveal Solution

Parallel section

1/R = 1/6+1/3 = 1/2 → R_parallel = 2 Ω

R_total = 1+2 = 3 Ω. I = 9/3

✓ R_total = 3 Ω, I = 3 A

⚠️ Common mistake: adding all resistors as if they were all in series, or all in parallel, without first identifying which specific resistors are grouped together. Always resolve the smaller sub-combination first, exactly as you would with brackets in an arithmetic expression.

Type 6 · Electric Power

6.1

P = VI

An appliance operates at 220 V and draws a current of 5 A. Find the power consumed.

Reveal Solution

P = VI = 220 × 5

✓ P = 1100 W

6.2

P = I²R

A current of 2 A flows through a 10 Ω resistor. Find the power dissipated.

Reveal Solution

P = I²R = 2² × 10

✓ P = 40 W

6.3

P = V²/R

A resistor of 55 Ω is connected to a 110 V supply. Find the power consumed.

Reveal Solution

P = V²/R = 110²/55 = 12100/55

✓ P = 220 W

6.4

P = VI

A torch bulb operates at 12 V and draws 0.5 A of current. Find its power rating.

Reveal Solution

P = VI = 12 × 0.5

✓ P = 6 W

6.5

P = I²R

A current of 3 A flows through a 20 Ω heating coil. Find the power dissipated.

Reveal Solution

P = I²R = 3² × 20

✓ P = 180 W

Type 7 · Electrical Energy & Cost

7.1

Energy Cost

A 1000 W heater is used for 5 hours a day for 30 days. Find the electrical energy consumed in kWh and the cost at ₹6 per unit.

Reveal Solution

Energy = (1000/1000) × 5 × 30 = 150 kWh

Cost = 150 × 6

✓ Energy = 150 kWh (units), Cost = ₹900

7.2

Energy Cost

A 60 W bulb is used for 6 hours a day for 30 days. Find the electrical energy consumed and the cost at ₹8 per unit.

Reveal Solution

Energy = (60/1000) × 6 × 30 = 10.8 kWh

Cost = 10.8 × 8

✓ Energy = 10.8 kWh, Cost = ₹86.40

7.3

Energy Cost

A 1500 W geyser is used for 4 hours a day for 20 days. Find the electrical energy consumed and the cost at ₹5 per unit.

Reveal Solution

Energy = (1500/1000) × 4 × 20 = 120 kWh

Cost = 120 × 5

✓ Energy = 120 kWh, Cost = ₹600

7.4

Energy Cost

A 100 W device is used for 10 hours a day for 30 days. Find the electrical energy consumed and the cost at ₹7 per unit.

Reveal Solution

Energy = (100/1000) × 10 × 30 = 30 kWh

Cost = 30 × 7

✓ Energy = 30 kWh, Cost = ₹210

7.5

Energy Cost

A 2000 W appliance is used for 3 hours a day for 25 days. Find the electrical energy consumed and the cost at ₹6 per unit.

Reveal Solution

Energy = (2000/1000) × 3 × 25 = 150 kWh

Cost = 150 × 6

✓ Energy = 150 kWh, Cost = ₹900

⚠️ Common mistake: forgetting to convert watts to kilowatts before multiplying by hours. The commercial "unit" of electricity is the kilowatt-hour, not the watt-hour — always divide power by 1000 first.

Type 8 · Heating Effect of Current

8.1

H = I²Rt

A current of 2 A flows through a 5 Ω resistor for 60 seconds. Find the heat produced.

Reveal Solution

H = I²Rt = 2² × 5 × 60

✓ H = 1200 J

8.2

H = I²Rt

A current of 5 A flows through a 4 Ω heating element for 2 minutes. Find the heat produced.

Reveal Solution

Convert time to seconds

2 minutes = 120 s

H = 5² × 4 × 120

✓ H = 12000 J

8.3

H = I²Rt

A current of 1 A flows through a 10 Ω resistor for 5 minutes. Find the heat produced.

Reveal Solution

Convert time

5 minutes = 300 s

H = 1² × 10 × 300

✓ H = 3000 J

8.4

H = I²Rt

A current of 3 A flows through an 8 Ω resistor for 100 seconds. Find the heat produced.

Reveal Solution

H = 3² × 8 × 100

✓ H = 7200 J

8.5

H = I²Rt

A current of 4 A flows through a 2 Ω resistor for 150 seconds. Find the heat produced.

Reveal Solution

H = 4² × 2 × 150

✓ H = 4800 J

⚠️ Common mistake: forgetting to square the current — H depends on I², not I, so doubling the current quadruples the heat produced, not just doubles it.

Which of These 8 Types Would You Actually Recognise Cold?

Reading 40 solved problems and being able to solve fresh ones without the formula named for you are different skills. The real exam test is knowing which of these 8 patterns a new question belongs to, and which formula variant fits the data given.

What a Genelis weak area map looks like after working through Electricity practice sets

Ohm's Law & resistivity

86%

Series & parallel circuits

69%

Power & energy cost

55%

Combination circuits

32%

Next session: combination circuits (32%) — not more basic Ohm's Law practice. Genelis tracks accuracy by type, not just by chapter, so it knows exactly which pattern needs more reps.

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💡 For chapter strategy and the complete Class 10 Science formula reference, see the complete Class 10 Science guide.

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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What numerical types appear in CBSE Class 10 Electricity?

Eight distinct types recur across CBSE Class 10 Electricity: direct application of Ohm's Law, finding resistance from resistivity and wire dimensions, resistors combined in series, resistors combined in parallel, combination circuits mixing both, electric power using its three equivalent formulas, electrical energy and cost calculations for household appliances, and the heating effect of current.

How does resistance change when a wire is stretched to a different length?

If a wire is stretched so that its length changes while its total volume stays constant, its cross-sectional area must change inversely to keep volume fixed. Since resistance is directly proportional to length and inversely proportional to area, stretching a wire to double its length (which halves its area, since volume is constant) increases its resistance by a factor of four, not just two — both the increased length and the decreased area work in the same direction to raise resistance.

Is current the same or different across resistors connected in series versus parallel?

In a series circuit, the same current flows through every resistor, since there is only one path for charge to flow, while the voltage divides across each resistor according to its resistance. In a parallel circuit, the situation is reversed — the same voltage appears across every resistor, since each is connected directly across the same two points, while the current divides between the branches according to each resistor's value.

What is the difference between the three formulas for electric power, and when should each be used?

P=VI is the most general formula, usable whenever both voltage and current are known. P=I²R is most useful when current and resistance are known but voltage isn't directly given, such as in a series circuit. P=V²/R is most useful when voltage and resistance are known but current isn't directly given, such as when calculating the power rating of a household appliance from its voltage and resistance. All three are mathematically equivalent and follow directly from combining P=VI with Ohm's Law.

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Class 10 Science Electricity Numericals: 40 Solved Problems for CBSE 2027

Every Numerical Type Covered

Type 1 · Ohm's Law

Find V

Find I

Find R

Find V

Find R

Type 2 · Resistance from Resistivity

R = ρL/A

Convert area to m²

R = ρL/A

Convert area

R = ρL/A

Convert area

Effect of Stretching

Volume is constant, so if length doubles, area halves

Ratio Problem

Resistance is inversely proportional to area, and area ∝ radius²

Type 3 · Resistors in Series

Series Circuit

Series Circuit

Series Circuit

Series Circuit

Series Circuit

Type 4 · Resistors in Parallel

Parallel Circuit

Parallel Circuit

Parallel Circuit

Parallel Circuit

Parallel Circuit

Same voltage (20V) across each resistor in parallel

Type 5 · Combination Circuits

Series + Parallel

Step 1 — Resolve the parallel section

Total current

Voltage across the parallel section

Series + Parallel

Parallel section

Series + Parallel

Parallel section

Series + Parallel

Parallel section

Series + Parallel

Parallel section

Type 6 · Electric Power

P = VI

P = I²R

P = V²/R

P = VI

P = I²R

Type 7 · Electrical Energy & Cost

Energy Cost

Energy Cost

Energy Cost

Energy Cost

Energy Cost

Type 8 · Heating Effect of Current

H = I²Rt

H = I²Rt

Convert time to seconds

H = I²Rt

Convert time

H = I²Rt

H = I²Rt

Which of These 8 Types Would You Actually Recognise Cold?

What a Genelis weak area map looks like after working through Electricity practice sets

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