Electricity Class 10 Numericals: Worked Problems
Solve current, resistance, series–parallel circuits and electrical energy numericals with units, step-by-step working and practice answers.
Begin every numerical by listing the given quantities and converting units. Select a formula before substituting numbers. The examples here are paper exercises; do not test calculations using household mains electricity.
Formulas and units
| Quantity | Relation | Unit |
|---|---|---|
| Current | I = Q/t | ampere (A) |
| Potential difference | V = IR | volt (V) |
| Series resistance | R = R₁ + R₂ + … | ohm (Ω) |
| Parallel resistance | 1/R = 1/R₁ + 1/R₂ + … | ohm (Ω) |
| Power | P = VI = I²R = V²/R | watt (W) |
| Energy | E = Pt | joule if P is W and t is seconds |
Ohm’s law applies to an ohmic conductor when physical conditions, including temperature, remain constant. One kilowatt-hour equals 3.6 million joules. A kilowatt is a unit of power; a kilowatt-hour is a unit of energy.
Current and resistance
1. A charge of 240 C passes a point in 2 minutes
- 1t = 2 × 60 = 120 s.
- 2I = Q/t = 240/120 = 2 A.
2. A 6 V source drives 0.5 A through a resistor
- 1R = V/I = 6/0.5 = 12 Ω.
- 2Power = VI = 6 × 0.5 = 3 W.
Series and parallel
3. Resistors of 4 Ω and 8 Ω are in series across 12 V
- 1Total resistance = 4 + 8 = 12 Ω. Current = 12/12 = 1 A.
- 2Voltage across 4 Ω = 1 × 4 = 4 V; across 8 Ω = 8 V. The drops add to 12 V.
4. Resistors of 6 Ω and 3 Ω are in parallel across 6 V
- 11/R = 1/6 + 1/3 = 1/2, so R = 2 Ω.
- 2Branch currents are 6/6 = 1 A and 6/3 = 2 A. Total = 3 A.
- 3Check total current from equivalent resistance: 6/2 = 3 A. The equivalent resistance is smaller than either branch.
Power and energy
5. A 60 W lamp operates for 5 hours
- 1Energy = 0.060 kW × 5 h = 0.30 kWh.
- 2In joules: 60 W × 18000 s = 1080000 J.
- 3At an assumed exercise tariff of ₹8 per kWh, the energy cost is ₹2.40. This is a made-up tariff for calculation, not a current electricity price.
Independent practice
- 6. A 10 Ω resistor carries 0.3 A. Find V and P. Answer: 3 V and 0.9 W.
- 7. Two 12 Ω resistors are in parallel. Find equivalent resistance. Answer: 6 Ω.
- 8. A 1000 W heater operates for 30 minutes. Find energy. Answer: 0.5 kWh = 1800000 J.
- 9. Find the heat in 10 Ω carrying 2 A for 60 s. Answer: I²Rt = 4 × 10 × 60 = 2400 J.
If a parallel equivalent is greater than the smallest branch resistance, recheck the reciprocal step. If an energy answer is 60 times too small, check the conversion between minutes and seconds. A unit beside every substituted number often reveals the error before the final line.