Electricity Class 10: Series and Parallel Numericals Solved

This post collects the Electricity Class 10 numericals you need, solved step by step with units at every line. You will first see all the formulas from the chapter, then 12 worked problems that go from easy to hard, with the focus on resistors in series and parallel. Three practice questions with answers come at the end.

Formulas you need

Two rules to keep in mind:

• Series: the same current flows through every resistor; the voltages add up: V = V1 + V2 + V3. Rs is greater than any single resistance.
• Parallel: every resistor gets the same voltage; the currents add up: I = I1 + I2 + I3. Rp is less than the smallest single resistance.

Warm-up: current, charge and voltage

Numerical 1. A filament draws 0.5 A for 10 minutes. How much charge flows?

• t = 10 min = 10 × 60 s = 600 s
• Q = I × t = 0.5 A × 600 s = 300 C

Numerical 2. How much work is done in moving 2 C of charge across a potential difference of 12 V?

• W = V × Q = 12 V × 2 C = 24 J

Numerical 3. The potential difference across a resistor is halved and its resistance stays the same. What happens to the current?

• I = V / R
• New current = (V/2) / R = I / 2
• The current becomes half its earlier value.

Numerical 4. A wire of length l and area A has a resistance of 4 Ω. Find the resistance of a wire of the same material with length l/2 and area 2A.

• R = ρ × l / A = 4 Ω
• R' = ρ × (l/2) / (2A) = (1/4) × ρ × l / A
• R' = (1/4) × 4 Ω = 1 Ω

Resistors in series

Numerical 5. A lamp of 20 Ω and a conductor of 4 Ω are joined in series to a 6 V battery. Find the current and the voltage across each.

• Rs = 20 Ω + 4 Ω = 24 Ω
• I = V / Rs = 6 V / 24 Ω = 0.25 A
• V across lamp = 0.25 A × 20 Ω = 5 V
• V across conductor = 0.25 A × 4 Ω = 1 V
• Check: 5 V + 1 V = 6 V, equal to the battery voltage.

Numerical 6. Resistors of 5 Ω, 8 Ω and 12 Ω are in series with a 6 V battery. Find the current and the voltage across each resistor.

• Rs = 5 Ω + 8 Ω + 12 Ω = 25 Ω
• I = 6 V / 25 Ω = 0.24 A
• V1 = 0.24 A × 5 Ω = 1.2 V
• V2 = 0.24 A × 8 Ω = 1.92 V
• V3 = 0.24 A × 12 Ω = 2.88 V
• Check: 1.2 V + 1.92 V + 2.88 V = 6 V.

Resistors in parallel

Numerical 7. Find the equivalent resistance of 5 Ω, 10 Ω and 30 Ω in parallel.

• 1/Rp = 1/5 + 1/10 + 1/30 (each in 1/Ω)
• LCM is 30: 1/Rp = (6 + 3 + 1)/30 = 10/30 = 1/3
• Rp = 3 Ω
• Check: 3 Ω is less than the smallest resistor, 5 Ω.

Numerical 8. The same three resistors (5 Ω, 10 Ω, 30 Ω) in parallel are connected to a 12 V battery. Find the current in each and the total current.

• Each resistor has the full 12 V across it.
• I1 = 12 V / 5 Ω = 2.4 A
• I2 = 12 V / 10 Ω = 1.2 A
• I3 = 12 V / 30 Ω = 0.4 A
• Total I = 2.4 A + 1.2 A + 0.4 A = 4.0 A
• Check with Rp: I = 12 V / 3 Ω = 4 A. Both methods agree.

Numerical 9. What are the lowest and highest resistances you can get from four 24 Ω coils?

• Lowest (all in parallel): 1/Rp = 4 × (1/24) = 4/24 = 1/6, so Rp = 6 Ω
• Highest (all in series): Rs = 4 × 24 Ω = 96 Ω

A mixed circuit

Numerical 10. A 3 Ω and a 6 Ω resistor are joined in parallel. This pair is connected in series with a 2 Ω resistor and a 12 V battery. Find the total current and the current in each parallel branch.

Step 1: the parallel pair.

• 1/Rp = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2, so Rp = 2 Ω

Step 2: total resistance.

• R = 2 Ω + 2 Ω = 4 Ω

Step 3: total current.

• I = 12 V / 4 Ω = 3 A

Step 4: voltage across the parallel pair.

• V = 3 A × 2 Ω = 6 V (the 2 Ω series resistor also takes 3 A × 2 Ω = 6 V; 6 V + 6 V = 12 V)

Step 5: branch currents.

• Through 3 Ω: 6 V / 3 Ω = 2 A
• Through 6 Ω: 6 V / 6 Ω = 1 A
• Check: 2 A + 1 A = 3 A.

Heating, power and energy

Numerical 11. Find the heat produced when 96000 C of charge is transferred in one hour through a potential difference of 50 V.

• H = V × I × t, and I × t = Q, so H = V × Q
• H = 50 V × 96000 C = 4,800,000 J = 4.8 × 10⁶ J

Numerical 12 (two parts).

(a) An electric motor takes 5 A from a 220 V line. Find its power.

• P = V × I = 220 V × 5 A = 1100 W (1.1 kW)

(b) A 400 W refrigerator runs 8 hours a day for 30 days. At Rs 3 per kWh, what does the energy cost?

• Energy = 400 W × 8 h/day × 30 days = 96,000 Wh
• In kWh: 96,000 Wh ÷ 1000 = 96 kWh
• Cost = 96 kWh × Rs 3 per kWh = Rs 288

Common mistakes in these numericals

• Forgetting to convert time. Minutes and hours must become seconds for Q = I × t and H = I² × R × t.
• Stopping at 1/Rp. After adding reciprocals, flip the fraction to get Rp.
• Using total voltage on one series resistor. In series, each resistor gets only its share: V1 = I × R1.
• Splitting voltage in parallel. In parallel, every branch gets the full voltage; it is the current that splits.
• Mixing W and kWh. Divide watt-hours by 1000 to get kWh.

A quick sense-check: a series total is always bigger than the biggest resistor, and a parallel total is always smaller than the smallest one. If your answer breaks this rule, recheck it.

Practice questions

1. Resistors of 6 Ω and 3 Ω are joined in series to a 9 V battery. Find the current. Answer: Rs = 9 Ω, so I = 9 V / 9 Ω = 1 A.
2. The same 6 Ω and 3 Ω resistors are joined in parallel to a 6 V battery. Find the equivalent resistance and the total current. Answer: 1/Rp = 1/6 + 1/3 = 1/2, so Rp = 2 Ω; I = 6 V / 2 Ω = 3 A.
3. A 1000 W appliance runs for 2 hours. How much energy does it use in kWh and in joules? Answer: 2 kWh = 2 × 3.6 × 10⁶ J = 7.2 × 10⁶ J.

To keep practising circuits with instant feedback, study this chapter with Joy on Learnijoy.