Samacheer Kalvi 10th Science Chapter 4 Numerical Problems — Electricity with Solutions

Numerical Problems — Chapter 4: Electricity

Important Formulas:

Problem 1. A potential difference of 12 V is applied across a resistor of 4 Ω. Find the current flowing through it.

• Voltage, V = 12 V
• Resistance, R = 4 Ω
• Current, I = ?

By Ohm's Law: V = IR → I = V/R

I = 12/4 = 3 A

Problem 2. Three resistors of 2 Ω, 4 Ω, and 6 Ω are connected in series to a 12 V battery. Find: (a) Total resistance (b) Current in the circuit (c) Voltage across each resistor

Part (a): Total resistance in series

R = R₁ + R₂ + R₃ = 2 + 4 + 6 = 12 Ω

Part (b): Current

I = V/R = 12/12 = 1 A

(In series, same current flows through all)

Part (c): Voltage across each resistor (using V = IR)

V₁ = I × R₁ = 1 × 2 = 2 V V₂ = I × R₂ = 1 × 4 = 4 V V₃ = I × R₃ = 1 × 6 = 6 V

Check: V₁ + V₂ + V₃ = 2 + 4 + 6 = 12 V ✓

Problem 3. Three resistors of 3 Ω, 6 Ω, and 9 Ω are connected in parallel to a 9 V battery. Find: (a) Equivalent resistance (b) Total current from the battery (c) Current through each resistor

Part (a): Equivalent resistance in parallel

1/R = 1/R₁ + 1/R₂ + 1/R₃ = 1/3 + 1/6 + 1/9

LCM of 3, 6, 9 = 18

1/R = 6/18 + 3/18 + 2/18 = 11/18

R = 18/11 ≈ 1.64 Ω

Part (b): Total current

I = V/R = 9 ÷ (18/11) = 9 × 11/18 = 99/18 = 5.5 A

Part (c): Current through each (In parallel, voltage is same = 9 V)

I₁ = V/R₁ = 9/3 = 3 A I₂ = V/R₂ = 9/6 = 1.5 A I₃ = V/R₃ = 9/9 = 1 A

Check: I₁ + I₂ + I₃ = 3 + 1.5 + 1 = 5.5 A ✓

Problem 4. An electric iron has a resistance of 40 Ω and is connected to a 200 V supply. Find: (a) Current drawn (b) Power consumed (c) Heat produced in 30 minutes

• R = 40 Ω; V = 200 V; t = 30 min = 30 × 60 = 1800 s

Part (a): Current

I = V/R = 200/40 = 5 A

Part (b): Power

P = VI = 200 × 5 = 1000 W = 1 kW

Or: P = V²/R = 200²/40 = 40000/40 = 1000 W ✓

Part (c): Heat produced

H = Pt = 1000 × 1800 = 1,800,000 J = 1.8 × 10⁶ J

Problem 5. A 60 W bulb and a 100 W bulb are both rated for 220 V. Which has higher resistance? Find the resistance of each.

• Bulb 1: P₁ = 60 W, V = 220 V
• Bulb 2: P₂ = 100 W, V = 220 V

Using P = V²/R → R = V²/P

R₁ = 220²/60 = 48400/60 ≈ 807 Ω

R₂ = 220²/100 = 48400/100 = 484 Ω

The 60 W bulb has HIGHER resistance.

Why: Lower wattage bulb draws less current → needs higher resistance to limit current at the same voltage. Brighter bulbs are "less resistant."

Problem 6. Calculate the cost of running a 1000 W electric heater for 5 hours, if the cost of electricity is ₹5 per kWh.

• Power, P = 1000 W = 1 kW
• Time, t = 5 hours
• Rate = ₹5 per kWh

Energy consumed = P × t = 1 kW × 5 h = 5 kWh

Cost = 5 kWh × ₹5 = ₹25

Problem 7. A current of 2 A flows through a resistance of 5 Ω for 10 minutes. Find the heat produced.

• I = 2 A; R = 5 Ω; t = 10 min = 600 s

H = I²Rt = 2² × 5 × 600 = 4 × 5 × 600 = 12,000 J

Problem 8. Two resistors of 6 Ω and 12 Ω are connected in parallel, and this combination is connected in series with a 4 Ω resistor. The whole circuit is connected to a 24 V battery. Find: (a) Equivalent resistance (b) Total current (c) Current through each parallel branch

Part (a): Equivalent resistance

Parallel combination: R_p = (6 × 12)/(6 + 12) = 72/18 = 4 Ω

Total (series with 4 Ω): R_total = R_series + R_p = 4 + 4 = 8 Ω

Part (b): Total current

I_total = V/R_total = 24/8 = 3 A

Part (c): Voltage across parallel combination

V_p = I_total × R_p = 3 × 4 = 12 V

Current through 6 Ω: I₁ = V_p/R₁ = 12/6 = 2 A

Current through 12 Ω: I₂ = V_p/R₂ = 12/12 = 1 A

Check: I₁ + I₂ = 2 + 1 = 3 A = I_total ✓

Problem-Solving Tips

1. Identify the circuit type first: Series or parallel or mixed? 2. In series: Same current everywhere, voltages add up 3. In parallel: Same voltage everywhere, currents add up 4. Always check your answer: Do the voltages/currents add up correctly? 5. Unit conversions to remember:

• Minutes → seconds: × 60
• Watts → kilowatts: ÷ 1000
• Hours → seconds: × 3600

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