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Electric Circuits & Ohm's Law Worksheet

Class 10 Science · about 40 minutes · open the simulation at jugnix.com/simulations/electric-circuits

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By the end you will: use Ohm's law, find the resistance of series and parallel circuits, and work out electric power.

A. Explore with the simulation

  1. Set 2 cells (3 V), 1 bulb and a resistance of 6 Ω. Close the switch and record the current. Now change only the resistance and fill in the table.

    Resistance (Ω)Current (A)V ÷ I (Ω)
    3
    6
    12

  2. Choose 2 bulbs of 6 Ω each, with 2 cells. Record the total resistance and current in series, then in parallel. Which way do the bulbs glow brighter?

    WiringTotal resistance (Ω)Battery current (A)Power of each bulb (W)
    Series
    Parallel

  3. Look at the paths the current takes in each wiring. If one bulb breaks, what happens to the other bulb in series? And in parallel?

  4. Set 4 cells (6 V) and 1 bulb. Lower its resistance step by step until the 3 A fuse blows. At what resistance did it blow? Show why with I = V ÷ R.

B. Calculate (then check in the simulation)

  1. Resistors of 4 Ω and 8 Ω are in series across 6 V. Find the total resistance, the current, and the voltage across each resistor.

  2. Resistors of 3 Ω and 6 Ω are in parallel across 6 V. Find the total resistance, the current from the battery and the current in each.

  3. An electric iron is rated 1000 W, 220 V. Find the current it draws and its resistance.

  4. A 1.5 kW geyser runs for 2 hours a day. How many units (kWh) does it use in 30 days? At ₹7 a unit, what does it cost?

C. Think it through

  1. How would you connect three 2 Ω resistors to get (a) 6 Ω, (b) 2/3 Ω, (c) 3 Ω?

Answer key (for teachers and parents)

  1. 3 Ω: 1.00 A · 6 Ω: 0.50 A · 12 Ω: 0.25 A. V ÷ I equals the resistance every time: that's Ohm's law, V = IR. Doubling R halves I.
  2. Series: 12 Ω, 0.25 A, 0.375 W each. Parallel: 3 Ω, 1.00 A, 1.5 W each. Parallel bulbs glow brighter: each gets the full 3 V.
  3. In series the circuit is broken, so the other bulb goes out too. In parallel each bulb has its own path, so the other stays lit. That's why houses are wired in parallel.
  4. It blows at 1 Ω: I = 6 ÷ 1 = 6 A, more than 3 A. At 2 Ω the current is exactly 6 ÷ 2 = 3 A, which the fuse just carries.
  5. R = 4 + 8 = 12 Ω. I = 6 ÷ 12 = 0.5 A. V₄ = 0.5 × 4 = 2 V, V₈ = 0.5 × 8 = 4 V. The two add up to 6 V.
  6. 1/R = 1/3 + 1/6 = 1/2, so R = 2 Ω. I = 6 ÷ 2 = 3 A. Through 3 Ω: 6 ÷ 3 = 2 A; through 6 Ω: 6 ÷ 6 = 1 A. 2 + 1 = 3 A.
  7. I = P ÷ V = 1000 ÷ 220 ≈ 4.5 A. R = V ÷ I = V² ÷ P = 220² ÷ 1000 = 48.4 Ω.
  8. Energy = 1.5 kW × 2 h × 30 = 90 kWh = 90 units. Cost = 90 × ₹7 = ₹630.
  9. (a) All three in series: 2 + 2 + 2 = 6 Ω. (b) All three in parallel: 2 ÷ 3 ≈ 0.67 Ω. (c) Two in parallel (1 Ω), then in series with the third: 1 + 2 = 3 Ω.

Free to print and share for teaching. Made with the interactive simulation at jugnix.com/simulations/electric-circuits.