Q1. A bulb draws a current of 0.5 A when connected to a 220 V supply. Find its resistance and power.
Given: I = 0.5 A, V = 220 V
Resistance: R = V ÷ I = 220 ÷ 0.5 = 440 Ω
Power: P = V × I = 220 × 0.5 = 110 W
Stop memorising and start seeing it. Drag the sliders below to watch Ohm's law come alive, build your own series and parallel circuits, and solve every type of board numerical step by step.
Change the battery voltage and the resistance. The current, power and bulb brightness update instantly — exactly as V = I R predicts.
Medium voltage and resistance — the bulb glows steadily.
Try this: keep the voltage fixed and slide the resistance up — the current (and brightness) fall. That inverse relationship is the heart of Ohm's law.
Enter three resistor values and switch between series and parallel. See the equivalent resistance, the working, and the total current from a 6 V battery.
Remember: series resistance is always bigger than the largest resistor; parallel resistance is always smaller than the smallest one. Home wiring is parallel so each appliance gets the full 220 V.
These are the exact patterns CBSE repeats. Read the question, try it yourself, then reveal the full solution.
Q1. A bulb draws a current of 0.5 A when connected to a 220 V supply. Find its resistance and power.
Given: I = 0.5 A, V = 220 V
Resistance: R = V ÷ I = 220 ÷ 0.5 = 440 Ω
Power: P = V × I = 220 × 0.5 = 110 W
Q2. Three resistors of 4 Ω, 6 Ω and 12 Ω are joined in parallel across a 6 V battery. Find the equivalent resistance and the current drawn from the battery.
Parallel: 1/R = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12
So R = 12 ÷ 6 = 2 Ω
Current: I = V ÷ R = 6 ÷ 2 = 3 A
Q3. An electric heater of 1000 W runs for 2 hours daily. Find the energy used in a 30-day month and the cost at ₹6 per unit.
Energy per day: 1000 W × 2 h = 2000 Wh = 2 kWh (units)
Per month: 2 × 30 = 60 units
Cost: 60 × ₹6 = ₹360
| Quantity | Formula | SI unit |
|---|---|---|
| Electric current | I = Q / t | ampere (A) |
| Potential difference | V = W / Q | volt (V) |
| Ohm's law | V = I R | — |
| Resistance of a wire | R = ρ L / A | ohm (Ω) |
| Resistors in series | R = R₁ + R₂ + R₃ | ohm (Ω) |
| Resistors in parallel | 1/R = 1/R₁ + 1/R₂ + 1/R₃ | ohm (Ω) |
| Electric power | P = VI = I²R = V²/R | watt (W) |
| Heat produced (Joule) | H = I²Rt | joule (J) |
| Electrical energy | E = P × t | kWh / joule |
At constant temperature, the current through a conductor is directly proportional to the potential difference across it: V = I R. On a V–I graph this gives a straight line through the origin.
So every appliance gets the same full voltage (220 V), each can be switched on/off independently, and if one fails the others keep working. A series connection would dim everything and one failure would break the whole circuit.
Resistance (R, in Ω) depends on the size and shape of the specific wire. Resistivity (ρ, in Ω·m) is a property of the material only and does not change with length or area.
One unit = one kilowatt-hour (kWh) = the energy used by a 1000 W appliance running for 1 hour = 3.6 × 10⁶ joules.
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