KinexiaLEARN PHYSICS

Ohm's Law — Current Through a Light Bulb (Worked Example)

Easy
Problem

An incandescent bulb has a hot resistance of 240 Ω and is connected to a 120 V household supply. What current flows through it, and what power does it dissipate?

Simple circuit with a 120 V battery and a 240 ohm bulb resistor showing 0.50 A current+ − 120 V240 Ω bulbI →I = 0.50 ACurrent flows from + to − through R
The bulb as a 240 Ω resistor on a 120 V supply.

Given

  • R = 240 Ω

  • V = 120 V

Find

Current I and power P.

Principle

Ohm's law relates voltage, current and resistance; electrical power is voltage times current.

Formula

Solution

  1. Current: I=V/R=120/240I = V/R = 120/240

  2. Calculate: I=0.50 AI = 0.50\ \mathrm{A}

  3. Power: P=VI=120×0.50P = VI = 120 \times 0.50

  4. Calculate: P=60 WP = 60\ \mathrm{W}

I = 0.50 A; P = 60 W

Unit check

V/Ω = A; V × A = W. Both match.

What this means physically

A classic 60 W bulb draws half an amp — which is why a 15 A household circuit can safely run many bulbs at once, and why the filament glows: 60 joules per second become heat and light.

Common mistake

Computing power as V/R (which is current, not power) or forgetting that the 240 Ω is the hot resistance — cold filament resistance is much lower.