Chapter 34, Problem 68A

A lightbulb is marked ' 120 V, 60 W.'

a. What current flows through the filament when the bulb is turned on?

b. Show that the electrical resistance of the lightbulb filament is $240\Omega$.

To determine

To find: The magnitude of current

Explanation of Solution

The magnitude of current is

  $I=\frac{P}{V}=\frac{60\text{W}}{120\text{V}}=0.5\text{A}$

To determine

To show: The magnitude of resistance is $240\Omega$

Explanation of Solution

The magnitude of resistance is

  $R=\frac{P}{I^2}=\frac{60\text{W}}{{\left(0.5\text{A}\right)}^2}=240\Omega$