Question

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. Part A What is the resistance of a 150 W bulb? R = nothing Ω Request Answer Part B How much current does the 150 W bulb draw in normal use?

Answer 1

To solve this problem we will apply the concepts related to the calculation of power, from the two electrical forms:

$P = VI$

$P = \frac{V^2}{R}$

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

$R = \frac{V^2}{P}$

Replacing,

$R = \frac{(120V)^2}{150W}$

$R = 96\Omega$

Resistance of bulb is $96\Omega$

PART B)

$I = \frac{P}{V}$

$I = \frac{150W}{120V}$

$I = 1.25A$

The bulb will draw 1.25A current