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 = (V^2)/(R)`

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

`R = (V^2)/(P)`

Replacing,

`R = ((120V)^2)/(150W)`

`R = 96\Omega`

Resistance of bulb is
`96\Omega`

PART B)

`I = (P)/(V)`

`I = (150W)/(120V)`

`I = 1.25A`

The bulb will draw 1.25A current