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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?

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cjmejiab

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

[tex]P = VI[/tex]

[tex]P = \frac{V^2}{R}[/tex]

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

[tex]R = \frac{V^2}{P}[/tex]

Replacing,

[tex]R = \frac{(120V)^2}{150W}[/tex]

[tex]R = 96\Omega[/tex]

Resistance of bulb is [tex]96\Omega[/tex]

PART B)

[tex]I = \frac{P}{V}[/tex]

[tex]I = \frac{150W}{120V}[/tex]

[tex]I = 1.25A[/tex]

The bulb will draw 1.25A current

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