Question
A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is $1.5 \mathrm{A}$a. What is the power rating of the bulb?b. How much electric energy does the bulb convert in 11 min?
Step 1
In this case, V = 3.0 V and I = 1.5 A. Substituting these values into the formula, we get: \[ P = 3.0 V \times 1.5 A = 4.5 W \] So, the power rating of the bulb is 4.5 W. Show more…
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