Question

We will determine the amount of electric energy stored in a capacitor by discharging it through a light bulb. Light bulbs are rated according to their power output at a given voltage. Considering that power is the rate that energy is converted from one form to another (or, equivalently, work is done) per unit time, the energy stored in an initially-charged capacitor that is hooked up to the light bulb through which the capacitor discharges is approximately
A. the power rating of the light bulb divided by the time that the bulb remains lit.
B. simply the time that the light bulb remains lit.
C. the product of the power rating of the light bulb and the time that it remains lit.
D. the time that the light bulb remains lit divided by the power rating of the bulb.

Answer 1

Answer:

C. the product of the power rating of the light bulb and the time that it remains lit.

Explanation:

The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,

$P = \frac{E}{t}\\\\E = Pt$

where,

E = Energy Supplied to the bulb = Energy stored in capacitor = ?

P = Power rating of the bulb

t = time the bulb is lit up

Hence, the correct option is:

C. the product of the power rating of the light bulb and the time that it remains lit.