More dying: You connect a particular capacitor to a 5 V DC power source, and let

Question

More dying: You connect a particular capacitor to a 5 V DC power source, and let it charge fully. You then disconnect it, and connect it to a light bulb in series with an ideal ammeter. The light bulb glows brightly, but dims and then goes out. When you first connect the capacitor to the light bulb, the ammeter shows a current of 0.25 A. Twenty seconds later, however, it only shows a current of 0.05 A. Assume the bulb is ohmic (which is probably a terrible assumption, but oh well). What is the bulb's resistance? What is the capacitor's capacitance? Bonus challenge for the ambitious: Prove that the total energy initially stored in the electric field inside the capacitor is equal to the total electrical energy eventually dissipated by the bulb.

Explanation / Answer

given,

voltage = 5 V

initial current = 0.25 A

by ohm's law

V = IR

5 = 0.25 * R

R = 20 ohm

bulb's resistance = 20 ohm

current after 20 second = 0..05 A

I = (Vo / R) * e^(-t/RC)

0.05 = (5 / 20) * e^(-20 / (20 * C))

capacitor's capacitance C = 0.621 F

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