A 120 Ohm resistor and a 15.0 mu F capacitor are connected in series. Suppose a

Question

A 120 Ohm resistor and a 15.0 mu F capacitor are connected in series. Suppose a 12.0 V battery is connected across the series combination. a. Immediately after the battery is connected, how much current Bows in the circuit? What is the potential difference V_c across the capacitor? What is the potential difference V_R across the resistor? b. What is the RC time constant in seconds? c. How much time in seconds elapses before the current decreases to half of its initial value? When the current is half its initial value, what is common potential difference across the capacitor and the resistor? d. After a very long time has elapsed, how much current flows in the circuit? What is the potential difference V_c across the capacitor? What is the potential difference V_R across the resistor?

Explanation / Answer

a)immediately after the battery is connected:

I = V/R = 12/120 = 0.1 A

Hence, I = 0.1 A

The voltage across capacitor is zero at immediately after the battery connected.

Vc = 0

b)Time constant is given by

tau = RC = 120 x 15 x 10^-6 = 1.8 x 10^-3 s

Hence, tau = 1.8 x 10^-3 s = 1.8 ms

c)I' = I/2 = 0.1/2 = 0.05 A

We know that

I = I0 e^-t/RC

0.05 = 0.1 e^-t/RC

e^-t/RC = 0.5

taking natural log both sides

-t/RC = -0.693

t = 0.693 x RC = 0.693 x 120 x 15 x 10^-6 = 1.25 x 10^-3 s

Hence, t = 1.25 ms

d)After very long time,

I = 0

Current becomes zero in the circuit and the voltage across the capacitor is equal to the battery voltage.

Vc = 12 V

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