A 13.5-F capacitor is charged to a potential of 40.0 V and then discharged throu

Question

A 13.5-F capacitor is charged to a potential of 40.0 V and then discharged through a 65.0­ resistor.

1) How long after discharge begins does it take for the capacitor to lose 90.0% of its initial charge? (Express your answer to three significant figures.)

2) How long after discharge begins does it take for the capacitor to lose 90.0% of its initial energy? (Express your answer to three significant figures.)

3) What is the current through the resistor at the time when the capacitor has lost 90.0% of its initial charge? (Express your answer to three significant figures.)

4) What is the current through the resistor at the time when the capacitor has lost 90.0% of its initial energy? (Express your answer to three significant figures.)

Explanation / Answer

Given that,

C = 13.5 uF

V = 40 V

R = 65 ohm

(1)

lnitial Charge on capacitor, q = CV

q = 13..5*10^(-6)*40 = 5.40*10^(-4) C

New charge = (100 - 90 / 100)*5.40*10^(-4) C

q' = 5.4*10^(-5) C

New voltage, V' = 5.4*10^(-5) / 13.5*10^(-6)

V' = 4 V

V' = V * e^(-t / (RC))

4 = 40 * e^(-t / 65*13.5*10^(-6))

t = 0.002 s

(2)

energy stored in a capacitor = (1/2)CV^2

E = (1/2)*13.5*10^(-5)*40^2 = 0.0216 J

new energy, E' = (100 - 90 / 100)*0.0216

E' = 0.00216 J

E' = (1/2)CV'^2

0.00216 = (1/2)*13.5*10^(-6)*V'^2

V' = 17.88 V

V' = V * e^(-t / (RC))

17.88 = 40 * e^(-t / 65*13.5*10^(-6))

t = 0.0007 s

(3)

current through resistor, l = V' / R

l = 4 / 65

l = 0.061 A

(4)

l = 17.88 / 65

l = 0.275 A

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