Question

A 16.5-μμF capacitor is charged to a potential of 60.0 V and then discharged through a 75.0­ ΩΩ 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.)

answer .....ms

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.)

answer .....ms

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.)

answer .....mA

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.)

answer ....mA

Answer 1

capacitance = 16.5 * 10^-6 F

potential = 60 v

resistance = 75 ohm

charge = capacitance * voltage

charge = 16.5 * 10^-6 * 60

charge = 9.9 * 10^-4 C

when the capacitor will lose 90% of its initial charge so,

10% charge left

new charge = 10% of 9.9 * 10^-4

new charge = 9.9 * 10^-5

9.9 * 10^-5 = 16.5 * 10^-6 * new voltage

new voltage = 6 V

Vn = V * e^(-t/(RC))

6 = 60 * e^(-t/(75 * 16.5 * 10^-6))

ans.1) t = 0.002849 sec = 2.849 ms

energy stored in a capacitor = 0.5 * C * V^2

energy = 0.5 * 16.5 * 10^-6 * 60^2

energy = 0.0297 J

now 10% of energy is left

new energy = 10% of 0.00297

new energy = 0.00297 J

0.00297 = 0.5 * 16.5 *10^-6 * v^2

new voltage = 18.973 V

18.973 = 60 * e^(-t/(75 * 16.5 * 10^-6))

ans.2) t = 0.001424 sec = 1.424 ms

current through resistor = 6 / 75

ans. 3) current through resistor = 0.08 A = 80 mA

current through resistor = 18.973 / 75

ans.4) current through resistor = 0.252 A = 252 mA

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