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Please provide full reasoning: A capacitor is charged for a long time with a 10 V...

Please provide full reasoning:

A capacitor is charged for a long time with a 10 V battery. The battery is removed and the RC circuit is closed at time t=0. At time t=10ms the voltage across the capacitor is measured to be 6.1 V. What is the time constant of the circuit? Express your answer in milliseconds. (0 decimal places)

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Answer #1

Here ,

E = 10 V

time , t = 10 ms

let the time constant is T

for discharging of the capcitor

V = E * e^(-t/T)

6.1 = 10 * e^(-10/T)

solving for T

T = 20.23 ms

the time constant of the circuit is 20.23 ms

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