Posted by **Lanise** on Wednesday, February 10, 2010 at 11:02pm.

A capacitor in a single-loop RC circuit is charged to 64% of its final voltage in 1.5 s.

Find the time constant for the circuit.

Find the percentage of the circuit's final voltage after 3.9 s .

Vc/Vi= %

- College physics -
**drwls**, Thursday, February 11, 2010 at 1:01am
The voltage across the capacitor is

Vi * [1 - e^-(t/RC)]

in this type of circuit. Vi is the power supply voltage. RC is the time constant.

The final voltage is Vi.

When t = RC (the time constant), the voltage is

Vc = Vi*[1 - e^-1] = 0.632 Vi

When t = 1.5 s,

0.64 Vi = Vs * [1 - e^(-t/RC)]

e^(-t/RC)= 0.36

t/RC = 1.02

RC = 1.5/1.02 = 1.47 seconds

That is the time constant.

After 3.9 s

Vc = Vi* [1 - e^(-3.9/1.47)]

= 0.930 Vi

Unless you show some work on future posts here, I will not be assisting you again. Perhaps someone else will.

Vc = Vi* [1 - e^(-3.9/1.47)]

= 0.930 Vi

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