Question

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= %

Answer 1

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