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## Homework Statement

A circuit is powered by a 10 V power supply and has a resistor of 500 ohms in series with a capacitor. After 4 seconds, the value of ln(1-(Vc/Vo)) is -2, where Vc is the voltage in the capacitor at a given time and Vo is 10 V.

1) Find the capacitance.

2) Find how long it takes for the potential difference across the capacitor to rise from 0 V to 4 V.

## Homework Equations

Vc = Vo(1 - e^(-t/RC))

Tau = RC

## The Attempt at a Solution

ln(1-(Vc/Vo))/t = -2/4 = 0.5

Let's try to figure out what this represents

Vc = Vo(1 - e^(-t/RC))

ln Vc = ln Vo(1 - e^(-t/RC))

ln Vc = ln Vo + ln(1 - e^(-t/RC))

ln Vc - ln Vo = ln(1 - e^(-t/RC))

Vc/Vo = 1 - e^(-t/RC)

e^(-t/RC) = 1 - Vc/Vo

-t/RC = ln(1 - Vc/Vo)

-1/RC = ln(1 - Vc/Vo)/t = -0.5

So I have 1/RC = 0.5, therefore the time constant Tau = RC = 2 seconds.

1) (500)(C) = 2 -> Capacitance = 0.004 F

2) If -t/RC = ln(1 - Vc/Vo) then

t = -RC*ln(1 - Vc/v0) = -2*ln(1-4/10) = 1.021 seconds.

Is this correct? I've been told that I have a mistake somewhere, but I don't know where.