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Starting with V=Voe^(-t/RC) show that t(.5)= RCln2. This is an analysis question for a lab on RC circuits w/ AC. I don't quite understand the question.
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It's asking you for what time, t, is \( V = V_0/2 \).
So what I plug in the Voltage, Vo/2 and it should be the same as RCln(2)?
No, you want the value of \( t \) for which \( V = V_0/2 \). In other words, the value of \( t \) for which \[ \frac{V_0}{2} = V_0 e^{-t/RC} \] Solve that equation for \( t \).
So something like .5= e^-t/RC -> ln(.5) =-t/RC -> -(1/t) = 1/RCln(.5) -> t=RCln(.5)?
From here, ln(.5) =-t/RC remember that ln(1/2) = -ln(2)
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Oh yea, thanks got it.
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