# RC circuit and a capacitor

caramelgt
In the following circuit R1 = 12.9 Ohms, R2 = 3 Ohms, V = 39 Volts, and C = 10.1 micro-Farads. If one waits 2 time constants after the battery has been connected (charge in capacitor is zero before the battery is connected), what is the voltage across R1 in volts? Hint: first find the charge in capacitor after this amount of time, and then find the voltage across the capacitor. Use a loop rule to find voltage across the equivalent resistor and use this to find the current in the circuit at this time.

I attached a picture.

I tried solving it by first using Q = CV to find Qmax. My Qmax was 393.9 micro coulombs. I then multiplied this number by 0.63 because 1-e^(-2/2) is 0.63. This number was 248 micro coul. After finding this number i plugged it back into the Q = CV equation to solve for the new V and I got 24.57. I thought I could subtract the V's and get the answer, but its not working out. Can someone help me out?

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Homework Helper
caramelgt
Actually, I'm looking for the new voltage. The answer is 4.28, but I cant seem to find the right steps to get that.

Homework Helper
Once you have that 34 V (34 is approximate) on the capacitor, you can find the current in the circuit at time 2RC using I = V/R = (39-34)/15.9 = 0.31 Amps. The V across the R1 is then 4 Volts or so.

Homework Helper
While the charging of the capacitor
Q = CV[1 - e^-(t/RC)]
I = dQ/dt = CV*(1/RC)[e^-t/RC] = (V/R)[e^-t/RC]
Put t = 2RC and R = R1+R2 and find I. Then voltage across R1.