Discharging a capacitor through a resistor


by callawee
Tags: capacitor, discharging, resistor
callawee
callawee is offline
#1
Mar7-09, 03:39 PM
P: 2
In the circuit below, the time required (s) for the charge on the capacitor to drop to 47% of its value after the switch is closed is 2.83 s. If R = 1340 ohms, what is C in μF (microfarad)?


(dq/dt) + q/(RC) = 0

(.47q)/(t) = q/(RC)
t/(.47R)= C
Convert C from F to uF by multiplying by 10^6

I am getting 4993 uF, but that is the wrong answer and I'm not sure where I'm going wrong. Thanks for any help.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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LowlyPion
LowlyPion is offline
#2
Mar7-09, 04:10 PM
HW Helper
P: 5,346
Quote Quote by callawee View Post
In the circuit below, the time required (s) for the charge on the capacitor to drop to 47% of its value after the switch is closed is 2.83 s. If R = 1340 ohms, what is C in μF (microfarad)?

(dq/dt) + q/(RC) = 0

(.47q)/(t) = q/(RC)
t/(.47R)= C
Convert C from F to uF by multiplying by 10^6

I am getting 4993 uF, but that is the wrong answer and I'm not sure where I'm going wrong. Thanks for any help.
Welcome to PF.

Maybe try starting with an equation that describes the discharging of a capacitor?

Q = C*Vo*e-t/rc

Plugging the values and dividing by Q = CVo at t=0

.47 = e-2.83/1340*C
callawee
callawee is offline
#3
Mar7-09, 04:27 PM
P: 2
Thank you, the equation worked!


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