
#1
Mar709, 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 



#2
Mar709, 04:10 PM

HW Helper
P: 5,346

Maybe try starting with an equation that describes the discharging of a capacitor? Q = C*V_{o}*e^{t/rc} Plugging the values and dividing by Q = CV_{o} at t=0 .47 = e^{2.83/1340*C} 



#3
Mar709, 04:27 PM

P: 2

Thank you, the equation worked!



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