1.35 F capacitor is charged through a 124 resistor and then discharged through t

Question

1.35 F capacitor is charged through a 124 resistor and then discharged through the same resistor by short-circuiting the battery.

Part A While the capacitor is being charged, find the time for the charge on its plates to reach 1e of its maximum value.

Part B While the capacitor is being charged, find the current in the circuit at the time when the charge on its plates has reached 1e of its maximum value.

Part C During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1e of its initial value.

Part D Find the time for the current in the circuit to decrease to 1e of its initial value.

Explanation / Answer

a)
Capacitor charge equation .. Qt = Qo [1 - e^(-t/CR) ]
Qt/Qo = [1 - e^(-t/CR) ] = 1/e = 0.3679

1 - 0.3679 = e^(-t/CR)
Ln(0.6321) = -t/CR
t = -CR Ln(0.6321) .. - (1.35^-6F x 124 x -0.4587)

t = 7.678 x 10^-5 sec………Ans.

b)
When charge on C has reached Q/e the pd across C and R = Vo/e (VQ)
Current .. i = pd/R = (Vo/e) / R

i = Vo/(eR) = V0/2.718*124 =0.00296V0………..Ans. (battery pd(Vo) required, e= 2.718)

c)
Discharge equation .. Qt = Qo e^(-t/CR)
Qt/Qo = e^(-t/CR) = 1/e = e¹ therefore ..

(-t/CR) = -1
t = CR .. .. 1.35^-6F x 124

t = 1.674 x 10^-4 sec……..Ans.

d)
Current decay follows same decay curve as Q and V .. so the time for IoIo/e = time for QoQo/e which is calculated in part (c)

t = 1.674 x 10^-4 sec……..Ans.

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