A voltage source of 10.0 V is connected to a series RC circuit where R = 2.80 x 100 2, and C = 3.50 uF. Find the amount of time required for the current in the circuit to decay to 4.00% of its original value. Hint: This is the same amount of time for the capacitor to reach 96.0% of its maximum charge.

a. 31.5 s
b. 309 s
c. 0.143 s
d. 13.6 s
e. 19.1 s

The correct answer is option (a) 31.5 s

Explanation:

Given Data:

R = 2.8*10^6

C = 3.5 uF = 3.5*10^6 F

Calculating the time constant, we have;

T = R*C

= 2.8*10^6 * 3.5*10^-6

= 9.8 seconds

Calculating the amount of time required for current in the circuit to decay using the formula;

Q = Qmax*(1 - e^(-t/T) )

But  Q = 96% of Qmax

The equation becomes;

0.96*Qmax = Qmax*(1 - e^(-t/T))

0.96 = 1 - e^(-t/T)

e^(-t/T) = 1 - 0.96

t = -T*ln(0.04)

= -9.8*ln(0.04)

= 31.54 seconds

see earlier answer