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 dа ta:

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