Question

A 12.2??F capacitor is connected through a 0.885?M? resistor to a constant potential difference of 60.0 V.

Part A. Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0

s, 10.0 s, 20.0 s, and 100.0 s.

Part B. Compute the charging currents at the same instants.

Answer 1

Part 1) The basic equation is:

I (t) = (I at t=0) (e to the -t/tau)

The two times of interest are 0 sec and 4 sec:

I (at t= 0 sec) = (12.0 V) / 3.4M ohms = 3.53 uAmps

I (at t= 4 sec) = 3.0 V / 3.4M ohms = 0.882 uAmps

Use the basic equation above, with t = 4 sec, to solve for tau:

(I at t = 4 sec)/(I at t = 0 sec) = (e to the -t/tau) =

0.882 / 3.53 = 0.250 = (e to the -1.386)

Therefore 4/tau = 1.386

tau = 2.88 sec = RC

R = 3.4M ohms, therefore:

C = tau/R = 0.847 uf



Part 2):

The charge = Q = CV

C = 12.4 uF

tau = RC = (0.895)(12.4) = 11.0 sec

To find the charge Q(t), we need to find V(t) using the basic equation:

V(t) = (60 volts)(1 - 1/e to the t/tau)


For example, at t = 5.0 sec:

- t/tau = - 5.0/11.0 = - 0.4545

e to the -0.4545 = 1.72, and plugging these into the basic equation above:

V = (60.0)(1 - 1/1.72) = 60(0.42) = 25.2 volts

Q = CV = (12.4 uF)(25.2 V) = 312 uCoulombs

The charge on the capacitor at t= 10, 20, 100 etc can be calculated in the same way.