Question

1A)

A circuit consists of a 12.0 V battery, a 100 kΩ resistor, a 20.0 μF capacitor in series with a switch which is initially in the open position. The capacitor is initially uncharged. Calculate the charge on the capacitor 6.00 seconds after the switch is closed. Calculate the current through the resistor 6.00 seconds after the switch is closed.

1B)

A 20 μF capacitor has previously charged up to contain a total charge of Q=100 μC on it. The capacitor is then discharged by connecting it directly across a 100−kΩ resistor. At what point in time after the resistor is connected will the capacitor have 13.5 μC of charge remaining on it?

Answer 1

1A)

Given

V = 12.0 V

R = 100 x 10³ ohm

C = 20.0 x 10^-6 F

t = 6.00 s

Solution

Q_max = CV = 20.0 x 10^-6 x 12.0

Q_max = 240 x 10^-6 C

I_max = V/R = 12/100 x 10³ = 1.2 x 10^-4 A

Time constant

τ = RC = 100 x 10³ x 20.0 x 10^-6

τ = 2 s

Q_t = Q_max [1-e^-t/τ]

Q₆ = 240 x 10^-6 [1-e^-6/2]

Q₆ = 2.28 x 10^-4 C

I_t = I_max e^-t/τ

I₆ = 1.2 x 10^-4 x e^-3

I₆ = 5.97 x 10^-6 A

1B)

Given

C= 20 x10^-6 F

R = 100 x 10³ ohm

Q_o = 100 x 10^-6 C

Q = 13.5 x 10^-6 C

Solution

Q_t  = Q_o [e^-t/τ]

13.5 x 10^-6 = 100 x 10^-6 [e^-t/2]

0.135 = e^-t/2

e^t/2 = 1/0.135 = 7.407

t/2 = ln(7.407)

t = 4.00 s