The switch below has been closed for a long period of time and is opened at t=0.

Question

The switch below has been closed for a long period of time and is opened at t=0. a) Find the currents in each resistor and the capacitor just before the switch is opened, (assume the capacitor is fully charged) b) Find the Energy stored in the capacitor just before the switch is opened c) Find the initial value of the discharge current (magnitude and direction (out of top or out of bottom of the capacitor) when the switch is opened at t = 0. d) Find the time constant of the capacitor discharge e) Find the time(from t = 0) at which the charge on the plates is 13.5 % of the maximum value. Haw charge remains at this time?

Explanation / Answer

(A) current 2kohm near capacitor = 0

current through 1kOhm = current through 2 kohm = 12 / (1 + 2)

= 4 mA

(b) V = 4mA x 2 kohm = 8 Volt

U = C V^2 /2 = (1 x 10^-6) (8^2) / 2

U = 32 x 10^-6 J

(c) when switch is opened,

I = 8 / (2 + 2) = 2 mA

counterclockwise

(d) T= R C = (1 x 10^-6) (2000 + 2000)
  
= 4 x 10^-3 s

(e) V = Vmax [e^(-t/T)]

0.135 Vmax = Vmax [ e^(-t/T)]

- t / T = ln(0.135)

t = 2 T = 8 x 10^-3 s .........Ans

V = 0.135 x 8 = 1.08 Volt

Q = 1.08 x 1 x 10^-6 = 1.08 x 10^-6 C ..........Ans

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