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

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 twitch is opened. (assumed 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. How much charge remains at this time?

Explanation / Answer

a)

just before the switch was opened , the circuit was in steady state and capacitor was fully charged. hence Capacitor behaves as open circuit and no current flows through it.

Rtotal = Total resistance = 1 + 2 = 3 k ohm = 3000 ohm

V = battery Voltage = 12 volts

i = total current coming from the battery = V/Rtotal = 12/3000 = 0.004 A

i1 = current in 1 kohm resistor in series with battery = 0.004 A

i2 = current in 2 kohm resistor in parallel with capacitor = 0.004 A

i3 = current in 2 kohm resistor in series with capacitor = 0 A

b)

V2 = Voltage across the resistor 2 ohm in parallel with capacitor = Voltage across the capacitor = i2 R2

V2 = 0.004 x 2000 = 8 volts

energy stored by the capacitor is given as

E = (0.5) C V22 = (0.5) (1 x 10-6 ) (8)2 = 3.2 x 10-5 J

c)

io = initial current = Voltage across capacitor / total resistance = 8/3000 = 2.67 x 10-3 A

d)

T = Time constant = RC = (3000) (1 x 10-6) = 0.003 sec

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