You have an uncharged capacitor (12 x 10 -6 F) and a resistor (1000 ohms) connec

Question

You have an uncharged capacitor (12 x 10-6 F) and a resistor (1000 ohms) connected to a battery of 200 V at t=0 (the switch is closed at this time).

a) What is the time constant? d)What is the final current (after a very long time?

b)what is the initial current in the circuit? e) What is the charge at 0.01s?

c) what is the current at 0.01s? f) What is the final charge on the capacitor (after a very long time)? g) After the capacitor is fully charged, you disconnec tht e battery and allow the capacitor to discharge. What is the initial current in the circuit for the discharging capacitor? h) what is the current after 0.005 s of discharging? What is the charge on the capacitor at this time?

Explanation / Answer

C =12x10^-6 F, R =1000 ohms, Vo =200 V

(a) Time constant = RC = (1000)(12x10-6) = 0.012 s

(b) i =Vo/R = 200/1000 =0.2 A

(c) t =0.01s

For charging circuit Q =Qo (1-e-t/RC)

i =dQ/dt

i = (Qo/RC) (e-t/RC)

Qo /C=Vo

i = (Vo/R)(e-t/RC) = (200/1000)[(e-0.01/0.012)

i =0.0869 A

(d) Zero

(e) Q =Qo (1-e-t/RC)

Q = (VoC)(1-e-t/RC)

Q = (200)(12x10-6)[1 -e-0.01/0.012]

Q =1.92x10-4 C

(f) Q =VoC =(200)(12x10-6)

Q =2.4x10-3 C

(g) i =Vo/R = 200/1000 =0.2 A

(h) t =0.005

For discharging circuit Q =Qo (e-t/RC)

i = -(Qo/RC) (e-t/RC)

Qo /C=Vo

i = - (Vo/R)(e-t/RC) = - (200/1000)[(e-0.005/0.012)

i = -0.132 A

Q =Qo (e-t/RC)

Q =(200)(12x10-6) [(e-0.005/0.012)

Q= 2.302x10-3 C

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