For Figure 4. E = 35 kV, R = 5.5 k Ohm, and C = 6800.0 mu F. Determine the time

Question

For Figure 4. E = 35 kV, R = 5.5 k Ohm, and C = 6800.0 mu F. Determine the time constant for the circuit. (Give your answer in the form of "ab s".) Determine the maximum charge that will appear on the capacitor (Give your answer in the form of "abc" C) Assuming that the capacitor is initially uncharged, how long will it take for the capacitor to charge to half the maximum charge? (Give your answer in the form of "ab.c s".) Assume that the capacitor has been fully charged, It is then disconnected from the circuit and a 470 ohm resistor connected in parallel with it to discharge it. How long will it take to be 99% discharged7 Give your answer in the form "ab.c" s.

Explanation / Answer

E = 35 kV , R =3.5 kiloohms, C =6800 uF

Time constant T = RC = 3.5*1000*6800*10^-6

T =23.8 s

Qo = ECo = 35*1000*6800*10^-6 =238 C

For charging RC circuit

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

Qo/2 =Qo (1-e^-t/RC)

t =0.693RC = 0.693*23.8

t = 16.5 s

R =470 ohms, C = 6800 uF

T = 3.196 s

For discharging circuit

Q =Qo (e^-t/RC)

0.99Qo =Qo (e^-t/RC)

t =0.01*RC =0.01*3.196

t =0.032 s

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