Thanks! I believe (i) is zero for both. But I a confused on what happens in (ii)

Q: Thanks! I believe (i) is zero for both. But I a confused on what happens in (ii)

i initial charge Q = 0 initail current i = v/R = 9/200000 = 4.5*10^-5 A ii) apply Vc = Vo(1-e^-t/RC) Vc = 9*(1-e^-100/(500*10^-6*200*10^3) Vc = 5.689 volts Vr = V e^-t/RC Vr = 9 * e^-100/(500*10^-6*200*10^3) Vr = 3.31 volts

ID: 3895041 • Letter: T

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Thanks!

I believe (i) is zero for both. But I a confused on what happens in (ii) to figure this out. And does it matter if the resistance is after or before the Capacitor?

In the circuit shown a resistor of 200 k Ohm is in series with a capacitor of 500 muF and a 9.0 V battery. The switch S is closed at time t = 0. (i) What is the charge on the capacitor and the current in the circuit at time t = 0 (just after the switch has been closed)? (ii) At the time t = 100 s what is the voltage drop across the capacitor Vc and resistor Vr?

Explanation / Answer

i initial charge Q = 0

initail current i = v/R = 9/200000 = 4.5*10^-5 A

ii) apply Vc = Vo(1-e^-t/RC)

Vc = 9*(1-e^-100/(500*10^-6*200*10^3)

Vc = 5.689 volts

Vr = V e^-t/RC

Vr = 9 * e^-100/(500*10^-6*200*10^3)

Vr = 3.31 volts

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Thanks! I believe (i) is zero for both. But I a confused on what happens in (ii)

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