1.) In a discharging circuit, it takes 1.67 s for the voltage across a capaictor

Question

1.) In a discharging circuit, it takes 1.67 s for the voltage across a capaictor to drop from 6.0 V to 4.0 V. If the circuit resistance is 1000 Ohms, what is the capacitance?

2.) After this 1.67 s has passed, what is the charge on the capacitor?

3.) How long will it now take for the above circuit to now discharge down to 2.0 V?

Please show work and equation used for my understanding and when someone provides an answer, can someone please confirm that it is correct? I need to understand it not just get an answer, but to do so it has to be correct! lol

Explanation / Answer

1) capacitor discharging

V = Vo*e^(-t/RC)

4 = 6*e^(-1.67/1000*C)

-0.405 = -1.67/1000*C

C = 4.118x10^-3 F

2) Q = CVo*e^(-t/RC)

Q = 4.118x10^-3*6*e^(-1.67/1000*4.118x10^-3)

Q = 16.47x10^-3 C = 16.47 mC

3) V = 2

V = Vo*e^(-t/RC)

2 = 6*e^(-t/1000*4.118x10^-3)

-1.0986 = -t/4.118

t = 4.5248 s

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