A 12.8 F capacitor is connected through a 0.890 M resistor to a constant potenti

ID: 1436509 • Letter: A

Question

A 12.8 F capacitor is connected through a 0.890 M resistor to a constant potential difference of 60.0 V.

Part A

Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Part B

Compute the charging currents at the same instants.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Q = C* Vb * (1- e^-t/RC)

so Q at 0 = 12.8*10^-6 * 60 * (1 - e^0) = 0

Q at 5 = 12.8*10^-6 * 60 * (1 - e^-5/0.89*12.8) = 272.84 uC

Q at 10 = 12.8*10^-6 * 60 * (1 - e^-10/0.89*12.8) = 448.748 uC

Q at 100 = 12.8*10^-6 * 60 * (1 - e^-100/0.89*12.8) = 767.877 uC

I = Vb/R * (1 - e^-t/RC)

R = 0.89 Mohm, C = 12.8 uF , and t is the time =?

I at 10 s = Q at 10 / RC = 448.748uC / 0.89*12.8 = 39.39 uA

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.