A 2.00-nF capacitor with an initial charge of 4.54 HC is discharged through a 1.

Question

A 2.00-nF capacitor with an initial charge of 4.54 HC is discharged through a 1.34-k2 resistor. (a) Calculate the current in the resistor 9.00 s after the resistor is connected across the terminals of the capacitor. (Let the positive direction of the current be define such that dQ > o.) -58.95 Need Help? dt (b) What charge remains on the capacitor after 8.00 s? 2.29E-7 what is the equation describing the time dependence of the charge on a capacitor that is being discharged in an RC circuit? pC (c) What is the (magnitude of the) maximum current in the resistor? 1.69

Explanation / Answer

a) Current after t sec is i(t) = I*e^(-t/T)

T is the time constant = R*C = 2*10^-9*1.34*10^3 = 2.68*10^-6 sec

then

I = Q/(RC) = (4.54*10^-6)/(1.34*10^3*2*10^-9) = 1.69 A

then

i(t) = I*e^(-t/T)

i(t) = 1.69*e^(-9*10^-6/(2.68*10^-6)) = 0.0588 A

b) q(t) = Q*e^(-t/T)

q(t) = 4.54*10^-6*e^(-8/2.68) = 0.229*10^-6 C

q(t) =0.229 uC

you have done correct,but you have to enter in micro coulombs bro

c) I = Q/(RC) = (4.54*10^-6)/(1.34*10^3*2*10^-9) = 1.69 A

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