Question

A 2.00-nF capacitor with an initial charge of 4.84 µC is discharged through a 2.63-kΩ resistor. (a) Calculate the current in the resistor 9.00 µs after the resistor is connected across the terminals of the capacitor. (Let the positive direction of the current be define such that dQ dt > 0.) mA (b) What charge remains on the capacitor after 8.00 µs? µC (c) What is the (magnitude of the) maximum current in the resistor? A

Answer 1

Here ,

C = 2 nF

Q = 4.84 uF

a) voltage across the capacitor = Q/C = 4.84 *10^-6/(2 *10^-9)

V = 2420 V

time constant , T = R * C = 2 *10^-9 * 2.63 *10^3 = 5.26 *10^-6 s

T = 5.26 µs

Now, at t = 9 us

I = 2420/(2.63 *10^3) * (e^(-9/5.26))

I = 0.17 A

the current in the resistor is 0.17 A

b)

Q = 4.84 * (e^(-8/5.26))

Q = 1.05 uC

the charge is 1.05 uC

c)

maximum current in capacitor = V/R

maximum current in capacitor = 2420/(2.63 *10^3)

maximum current in capacitor = 0.92 A