Question

A 2.04 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.28 kΩ resistor. (a) Calculate the current in the resistor 9.00 µs after the resistor is connected across the terminals of the capacitor.

Answer 1

The current in the resistor is 56.44 mA.

Step-by-step explanation:

Given that,

Capacitor Q= 2.04 nF

Initial charge
`q= 4.55\ \mu C`

Resistor
`R= 1.28 k\omega`

Time
`t = 9.00*10^(-6)`

We need to calculate the current

Using formula of current

`I=(Q)/(RC)e^{(-t)/(RC)}`

Where, C = capacitor

R = resistor

t = time

Q = charge

Put the value into the formula

`I=(4.55*10^(-6))/(1.28*10^(3)*2.04*10^(-9))e^{(-9.00*10^(-6))/(1.28*10^(3)*2.05*10^(-9))}`

`I=0.05644\ A`

`I=56.44\ mA`

Hence, The current in the resistor is 56.44 mA.