For the circuit shown in the figure below, switch S has been open for a long tim

Question

For the circuit shown in the figure below, switch S has been open for a long time. At time t = 0 the switch is then closed. Let R = 1.70 M Ohm, V = 60.0 V, and C = 3.00 mu F. What is the battery current just after switch S is closed? What is the battery current a long time after switch S is closed? The switch has been closed for a long time. At time t = 0 the switch is then opened. Find the current in mu A through the 600 k Ohm resistor as a function of time. (Use the following as necessary: t.)

Explanation / Answer

when switch ahs been closed,
voltage drop across R = i thorugh battery * R
= 26.09*10^-6 * 1.70*10^6
                   = 44.4 v
                  
So, Volatge drop across capacitor,
Vc = 60 - 44.4 = 15.6 V

Qo = Vc*C
= 15.6*3*10^-6
= 4.68*10^-5 C

Now when switch has been open, caapcitor will start discharging
Io = Qo/(R*C)
= (4.68*10^-5)/(600*10^3 * 3*10^-6)
= 2.6*10^-5 A

Current through resistor at any time t,
I(t) = Io * e^-t/(RC)
= 2.6*10^-5 * e^(-t/(600*10^3 * 3*10^-6))
   = 2.6*10^-5 * e^(-0.56*t)
     
This is the expression you are looking for

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