In the figure R 1 = 10.1 k?, R 2 = 14.4 k?, C = 0.367 ?F, and the ideal battery

Question

In the figureR1= 10.1 k?,R2= 14.4 k?, C = 0.367 ?F, and the ideal battery has emf ? = 17.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at timet= 0. What is the current (in A) in resistor 2 att= 3.70 ms?

I can't find the potential difference across the capacitor and the resistor R2 at the given time which I need to find the current I = V / R

The time constant is T =0.0052848 s

What is the equation I need to find the potential difference at the given time?

Explanation / Answer

In steady state condition voltage across capacitor is equal to Voltage across R2

Vo =E*(R2/R1+R2) =17*(14.4/10.1+14.4)

Vo=9.99 V

Voltage across capacitor after the switch is opened at t=0 and thus t=3.7ms

V=Voe^(-t/RC)

V=9.99*e^(-3.7*10^-3/14.4*10^3*0.367*10^-6)

V=4.965 V

current across R2 is

I(R2) =V/R2 =4.965/14.4*10^3

I(R2)=3.45*10^-4 A

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