Consider the circuit given below, where V = 55 V, R = 100 k, and C = 0.3 F. Assu

Question

Consider the circuit given below, where V = 55 V, R = 100 k, and C = 0.3 F.

Assume that the capacitor in the circuit is fully discharged with S1 in Position 2.

A) How much is the capacitor voltage VC exactly one time constant after S1 is moved to Position 1? Round the final answer to two decimal places. V

B) How much is the capacitor voltage VC five time constants after S1 is moved to Position 1? Round the final answer to the nearest whole number. V

C) How much is the capacitor voltage VC 1 week after S1 is moved to Position 1? Round the final answer to the nearest whole number. V

Explanation / Answer

When switch S1 is moved to the position 1, the circuit act as RC circuit in which voltage across capacitor increases till it becomes fully charged.

We know that the equation of the voltage in the RC circiut is VC(t) = Vo(1- e-t/RC) where RC is the time constant of the circiut and Vo is the final voltage across the capacitor at t = infinity.

So here, RC (time constant) = .030 sec and Vo = 55V

Now we would solve the question with the above concept

A. Volatge VC across the capacitor at 1 time constant i.e. t = RC

VC(t=RC) =Vo(1-e-RC/RC)= Vo(1-e-1) = Vo(1-.368) =Vo * .632 = 55*.632 =34.76 V

B. Volatge VC across the capacitor at 5 time constant i.e. t = 5RC

VC(t=5RC) =Vo(1-e-RC/5RC)= Vo(1-e-1/5) = Vo(1-.006) = Vo* .993 = 55*.993 = 54.63 V

C. Since time constant of this circuit is .30 sec (value of RC) so it means 1 week period would be ideally act as infinite time. Hence, at t= infinity, capacitor would be fully charged and its value is 55 V.

So VC= 55V

Note: At t=infinity, no current would flow in the circuit as capacitor is fully charged and it has same voltage as applied in the circuit.

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