The capacitor in the following circuit it is initially uncha

The capacitor in the following circuit it is initially uncharged:

A. At t=0, s1 closed and s2 is open. Find the time constant of the circuit?

B. Find the charge in the capacitor at t=7ms?

C. After capacitor has begun charging s1 is open and s2 closed. How long must C be allowed to charge before adjusting the switch for maximum current in the circit to be 5mA.

Solution

For capacitor charging, we know that the voltage on the capacitor at any point of time, t is given as:

V(t) = (Q_max/C)[1 - e^-t/RC] where R is the resistance in the circuit and C is the capacitance. We will make use of this to solve the given problems.

Part a.) The time constant of an RC circuit is given as: T = RC

That is, T = 10 x 10^3 x 1 x 10^-6 = 0.01 seconds

Part b.) Now, the charge on the capacitor at any time t would be given as:

Q(t) = Qo[1 - e^-t/RC]

So, the charge on the capacitor at t = 7 ms would be:

Q(t) = (10 x 10^-6)[1 - e^{-7x10^-3/0.01}] = 5.036 C is the required charge.

Part C.) For a capacitor discharging through an inductor, the maximum current is given as:

Io = Qo / sqrt(LC)

Here, Io = 5 m A

That is, Qo = 5 x 10^-3 x sqrt(10 x 10^-6) = 5 x 10^-3 / 316.22 = 15.8 C

Now this is larger than the maximum charge which the capacitor can have, hence such a current is not possible in the second circuit.

NOTE: You might want to check the data for the inductor again. It might be in mH and not H. If that is the case, you can calculate the charge as above, and then use the capacitor charging equation to find at what time will the capacitor have required charge.