In the circuit shown in the figure switch S1 is closed at t

In the circuit shown in the figure, switch S1 is closed at t = 0. Find the current in the 8 M Ohm resistor immediately after the switch is closed. Find the maximum charge that will reside on the 12 muF capacitor. How much energy is stored in the capacitor when it\'s fully charged? After a long time, S1 is first opened and then S2 is closed. Find the current in the 2 M Ohm resistor just after S2 is closed. Letting q be the charge on the 4 muF capacitor at any time, write a differential equation that could be used to determine q as a function of time. Find the current in the 2 M Ohm resistor 1 minute after S2 is closed. Determine the final energy stored in the capacitors.

Solution

As mentioned, we will assume that the voltage of battery as 20 V.

Part a.) The moment the switch is closed with a capacitor in it, the capacitor starts getting charged, at this moment the capacitor acts just like a closed circuit and the current is determined only by the resistor in the circuit.

So, the current in the circuit = Voltage / Resistance = 20 / 14 x 10⁶ = 1.429 x 10^-6amperes.

Part b.) The maximum charge would be given as Q = Cv = 12 x 20 x10^-6 = 240 x 10^-6 C

Part c.) The energy stored in the capacitor is given as: E = 0.5 cV²

That is, E = 0.5 x 12 x 10^-6 x 400 J = 2400 x 10^-6 J

Part d.) The moment S1 is opened and the S2 is closed, the capacitor will again act like a closed circuit

Current = Voltage generated by the capacitor / Resistance = 20 / 10 x 10^6 = 2 x 10^-6 amperes

Part e.) Current in a circuit is defined as the rate of flow of charge.

That is, I = dQ/dt

Now, using the loop law, we can write: RI + Q/C₁ + (Qo - Q) / C₂ = 0

or, RdQ/dt = - Q/C₁ - (Qo - Q) / C₂ is the required differential equation

Part f.) Current in a discharing circuit is given as:

I = (Vo/R)e^-t/RCnet

So current at t = 1 minute would be given as:

I = (20 / 10 x10^6) e^-60/30 = 2 x 10^-6 x e^-2 = 0.27 x 10^-6 amperes

Part g.) We know that for a capacitor being discharded through a capacitor, half of the energy is lost, and one half is stored in the capacitors.

Hence the energy stored in the two capacitors would be half of energy originally stored on the first capacitor

Therefore the required energy = 2400 x 10^-6 / 2 = 1200 x 10^-6 J