Electrical – Adding short circuit to circuit

The voltage across a capacitor discharging into a fixed resistance decays exponentially. The time constant is RC, where C is the capacitance, and R is the resistance between the terminals of the resistor.

$$ V(t)=V(0)e^{\frac{-t}{RC}} \\ $$

You're right that the capacitor starts at 20V, because that's the voltage across the 200 ohm resistor after the capacitor is charged. So you know V(0), and you know C. All you're missing is R. To analyze that properly, think of the switch as a resistor of 0 ohms when closed, and don't worry about the current source. The parallel combination of a 200 ohm resistor and a 0 ohm resistor is 0 ohms. The series combination of a 50 ohm resistor and a 0 ohm resistor is 50 ohms. So the 50 ohm resistor is the only one that matters when determining your discharge constant.

The current through the 200 ohm resistor depends on the voltage across the 200 ohm resistor. The voltage across the 200 ohm resistor is the same as the voltage across the 0 ohm resistor (the closed switch). V=IR, so what's the voltage across that pair of resistors when the switch is closed?

And yes, the voltage across an ideal resistor can change instantaneously. Keep in mind though, there's no such thing as an ideal resistor in the physical world. Everything has capacitance to everything else.

In this case, you first want to reduce the circuit. I'm going to have to explain this without pictures, sorry. But I think it'll be pretty clear anyway!

1. The first thing to do is to ignore the small arrow marks that are showing the branch currents - they are no longer accurate once the circuit changed. Neither are V1 and V2.

2. Look at the 6k resistor. It has effectively been removed from the circuit. That is, there will be no current through this resistor. There are a few ways to see why this is so. Here are two of them:

   • The resistor is in parallel with a zero-Ohm path. Any current flow will happen through this short, and nothing will flow through the more difficult path of the resistor.

   • There can be no voltage that exists across the short. Therefore, no voltage will be presented across the 6k resistor, either, since they are in parallel.

3. So, now, you are left with the 1mA source, flowing though a 6k resistor in parallel with 12k resistor. Easy to analyse with either of the techniques you mentioned.

Good luck :)