# Resistor in parallel with a short circuit. Will current flow through it?

I'm trying to find the Isc of this circuit seen by the capacitor at t > 0 so I could find the Rth and get the circuit's time constant. If I make the capacitor a short circuit, it will be in parallel with the resistor. Is it right to assume that no current will flow through the 5ohm resistor?

Then I have the KCL equation:

V1 would be the voltage in the dependent source's node.

Isc = (Vsource - V1)/R1 + 0.5(Vsource - V1)

I already found V1 which is 2.25 V.

Am I doing this right?

• Your Vth = 2.25V voltage looks good. As for the Isc current it will be Isc = 3V/10Ω + 3V * 0.5S = 1.8A. Do you see the mistake you made? If you short the capacitor terminals the V1 voltage will now be equal to 0V.
– G36
Commented Sep 28, 2020 at 13:16
• May I ask why would it also be 0V?
– user263783
Commented Sep 28, 2020 at 14:51
• Due to the short circuit, there will be no voltage difference between points A and B. The points A and B are now at the same Voltage potential thus, 0V difference in voltages. i.sstatic.net/K80dr.png
– G36
Commented Sep 28, 2020 at 15:02