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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?

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    \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Commented Sep 28, 2020 at 13:16
  • \$\begingroup\$ May I ask why would it also be 0V? \$\endgroup\$
    – user263783
    Commented Sep 28, 2020 at 14:51
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    \$\begingroup\$ 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 \$\endgroup\$
    – G36
    Commented Sep 28, 2020 at 15:02

1 Answer 1

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If the short circuit is perfect, the two leads of the resistor are at the same Voltage potential. In that case, no current can possibly flow through the resistor, no matter how much current you run down that shorted node.

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