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This is the circuit I'm given:

enter image description here

And after disabling all of the sources, I'm now trying to calculate the equivalent impedance (which would be the Thevenin impedance):

enter image description here

But I'm stumped on how to combine the capacitor and the 2 kohm resistor. Seems like a dumb thing to ask but are they in series or parallel? I've debated myself on both counts. There's a trivial loop there so they are in parallel. But there's a trivial node connecting these two so they are in series? Am i just being dumb here and there's an obvious answer?

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I think capacitor and resistor are in series but then they combined are in parallel with a piece of wire which has least resistance(impedance in AC) so i think total impedance would be 2k

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  • \$\begingroup\$ So are all elements in series with each other then? The capacitor and both the resistors? \$\endgroup\$ – samz_manu Mar 20 '16 at 3:26
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    \$\begingroup\$ The cap and resistor are shorted by the wire on the bottom left corner. Therefore they do not contribute to the Thevenin impedance. The only value left is the left most 2kOhm resistor \$\endgroup\$ – Josh Jobin Mar 20 '16 at 3:48
  • \$\begingroup\$ @JoshJobin I don't understand how they wouldn't contribute to the overall impedance. The wire provides the least resistance so all the current would flow through the short circuit rather than go to the cap and resistor combination. Is that right? \$\endgroup\$ – samz_manu Mar 20 '16 at 4:12
  • \$\begingroup\$ I think Josh Jobin explained well that i did \$\endgroup\$ – Rahul Mar 20 '16 at 6:03
  • \$\begingroup\$ @samz_manu yes that is right. When anything is in parallel with a short circuit, you can essentially ignore it. \$\endgroup\$ – Josh Jobin Mar 20 '16 at 10:52

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