I'm on this problem:

enter image description here

And I'm trying to use KCL to find Vth, my approach is:

enter image description here

Am I on the right path regarding KCL?

Thanks in advance!


Hint 0: No your formula is not correct. The capacitor is not connected to the Vth node, so it won't be involved in the KCL equation for that node.

Hint 1: Anything in parallel with a voltage source has no effect on the rest of the circuit.

Hint 2: The Thevenin voltage can be calculated from the voltage divider formula.

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  • \$\begingroup\$ Hello, could I ask a different question? If we would consider the resistor to be between V0 and Zc, would Zc still have no effect on the rest of the circuit? \$\endgroup\$ – Vetenskap May 12 at 19:15
  • \$\begingroup\$ @Vetenskap, would the capacitor still be in parallel with the source? \$\endgroup\$ – The Photon May 12 at 19:20
  • \$\begingroup\$ No, so it would. Okay:) \$\endgroup\$ – Vetenskap May 12 at 19:21
  • \$\begingroup\$ Could I ask a question about Rth or do I have to post another question again? \$\endgroup\$ – Vetenskap May 12 at 19:25
  • \$\begingroup\$ @Vetenskap, you should search and see if anybody's asked the same question before. Then post a new question if not. \$\endgroup\$ – The Photon May 12 at 19:26

You're almost there. I would reconsider the equation itself and more specifically the Vth/Zc part. It is also helpful to add +/- markers on the components to make sure you write the equation in a consistent manner. You are also using KCL, which KVL is much more applicable, and not even needed in this problem.

Using a voltage divider, you notice the Vth is simply (Zl V0)/(R+Zl) because V0 is in parallel to the entire branch.

Using KVL, you can write RI1 + ZlI1 - ZcI2 = 0, then Vth = ZlI1 where I1 and I2 are the currents in each branch.

KCL will give you the same result with more work.

This picture should help as well: notes

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  • \$\begingroup\$ hello Nate, I'm trying to calculate Vth in rectangular form, and I got it to be 5sqrt(3)/2 + 5j/2 V. However your approach made me a confused. Would you mind showing med how you would go about it with KCL? \$\endgroup\$ – Vetenskap May 12 at 18:47
  • \$\begingroup\$ With KCL the only node to look at is the V0, R, Zc node where you can find the currents I1 and I2. You have to be careful because the V0 creates a supernode (because no resistive elements are in the branch) and you would use the Vth=ZL * I1 equation again. This link explains more of it: electricaltechnology.org/2015/02/… \$\endgroup\$ – nate May 12 at 19:00

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