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The solution to the question is:

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Why is the thevenin's equivalent Zth = R and not R-j as they are both in series?

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2 Answers 2

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Why is the thevenin's equivalent Zth = R and not R-j as they are both in series?

They are not in series when you look at the circuit from Thevenin perspective.

Replace the sources with their equivalents (voltage sources are shorted because they have zero series resistance, current source are left open because they have infinite parallel resistance) and find the impedance seen.

As you can see from the solution, the 3V source is replaced with a short circuit so it shorted the capacitor, yielding the net impedance to be only R.

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  • \$\begingroup\$ Thanks a lot. I have one more doubt regarding the calculation of Power. Why is there the additional 3i1 term? Where did that come from? \$\endgroup\$
    – Sam1470
    Jun 23 at 10:13
  • \$\begingroup\$ @DevyanshiSinghShekhawat it's coming from the 3V voltage source. The circuit B (series combination of the voltage source and the resistor R) is seen as a load from circuit A. So the transferred power to the load will be the sum of the dissipation of the resistor and the voltage source. The resistor's dissipation is resistance times current-squared, and the voltage source's dissipation (note that current flows into the source, not out of the source) is voltage times current i.e. 3 x i1. \$\endgroup\$ Jun 23 at 10:43
  • \$\begingroup\$ Thanks again. This really helped me. \$\endgroup\$
    – Sam1470
    Jun 24 at 5:28
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Why is the thevenin's equivalent Zth = R and not R-j as they are both in series?

The red-herring is the capacitor; it is totally in parallel with the right-hand voltage source and therefore, it plays no role in any analysis for Thevenin resistance.

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  • \$\begingroup\$ ohhk, thank you. \$\endgroup\$
    – Sam1470
    Jun 24 at 5:28

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