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I have the circuit below and I have to calculate complex current i1using thevenin equivalent circuit method:

I tried to follow the rules and open the terminal AB across the resistor: enter image description here

I want to ask if this is a correct way to convert the circuit to thevenin. Also I am not sure is capacitor's reactance -4iis same as -4j or 4, 90 degree ?

Then The voltage across AB terminal would be equal to voltage across the capacitor? Is it ok to in this case see the voltage source as a short?

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  • \$\begingroup\$ In addition to thinking about Thevenin equivalent circuits, you might also want to think about how the superposition principle applies to this problem. \$\endgroup\$
    – The Photon
    Jul 2, 2012 at 23:17
  • \$\begingroup\$ Yeah but this is an exam example and in exam it should be solved by the way teachers wants it :P \$\endgroup\$
    – Sean87
    Jul 2, 2012 at 23:20
  • \$\begingroup\$ Actually, superposition is very useful in finding Thevenin equivalents. \$\endgroup\$
    – Alfred Centauri
    Jul 3, 2012 at 1:38
  • \$\begingroup\$ In math i is commonly used to work with complex (imaginary) numbers. In electronics i has it's own very common meaning as current. That is why most electronics friendly/aware textbooks use j instead of i. They are exactly equivalent, just pick one and use that throughout. You're in electronics, so pick j. A capacitor is represented by \$\frac{1}{j\omega C}\$ which is equal to \$\frac{-j}{\omega C}\$ \$\endgroup\$
    – jippie
    Jul 3, 2012 at 7:07
  • \$\begingroup\$ What I meant to write with the -j is: -90 degrees relates to -j \$\endgroup\$
    – jippie
    Jul 3, 2012 at 7:24

1 Answer 1

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The Thevenin voltage looking into the terminals A & B can be found by inspection using superposition.

\$V_{th} = (-j2A) \cdot (-j4\Omega) + j4V \$

The first term is found by zeroing the voltage source on the right and the second term is found by zeroing the current source on the left.

The Thevenin impedance is easy too; simply zero both sources to get:

\$ Z_{th} = -j4 \Omega\$

The current is then found to be:

\$I_{resistor} = \dfrac{V_{th}}{Z_{th} + 8 \Omega}\$

I highly recommend reviewing the late Dr. Leach's notes on using superposition. I've taught it this way for years. If you practice using superposition, you will amaze your friends and professors alike by solving many circuits by inspection.

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  • \$\begingroup\$ Thanks, Shouldn't we OPEN the current source instead of shorting it for finding \$Z_{th}\$ ? \$\endgroup\$
    – Sean87
    Jul 3, 2012 at 8:19
  • \$\begingroup\$ Also can you tell me please if the arrow in the current source was upside down we would Substract it with voltage source instead of summing? \$\endgroup\$
    – Sean87
    Jul 3, 2012 at 8:28
  • \$\begingroup\$ We zeroed each and then both sources. A current source with zero current is open. A voltage source with zero voltage is shorted. And yes, the direction of the arrow matters. Reversing the arrow changes the sign of the voltage across the capacitor due to the current source. \$\endgroup\$
    – Alfred Centauri
    Jul 3, 2012 at 11:00
  • \$\begingroup\$ You're welcome! I just took another look at your schematic and I notice you have I1 and i1 where i1 is the current through the resistor. However, when I do AC problems, I always use a capital variable and lower case subscript so I'm going to edit my answer so that it's clear which current I'm solving for. \$\endgroup\$
    – Alfred Centauri
    Jul 3, 2012 at 11:36

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