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schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit, how do I find the Thevenin resistance and its corresponding voltage?

I know the technique of shorting voltage sources and making current sources open like so:

schematic

simulate this circuit

I don't where to go from there.

Is the 16 ohm not part of the resistance calculations anymore? If so, do I just add R3, R5, R2, and R4 in series?

Secondly, when finding its Thevenin voltage, I'm confused since I know that a voltage source and a resistance in series can be transformed into a current source that has a resistor in parallel, but in this case, the voltage source only has a resistor in parallel, not in series, so I don't know where to go from there.

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  • \$\begingroup\$ 1) Notice that R3 is shorted when calculating Thevenin resistance 2) For calculating Thevenin voltage it might be beneficial to use superposition method \$\endgroup\$
    – AlexVB
    Mar 26, 2021 at 4:37
  • \$\begingroup\$ @AlexVB so does that mean that the series is just equivalent to r2+r5+r4? For superposition method, do I check for node b and a to calculate its voltage? \$\endgroup\$ Mar 26, 2021 at 5:26
  • \$\begingroup\$ You are right about the resistance. Yes, you need to calculate Vab with r6 removed separately from each of two sources and then sum them. Check the Internet or your book for detailed explanation. \$\endgroup\$
    – AlexVB
    Mar 26, 2021 at 5:56
  • \$\begingroup\$ @OmarWalton Your R4 seems to have two different values in your two circuits. Just FYI. Aside from that, your I1 has infinite impedance so you can short R1, leaving I1:R2 as a Thevenin -40 V + 40 Ohm, which subtracts 40 V from the 520 V, leaving 480 V as your Thevenin voltage. The Thevenin resistance is then 40+60+4 = 104 Ohm. That's it. (R3 is bypassed by V1.) \$\endgroup\$
    – jonk
    Mar 26, 2021 at 9:19

2 Answers 2

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Is the 16 ohm not part of the resistance calculations anymore

You can disregard it, because the other end is open, it doesn't let any current flow through it.

do i just add r3 r5 r2 and r4 in series

Can you see that R3 is shorted? Current will hence flow through the short and R3 has no effect.

Finding Thevenin's voltage

Pretty straight forward to find \$ V_{AB} \$:

  • Open terminals A, B in the first circuit.
  • R5 and R4 go out of the picture.
  • That leaves you with R1, R2, R3 and two sources.
  • Assume all sources are ideal.
  • Take one source at a time and apply Superposition theorem.
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First step is to realize that R1 and R3 don't do anything. They could be replaced with 22 megohms and it wouldn't be detectable at A,B. Replace R1 with a short and open (remove) R3. Then do a Norton/Thevenin transformation on I1 R2. Then you'll have two voltage sources in series with a few resistors. Answer is obvious at that point.

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