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My objective is to find v (voltage drop across 8 ohm) in the below circuit. I was able to thevenize the circuit successfully (V thevenin = 12 V and R thevenin = -2 ohm) and find v. But I find it difficult to apply Norton's Theorem in the below circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The Norton equivalent resistance is easy to find (-2 ohm). To find the short circuit current, I shorted the 8 ohm resistor. This will also short the 4A current source and will hence become redundant (No interaction with left side of circuit). This reduces to the following circuit:

schematic

simulate this circuit

From the above circuit I wrote the following:

\$i_{N}=-(10-i)=i-10\$

\$\Longrightarrow i=10+i_{N}\$

Applying KVL to the mesh:

\$-2i+4i=0\$

\$i=0\$

\$\Longrightarrow i_{N}=-10 A\$

This is not the same as:

\$\dfrac{V_{th}}{R_{th}}=-\dfrac{6}{7}A\$

What have I done wrong? Have I overlooked something?

Edit:

I will show here how I calculated R thevenin. I first detached the 8 ohm resistor from the circuit and nullified all independent sources. Since the circuit has dependent sources, I attach a 1 V voltage source across the terminals AB.

schematic

simulate this circuit

\$R_{Th}\$ then becomes \$\dfrac{1}{i_{0}}\$. The idea is to find \$i_{0}\$ which is the same as \$i\$.

You can read more here: https://www.allaboutcircuits.com/technical-articles/thevenin-theorem-dependent-source-circuits/

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    \$\begingroup\$ Well done with the equation formatting. This site supports MathJAX directly so you should have no trouble using it for future posts. Use $$ tags for whole-line equations and \$ tags for inline. \$\endgroup\$ – Transistor Feb 17 at 9:14
  • \$\begingroup\$ @Transistor Oh! I wasn't aware about that. Thanks for telling! \$\endgroup\$ – Aditya Feb 17 at 9:22
  • \$\begingroup\$ Can I ask how did you get Rth = -14Ω. Because I'm getting different value Rth = -12V/6A = - 2Ω. Also, you cannot remove the shorted current source from the circuit is you are looking for Isc current. \$\endgroup\$ – G36 Feb 17 at 10:15
  • \$\begingroup\$ @G36 I have edited the question to address your issue. Yes, I realize that I cant remove the shorted current source. \$\endgroup\$ – Aditya Feb 17 at 10:34
  • \$\begingroup\$ @ADITYAD.S. Your method of finding Rth do not work in this case. And will give you the wrong result. \$\endgroup\$ – G36 Feb 17 at 10:41
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Guidances only:

1) Shorting the 8 Ohm resistor doesn't make the 4 amperes source redundant. 4 A is a part of the short circuit current. The rest comes from the controlled source.

2) Norton's theorem isn't a method how the equivalent should be calculated, Norton's theorem states that the equivalent current source with parallel resistor exists if the DC circuit is linear. It's possible that your teacher has declared "If I one day happen to order you to apply Norton's theorem to solve a certain voltage or current, then you should calculate the open circuit voltage and the short circuit current!". I do not believe such declaration really exists, but that's a belief.

3) when the circuit has controlled sources, you have a danger to use for ex. same i in different situations where the circuit is different due the needed opens and shorts to derive the equivalents. Your i is totally different when the circuit is the original, 8 Ohm is taken away or when 8 Ohm is replaced with a wire.

I recommend you to stop with equivalent circuits to keep every controlled current and voltage in unique state. You get 2 equations with 2 unknowns i and v easily. Simply write i=10A+4i and the other equation for the currents at the right end of your 4i source. With Norton's or Thevenin's equivalent sources - when applied right - you often generate only new unknowns to a circuit which has controlled sources and you will not advance at all.

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  • \$\begingroup\$ I find your answer and this answer to be contradictory. This answer says it is okay to get rid of the shorted current source. electronics.stackexchange.com/a/330842/213056 \$\endgroup\$ – Aditya Feb 17 at 9:46
  • \$\begingroup\$ @ADITYAD.S. no disagreement here. As shorted 4A source doesn't cause a voltage which excites the rest of the circuit. But 4A is still a part of the short circuit current when you derive the Norton equivalent for the circuit which feeds the 8 Ohm resistor. \$\endgroup\$ – user287001 Feb 17 at 10:00
  • \$\begingroup\$ So the 4A current is cycling only through 8 ohm and not the rest of the circuit right? So even if i short the 4A source, there is still a 4A cycling through the short circuit. Is this what you are trying to tell? \$\endgroup\$ – Aditya Feb 17 at 10:07
  • \$\begingroup\$ NO! if there's a wire put to replace the 8 ohm resistor, just in that case the current in that wire is 4i+4A assuming your 4i means current. \$\endgroup\$ – user287001 Feb 17 at 10:14
  • \$\begingroup\$ Yes i meant that the 4A was a part of the short circuit current. \$\endgroup\$ – Aditya Feb 17 at 10:21

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