Need help with a basic thevenin circuit

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
I thought I were supposed to find the voltage drop across the 4 ohm resistor. But if my calculations are correct, that should give 0.8 amps, the correct answer is 1,2 amp, which is (assuming my calculations are correct) the voltage drop across the 6 ohm resistor. What did I do wrong?

IMG_0014.jpg
 

WBahn

Joined Mar 31, 2012
26,145
Do you want the voltage (which has units of volts) or the current (which has units of amps)?

Assuming you mean 0.8 V (and not 0.8 A), then that is correct. Where did you get the supposedly correct answer of 1.2 V (again, I'm assuming that it's supposed to be V and not A)?

Note that the 6V source and the left 6 Ω resistor has no influence on anything to the right of the 2 V source -- the 2 V source isolates the left side from the ride side by clamping the voltage at the top-center node. The means that the right side is nothing more than a simple voltage divider and the voltage across the 4 Ω resistor has to be less than the voltage across the 6 Ω resistor (since they have the same current flowing in them) and since the sum of the two has to be 2 V, the voltage drop across the 4 Ω resistor has to be less than half of 2 V. So an answer of 1.2 V would make no sense.
 

Ylli

Joined Nov 13, 2015
978
Keep it simple. If you have 2 volts from the junction of the 6 ohm resistors to 'ground', (the lower wire), then you have 2 volts at that point. Looking to the right that 2 volts sees 6 + 4 = 10 ohms. The current in that loop will be I = E/R = 2/10 amps.

The 6 volts on the left trying to drive the 2 volt source will see a voltage difference of 4 volts across the left 6 ohm resistor for a current in that resistor of I = E/R = 4/6 = 2/3 amp
 

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
Do you want the voltage (which has units of volts) or the current (which has units of amps)?

Assuming you mean 0.8 V (and not 0.8 A), then that is correct. Where did you get the supposedly correct answer of 1.2 V (again, I'm assuming that it's supposed to be V and not A)?

Note that the 6V source and the left 6 Ω resistor has no influence on anything to the right of the 2 V source -- the 2 V source isolates the left side from the ride side by clamping the voltage at the top-center node. The means that the right side is nothing more than a simple voltage divider and the voltage across the 4 Ω resistor has to be less than the voltage across the 6 Ω resistor (since they have the same current flowing in them) and since the sum of the two has to be 2 V, the voltage drop across the 4 Ω resistor has to be less than half of 2 V. So an answer of 1.2 V would make no sense.
It was a typo. I meant to say volts not amps. The solution says 1.2V by voltage division: 2V * 6/(4+6) =1,2V. It doesn't make sense to me either.
 

WBahn

Joined Mar 31, 2012
26,145
It was a typo. I meant to say volts not amps. The solution says 1.2V by voltage division: 2V * 6/(4+6) =1,2V. It doesn't make sense to me either.
Whoever did the solution made several mistakes. The didn't track their units properly (sadly that's way too common), they misapplied a common formula (we all make mistakes) and they didn't bother to ask if their answer made sense (which is a pretty major one, in my view).

As I said, we all make mistakes, so it's not fair to judge the entire body of work based on one botched solution, but it would give me cause to suspect (and expect) a certain level of sloppiness in the future from that source's material.
 

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
Whoever did the solution made several mistakes. The didn't track their units properly (sadly that's way too common), they misapplied a common formula (we all make mistakes) and they didn't bother to ask if their answer made sense (which is a pretty major one, in my view).

As I said, we all make mistakes, so it's not fair to judge the entire body of work based on one botched solution, but it would give me cause to suspect (and expect) a certain level of sloppiness in the future from that source's material.
That would be my university professor, the university is closed so I cant ask him about the solution. But do you agree that the thevenin equivalent circuit would consist of a 0.8 voltage source and a 2.4 ohms resistor, connected in series?
 

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
Is it possible to do the mesh current method without using source transformation, when there is a current source in the circuit? I know for node voltage, when we see a current source, we add the current based on direction. But for mesh current, we add the voltage based on direction, what do we do with the current source?

IMG_0017.jpg
 

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
I drew the circuit wrong. The terminals should be right before the RL resistor not after.

I tried to solve it using the node voltage method, I see two nodes we need to consider. The node connecting the 6,12, and 3 ohms resistors, and the node connecting the 3 and 2 ohms resistors and the 2A current source, but the way my professor solved it was with only the node connecting the 6,3 and 12 ohms resistors. I'm confused.

This is what I got:

IMG_0018.jpg
 
Last edited:

Thread Starter

Aria Nemati

Joined Jul 19, 2019
27
I think I see it. The last 2 ohm resistor is not connected to the circuit from both ends. Is this correct?

EDIT: I see I drew my reference node wrong. Nevermind that

It still doesnt make sense. Why did he skip the 3 ohm resistor then?

Skjermbilde 2019-07-21 kl. 21.09.26.png
 

RBR1317

Joined Nov 13, 2010
563
It still doesnt make sense. Why did he skip the 3 ohm resistor then?
That looks like the node equation for the node Vx. The 3Ω resistor is not relevant because the current through that branch is not affected by the actual value of that resistance - it depends only on the current source. Not exactly sure what the " |•12Ω " is supposed to mean.
 
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