KVL/KCL with two sources - can a loop have both sources?

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SUMMARY

The discussion centers on the validity of using two voltage sources in a circuit analysis involving Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). Participants confirm that a loop can indeed contain multiple sources, and the potential changes must be summed accordingly. The original poster (OP) initially misapplied current values in their equations, leading to incorrect results. The correct approach involves accounting for all currents in the circuit branches and ensuring proper application of KVL and KCL principles.

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Homework Statement


In the circuit shown, find (a) the current in each resistors (b)The power dissipated in each resistor.

Homework Equations


V=IR, P=VI, KVL and KCL

The Attempt at a Solution


See attached.

I was going to post this in the main section as its really a question about whether you can have two voltage sources in your equation, but I read the sticky on the main page for general physics and it directed all coursework type questions here.

In the drawing I have a section underlined. When I set up my equations, I got two voltages in the equation. I wanted to know if that is valid? If I should come across another equation like this, is there a rule for knowing not to do this? The solution to this problem is not the same as what I derived. Please give me some direction with this, thanks!
 
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Sure, a given loop can contain zero or more sources. For KVL their potential changes sum just as you've done.
 
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chopnhack said:
In the drawing I have a section underlined. When I set up my equations, I got two voltages in the equation. I wanted to know if that is valid? If I should come across another equation like this, is there a rule for knowing not to do this? The solution to this problem is not the same as what I derived. Please give me some direction with this, thanks!
Your second equation is incorrect.
There should be i2 instead of i1.
 
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gneill said:
Sure, a given loop can contain zero or more sources. For KVL their potential changes sum just as you've done.
Thanks mate! BTW - if you haven't seen the finale... jumping the shark abit... Still a fan though :-)

cnh1995 said:
Your second equation is incorrect.
There should be i2 instead of i1.
Hi CNH, I think I see the error now. The 5I1 should really be just I1 times the 1 ohm resistor. That makes it all work.
Thanks gents!
 
cnh1995 said:
Does it?

I1 flows through the 3 ohm resistance too. Plus, I2 also flows through the 3 ohm resistance. You should have all the three currents in that equation.
OP has 3 individual currents, each one for each branch of the circuit. He then uses a junction equation to relate them.
 
Last edited:
It was almost 12 am when I posted it and I was kinda sleepy:oops:. I guess I have to sleep now before I post anything stupider.
SammyS said:
OP has 3 individual currents, each one for each branch of the circuit. He then uses a junction equation to relate them.
Yes. I wasn't able to delete that because of network problems (I am traveling in a hilly area).
Fixed it now.
 
Last edited:

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