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

  • Thread starter chopnhack
  • Start date
  • #1
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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!
 

Answers and Replies

  • #2
gneill
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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.
 
  • #3
cnh1995
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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.
 
  • #4
52
3
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 :-)

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!
 
  • #5
SammyS
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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:
  • #6
cnh1995
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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.
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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