# do you think I can solve for v, I1, and I2?

#### PG1995

Joined Apr 15, 2011
832
Hi

Please have a look on the attachment. Do you think I can solve for v, I1 and I2? I know there are simpler ways to do but I'm interested doing it this way. So, please help me. Thank you.

Regards
PG

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#### joeyd999

Joined Jun 6, 2011
5,355
There is only 1 loop. Not 2.

#### DerStrom8

Joined Feb 20, 2011
2,390
There is only 1 loop. Not 2.
Joey is correct. Unless 'V' is another source, which I don't believe it is. I think it's just a test point. I am not sure if you can use KVL to find V in this circuit. Well, at least not easily. Why don't you just use the rules for a series circuit?

#### steveb

Joined Jul 3, 2008
2,436
Hi

Please have a look on the attachment. Do you think I can solve for v, I1 and I2? I know there are simpler ways to do but I'm interested doing it this way. So, please help me. Thank you.

Regards
PG
I agree with the above 2 posts, but your question can still be given a direct answer.

There are two issues with your (more complicated and confusing - no offense intended) approach.

First, you made a mistake with the polarity of V. You defined two opposite polarities in the loops, but then you tried to say the voltages are equal. The way you wrote it, V for the left loop equals negative V for the right loop.

The second mistake is that you need to assign a resistance value to the open connection across V. This resistance will be infinity, but you need it in order to get another relation that allows simultaneaous equations to be set up. You need to have a KCL equation at the node, which then allows you to determine that I1=I2 because no current flows through an infinite resistance.

I guess another way to say it is that your approach requires you to define what V is, if you want to solve the system that way. You can't just say V is anything and then use it to equate the two sides. The answer will change depending on what V is. Is it a source, or a resistance or something else? Clearly in this case, it is an infinite resistance, so you have to use that fact. The result of that fact is that I1=I2, and then you can use your equation (once corrected) to solve for I1 and I2 (which are equal to each other), and then the value of V is easy to figure out (note that you can't get V from ohms law because zero times infinity is undefined, but you could solve using the variable R for resistance and then take a limit as R goes to infinity).

EDIT: Doing it out, I get the following equations.

v=10-2*I1
v=6*I2-5
v=(I1-I2)/R

Solving for v, I1 and I2, I get the following.

v=25*R/(4*R+6)
I1=(15*R+60)/(8*R+12)
I2=(15*R+10)/(8*R+12)

Now, taking the limit as R goes to infinity, I get the following.

v=25/4 V
I1=15/8 A
I2=15/8 A

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