# Finding a voltage across 2 nodes [Check answer - work already done]

#### naivy

1. The problem statement, all variables and given/known data 2. Relevant equations

3. The attempt at a solution
I got V1=18V
Here's my work: Related Engineering and Comp. Sci. Homework News on Phys.org

#### Zryn

Gold Member
Can you repost the original question and then rewrite your solution with some brief explanations to show what you are doing? i.e KVL (Right): V1 - Vx - 24 = 0

#### rude man

Homework Helper
Gold Member
I got V1 = 37.33V, Vx= 13.33V. It checks when you plug the numbers into your schematic.

If V1 = 18V then Vx = 18 - 24 = -6V so i thru the 25K = -6/25 = 0.24 mA
whereas i thru the 5K would then be
(-3*6 - 18)/5 = -7.2 mA so that can't be right.

#### Zryn

Gold Member
Vab = Va - Vb. Assuming Va is at V1+ and Vb is at V1-, what is V1?

Given V1, and using KVL around the right hand loop, what is Vx?

#### rude man

Homework Helper
Gold Member
There are 2 loops? If so, V1 = 0!!

#### Zryn

Gold Member
Given the line drawn between V1+ and V1-, and the node 'dots', it seems a safe assumption that there are two loops, rather than just one with an errant line.

I got V1 = 37.33V, Vx= 13.33V. It checks when you plug the numbers into your schematic.
I did it your way as well though and got the same numbers, but wanted the original question reposted just to verify.

#### rude man

Homework Helper
Gold Member
Two loops were my firat assumption too, but then I realized that that would make the problem pathological (viz. V1 = 0). Also, the OP calculated a finite (albeit erroneous) V1. So I decided that extra line must indicate voltage rise gnd to V1, although I did wonder why there was no arrow at the top of that line. Also, if two loops, why did the OP have - and + signs next to that line?

#### Zryn

Gold Member
Also, if two loops, why did the OP have - and + signs next to that line?
It could possibly have been designed as a (trick) knowledge problem, rather than a circuit analysis problem. Needs more OP confirmation I suppose.

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