# help finding an unknown voltage source

#### luck

Joined Jan 25, 2008
8
attached is the problem, here is what I did and the answer I got.
Because the current supply produces 50 watts we know it is producing 10 volts. also we know that 5 amps across 4 ohms is a drain of 20 volts, thus we have -10 volts at the node between the two 4 ohm resisters. Then using KCL going into that node we have (0v- -10v)/2 ohms [from ground to node]+ (5v- -10v)/4 ohms [from 5 volt source to node) + 5 amps [from 5 amp source] = 13.75 amps. Now Vx=2 ohms * the current found =27.5 v

The correct answer is suppose to be 42.5 v So I ask you what have I done wrong?

thanks

#### Ron H

Joined Apr 14, 2005
7,014
I think your answer is correct. Before looking at either of your answers, I solved the problem, and I also got 42.5V. While double-checking all my voltages and currents, I discovered that I had labeled the voltage at the negative terminal of Vx as -20V, even though I had labeled the voltage on the current source as +10V. My error was that I had taken the drop across the 4 ohm resistor as 20V, but forgot to subtract the 10V on the other end. Seeing my error, I worked through the problem again, and this time got Vx=27.5V. I'm convinced this is correct.
See below.

EDIT: I found my errors. Vx=42.5 is the correct answer, as hgmjr solved below.
I have edited my schematic and reposted it.

#### hgmjr

Joined Jan 28, 2005
9,029
attached is the problem, here is what I did and the answer I got.
Because the current supply produces 50 watts we know it is producing 10 volts. also we know that 5 amps across 4 ohms is a drain of 20 volts, thus we have -10 volts at the node between the two 4 ohm resisters. Then using KCL going into that node we have (0v- -10v)/2 ohms [from ground to node]+ (5v- -10v)/4 ohms [from 5 volt source to node) + 5 amps [from 5 amp source] = 13.75 amps. Now Vx=2 ohms * the current found =27.5 v

The correct answer is suppose to be 42.5 v So I ask you what have I done wrong?

thanks
Greetings luck,

I decided to try my hand at the problem you gave to see what value I would get if I applied Millman's Theorem to the circuit.

I have attached my effort herein.

The value I obtained agreed with the +42.5 volt you stated in your original post.

Guys, please take a look over my effort and let me know if I took a wrong turn with any of my assumptions or equations.

hgmjr

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#### Ron H

Joined Apr 14, 2005
7,014
Greetings luck,

I decided to try my hand at the problem you gave to see what value I would get if I applied Millman's Theorem to the circuit.

I have attached my effort herein.

The value I obtained agreed with the +42.5 volt you stated in your original post.

Guys, please take a look over my effort and let me know if I took a wrong turn with any of my assumptions or equations.

hgmjr
You're correct. I made a couple of stupid errors in my original analysis. One was incorrectly summing the currents into the right-hand node, and this led me to completely ignore the current through the 5 volt source.  I have edited my schematic and re-posted it above.

#### hgmjr

Joined Jan 28, 2005
9,029
No problem. I had several false starts before I was able to unravel the network into something a bit less complicated.

It is easy to get tripped up on network problems that have a few extra components in them just to obfuscate the solution. Instructors really relish these problems since it helps them spot the students that are on their toes.

hgmjr