# Need help on simple circuit problem.

#### Dedos

Joined Sep 2, 2010
3
Hello, I am new to AAC forums and I have a question on one of the problems form my textbook. I have tried various methods to obtain the resistance of the unknown variable but have no luck so far. There was also a hint that suggested to find the current in the 25Ω first, but how? Current divider?

Any tips on how to start this?

The picture of the circuit is below.

Things I have tried

KCL, KCV, Ohms Law

Here are the equations that I have (un)successfully tried to use, but there seems to be too many variables.

Assuming the first node (v1) is above the 50Ω and the second node (v2) is above the 25Ω.

Rich (BB code):
V=IR
Known
I(total) = 8A
R1=50Ω
R2=Ro
R3=25Ω
V(Ro)=240V

KCL
node1: I(source) - V1/50Ω - (V1-V2)/Ro = 0
node2: V2-V1/Ro - V2/25Ω = 0

KVL
Mesh 1 on left and Mesh 2 on right
Mesh 1 = 50Ω(8A-I2)
Mesh 2 = 25Ω(I2) + 50Ω(I2-I1) + 240V

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#### Ghar

Joined Mar 8, 2010
655
For KCL you need to add the equation V1 - V2 = 240V

For KVL you have a mistake, the first loop you're missing a term. There are two voltage drops, the one across the 8A source and the one across the resistor.
0 = 50(8 - I2) - Vsrc

I1 = 8A, by the way, which helps your second loop equation.

For Ohm's law there just isn't enough.

Personally I would solve this using a Thevenin equivalent (aka source transform) but you might not have studied it yet.

#### Dedos

Joined Sep 2, 2010
3
Oh man, I didn't even notice the (I1=8A) 2nd part of the KVL equation ..

Ah, then figuring out I2 and finding V/I = R gave me the right answer.
I think I was just looking too hard at the problem and didnt realize the small parts.

Thanks Ghar!

#### hgmjr

Joined Jan 28, 2005
9,029
Oh man, I didn't even notice the (I1=8A) 2nd part of the KVL equation ..

Ah, then figuring out I2 and finding V/I = R gave me the right answer.
I think I was just looking too hard at the problem and didnt realize the small parts.

Thanks Ghar!

hgmjr

#### Dedos

Joined Sep 2, 2010
3
Yeah it worked.

Solving for I2 in mesh 2 yielded the current in the right hand side mesh which then allowed me to use Ohm's Law to solve for the resistance in the 240V component.

#### Jony130

Joined Feb 17, 2009
5,024
We can also write this nodal equation
8A = V1/50Ω + (V1 - 240V)/25Ω
Or use Thevenin equivalent for current source

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