# Need help solving a circuit question

#### Slowburn

Joined Jan 26, 2021
4
Good day

I hope everyone is doing well, I'm struggling with a question I received in my text book. It is probably very simple and the answer is staring me in the face but somewhere I'm missing an important step. If someone can please help me by explaining the steps and how you got to your answer. I'm still very new in the world of electronics. The question ask that i must find the value of R. See diagram below.

#### ericgibbs

Joined Jan 29, 2010
12,908
Hi Slowburn,
Welcome to AAC.
E

#### Slowburn

Joined Jan 26, 2021
4
Hi Ericgibbs

I honestly do not know where to start. I've tried dividing the circuit into two loops and applied KCL but still I'm not getting anywhere. I'm not looking for a answer I just want to understand how to solve these types of circuits

#### sagor

Joined Mar 10, 2019
439
Think a bit. Since the 2A is on only one wire leading to R, the other 7 ohm resistor has nothing to do with that current...

#### RBR1317

Joined Nov 13, 2010
631
If you were given this circuit, could you find V1?

#### sagor

Joined Mar 10, 2019
439
I don't see how finding V1 in that diagram helps, as in the original circuit you have 2 resistors in parallel and V1 cannot be determined without knowing the parallel resistance.
As drawn, V1 is the zero volts (70V return) since it connects directly to the 70V return

#### RBR1317

Joined Nov 13, 2010
631
and V1 cannot be determined without knowing the parallel resistance.
Why can't you just write the node equation for V1? (And the circle is a 2-amp current source.)

#### ericgibbs

Joined Jan 29, 2010
12,908
hi Slowburn,
E

#### Slowburn

Joined Jan 26, 2021
4
hi Slowburn,
E
Sure Thing

R= V/I^2
R= 70/2^2
R= 17.5 ohms

#### MrAl

Joined Jun 17, 2014
8,245
Hello,

When you have to find a value for a component like a resistor you just use "reverse osmosis" <chuckle>. You do a regular analysis keeping the unknown part as a variable like simply "R" or something, maybe "R1" or "R9", etc.
Once you write the equations, you then solve for that value of R rather than any voltage or current. It is really the same thing although you have to use algebra.