# Solving Circuit Diagram

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1. Sep 13, 2017

### TheRealBillNye

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

A screenshot with everything given
2. Relevant equations
No idea

3. The attempt at a solution
No idea

I'm not sure how to solve this or where to look for a solution. I was given this yesterday and have a test tomorrow. The professor has not gone over this at anytime, it's only the second week. Anything helpful would be great, even just what I need to google so I can attempt to solve this.

P.S. I do not know the solution.

2. Sep 13, 2017

### scottdave

The sum of voltages around any given loop equals zero. So if you go from - to + in an element, you add voltage. If you go from + to - in an element, you subtract the voltage. The diamond shaped source is a dependent voltage (it is 5 multiplied by the value of I0).

You might want to take a look at the Navy's NEETS modules, for electricity and electronics. The US Navy has published these and made them available to the general public. Davidson College hosts all 24 PDF's here: http://www.phy.davidson.edu/instrumentation/NEETS.htm

3. Sep 14, 2017

### cnh1995

@TheRealBillNye, your own attempt at a solution is required as per forum rules.
This much information is sufficient for solving this problem.

4. Sep 14, 2017

### TheRealBillNye

Thanks for the help, would someone mind checking this for me as I don't know the answer.

5. Sep 14, 2017

### cnh1995

That's not correct. Which current is I1?

How about you just consider the lefmost loop containing the 30V source? Apply KVL in that loop only.

6. Sep 14, 2017

### scottdave

What is this 18 volts across I1 ? Try a loop around the lower left-hand square of circuitry.

7. Sep 20, 2017

If you draw the schematic in this way you may calculate ia and ib as function of Vo and then Vo=18 V.
[5Io=5*2=10V]

8. Sep 20, 2017

### CWatters

It's easier than that. No need to calculate ib at all. Just apply KVL to the left hand loop.

9. Sep 28, 2017

### Ronald Channing

Hi.

It is a very easy question..One needs to apply KVL around the loop with the 30 V source and 12 V component.

KVL simply states that the sum of voltages around a closed loop is equal to zero. Equivalently, sum of voltage drops is equal to sum of voltage rises around the loop.

As a side note, please know that any device in circuit analysis is a model. Two terminal devices are models/abstractions, for which the:

1. Current entering any terminal is defined.
2. Voltage function between the positive and negative terminal can be defined.

We arrive at these models through the lumped circuit abstraction, please google it if you would like to know more. Anant Agarwals lectures on MIT Opencourseware nicely cover it. This is if you want a good introduction to circuit analysis and basic electronics:

So, you have the voltage functions for two devices, you know KVL, you simply must apply this law around this loop and find $V_{0}$

We can use a simple equation here, but usually when going around in a "loop" I learned to check the polarity of the terminal through which the curent I am tracking is entering, and assign this same coefficient to the voltage function, keeping this convention consistent is important.

So if we start from the battery:

$- 30 V + 12 V + V_{0} = 0$
$-18 V = -V_{0}$
$V_{0} = 18 V$