I need help relearning circuitry again (Physics/EE)

[Ricemarine]

After looking at my EE homework, I'm still in the dark as to how to do more complex multiloop circuits correctly.

Such as this example....
My attempt:
- Use KCL and KVL to try and get equations
- Each resistor is assigned a current in an arbitrary direction
- Four nodes, A,B,C,D from left to right, top to bottom. Same with resistors.

For KCL
+ A = R1 + R2 = 0
+ B = R3 + R4 - R2 = 0
+ C = R5 - R3 - R1 = 0
+ D = R6 - R4 = 0.

For KVL I'm at awe as to how to implement it properly in this. I know that if I can somehow solve for all the equations and use Ohm's law, I would be able to solve for the variating R.

I have been trying to fully understand how to do circuits properly, in my physics book, EE book, online, but it's really not clicking in my head. I'm asking ATOT to try and see if there is a way for it to click in my mind (I'm going to visit the TA's/professor tomorrow for help too). If someone is willing to give a tutorial/step by step process in more layman's terms with an example of a more complex circuit, that would be very, very, helpful.

Thanks.

 

[us3rnotfound]

Are you familiar with Mesh Analysis? (KVL)

If so:
You would setup 3 current loops. One loop around the inside of each "Rectangle." By definition of KVL each loop should have a total voltage of 0 when summing up the voltages of each circuit component in a loop. V=IR. So you go around the loop and add up the voltages and set them equal to 0. When you encounter a circuit element that has 2 Loops defined, you the add currents together and multiply by the resistance. If the Loops oppose each other, you basically subtract the currents, making the loop you are in positive.

You know that Va needs to be 60V, so you can determine the Current needed to be flowing through the 12 ohm resistor to be 5Adc.

So I'm using I1, I2, I3 for the 3 loops, and I set it up in a clockwise fashion around the circuit (Loop 1 in first rectangle with source, Loop 2 in top rectangle, and Loop 3 in bottom right rectangle where Va is), and each current Loop will be run in clockwise. Makes it easier.

Loop 1:
-240 + R(I1 - I2) + 180(I1-I3) = 0.

Loop 2:
45I2 + 10(I2-I3) + R(I2-I1) = 0.

Loop 3:
180(I3-I1) + 10(I3-I2) + 12(I3) + 18I3 = 0.

Constraint: I3 = 5A.

There you go. 3 Equations, 3 Unknowns. Simple Algebra should lead you to the right answer.

I tend to avoid Node analysis because I think Summing voltages around a loop is more convenient and easier to picture.

Please post a challenging problem, I can make a step by step tutorial of that as well and maybe more things will click. This all gets easier ONLY by repetition.
Edited for signs.

 

[esun]

So there's two ways to approach a problem like this. The first way is the naive way that always works but can be a pain. This method is basically: write down all equations associated with the circuit and solve the system. If you don't know where to start, most likely this is the method you're going to use.

The second method involves looking at what you have and solving small pieces of the circuit until you have what you need. This is generally much cleaner than a system of N equations with N unknowns. It looks like you already know how to just brute force the problem, so I've written a solution that is geared toward a more step-by-step approach to the answer that doesn't require solving any complex system.

http://eudean.com/misc/tmp.pdf

 

[TecHNooB]

Please post any circuit questions you have or PM me. I love answering these!

Okay, you can solve it two ways. Node or mesh analysis. Of course, you don't want to be doing anything legit! You wanna cheat your way to some easy variables 🙂

Let's establish this first:
Bottom left square = i_1 clockwise
Top square = i_2 clockwise
Bottom right square = i_3 clockwise

So the question wants you to adjust R so that V_a = 60V. Let's set V_a to 60V to see what kind of current we get for i_3. Turns out, i_3 is just 5 A(60 / 12).

We could solve right now with more equations or we can grab another free variable. Let's grab another free variable 🙂 How about a loop around the entire thing?
-240 + 45 * i_2 + (18 + 12) * 5 = 0 ~> i_2 = 2A

Now try going around the i_1 loop.
-240 + R * (i_1 - i_2) + 180 * (i_1 - i_3) = 0
* We have i_2 and i_3 but we are missing R and i_1.

Let's get i_1 by going around the i_3 loop.
i_3 * (12 + 18) + 180 * (i_3 - i_1) + 10 * (i_3 - i_2) = 0 ~> i_1 = 6A

Now you can go back to the previous equation and solve for R, which turns out to be 15 ohms.
---------------------------------------------------------

Another way to do this is by calling the center node (above the 180 ohm) V. I'll make this one quick.

*Summing currents going out
Center node equation:
(V - 240) / R + V / 180 + (V - (30*5)) / 10 = 0

Right node equation:
5 + (150 - V) / 10 + (150 - 240) / 45 = 0 ~> V = 180V

With V, you can solve for the center node equation and find R.