# Circuit Diagrams w/o Resistance

1. May 19, 2010

### moshrom

1. The problem statement, all variables and given/known data
Solve for the unknowns.
[PLAIN]http://img535.imageshack.us/img535/7494/24800354.png [Broken]
[PLAIN]http://img517.imageshack.us/img517/5765/33984576.png [Broken]
[PLAIN]http://img707.imageshack.us/img707/2511/75290122.png [Broken]
[PLAIN]http://img96.imageshack.us/img96/2593/40824092.png [Broken]

2. Relevant equations
No actual equations are necessary, our teacher just told us that:
1. In a pathway, Voltage in = Voltage out
2. The current that enters a junction must leave that junction.

3. The attempt at a solution
We've been assigned pages and pages of these, and these 4 are the ones I had trouble with. In the first diagram, v2= 90-70=20v. I'm stuck from there.

I normally look for online tutorials myself, but I couldn't find any of these ones- without resistance.
Thanks, all help is appreciated.

Last edited by a moderator: May 4, 2017
2. May 19, 2010

### PhaseShifter

Consider the junctions that are connected by R2.
The potential difference between these two junctions must be the same, regardless of whether you follow the branch with R2, the branch containing the battery, or the more complex branch on the right.

Given V2, V1, V5, and V3, what does that tell you about V4 (and V7)?

3. May 19, 2010

### moshrom

Er. I'm not sure. I'm sorry, but I'm not good at learning by having people tell me what to do. Could you tell me the unknown values in only the first diagram? Then I can see the values and the pattern. That's what I normally do.

4. May 19, 2010

### collinsmark

Hello Moshrom,

Don't forget about
In your first circuit, consider the top node. You know that there is 50 A going through the battery into the top node. You also know that there is 10 A going through R2. So how much current must be left over for R1?

5. May 19, 2010

### moshrom

40A. Which means I5=40A, I4=10A, I3=40A, and I6=50A.

I think I knew that before. I'm mainly having trouble grasping the voltages.

6. May 19, 2010

### collinsmark

There's something you mentioned before that makes me sort of leery. You said,

That's not the way I would phrase it. Voltages don't "flow" through a pathway. Voltage (aka electric potential) simply exists across different places in the circuit.

Saying that voltage goes in or out (or "flows") doesn't make much sense. Imagine a hill of height h. Asking a question such as "how much potential energy per unit mass flows down the hill?" doesn't make any sense. The potential energy per unit mass (from the top to the bottom of the hill) is simply gh. It doesn't flow anywhere. It just is.

[btw, voltage can sometimes be thought of as potential energy per unit charge.]

Across each resistor there is a "voltage drop." There is 0 voltage drop across wires. If two resistors are in series, you can add the voltage drops together to find the voltage difference from the top of the top resistor to the bottom of the bottom resistor.

If two resistors are in parallel (sharing common nodes at their ends) the voltage drops must be equal!

If you add up all the voltage drops in a loop, such that the loop starts and ends in the same place, all the voltages in the loop must add up to zero (you need to consider + and - signs when you go through the loop).

Imagine a house that has an attic, a main floor and a basement. In the basement there is a ladder that you can use to climb up to get outside through a window. From the outside, there is a fire escape that you can use to climb up to an attic window.

Suppose you start in the attic and go down the stairs to the main floor, then go down another set of stairs to the basement, then up the basement ladder to outside, then up the fire escape back into the attic. You end up with the same potential energy you had when you started. There are many things that can be said about this situation, but a couple of them are:

Potential drop of the attic stairs + potential drop of the basement stairs = potential drop of basement ladder + potential drop of fire escape

which is also equal to the potential difference between the attic and the basement.

In terms of adding everything up in a loop, you could say:
- (Potential drop of the attic stairs)
- (potential drop of the basement stairs)
+ (potential drop of basement ladder)
+ (potential drop of fire escape)
= 0

7. May 19, 2010

### moshrom

Yes, I understood all that except how voltage doesn't flow. Thank you. But I'd still like to know the unknown voltage values- I understand the concept, I just can't apply it to the diagram. Odd, because I normally find these things pretty easy.

V2=20v, V4=2.5v, V7=2.5v. Right? Then what?

Last edited: May 19, 2010
8. May 19, 2010

### PhaseShifter

Look at two different paths connecting the same junctions.

You can go across R2, or you can take the long way around through R1, R5, R4 (or R7), and R3.

The change in potential is the same either way.

9. May 19, 2010

### moshrom

Yes, I understand that. Please, I'd like to see the actual values.

10. May 19, 2010

### PhaseShifter

V2=V1+V5+V4+V3
All of these have known values except for V4. Calculate it from the others.

V7=V4, since they are in parallel.

11. May 19, 2010

### moshrom

Just to make sure, V2=20v and what are V4 and V7?

12. May 20, 2010

### PhaseShifter

As for the currents, take a given branch of the circuit and break it down to elements connected in series. Take I4 and I7 together, since they are in parallel.
I3=I5=I1=(I4+I7)
This branch of the circuit is in parallel with I2, so I2+I3=I0=I6

13. May 20, 2010

### moshrom

YES, if you read the thread you'd see I already figured out the current. I'm asking for voltage. What are the SPECIFIC voltages of V4 and V7?

14. May 20, 2010

### PhaseShifter

You got them back in post #7.

15. May 20, 2010

### moshrom

Thank you.

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