# Homework Help: Finding the current

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1. Nov 25, 2015

### Gbox

1. The problem statement, all variables and given/known data
All resistors have resistance of R and all voltage sources have voltage of V, Find the current I

2. Relevant equations

3. The attempt at a solution
At the leftmost corner I can connect the two resistors in parallel, but what should I do with the short on the top right?

2. Nov 25, 2015

### .Scott

I would do this one voltage source at a time - and then add the results up.
You have three resistors in parallel at the top left and two in parallel at the top right.

3. Nov 25, 2015

### Gbox

What does it mean each "voltage source at a time"?

in the top left I can see just 2 resistor in parallel and so in the right:

4. Nov 25, 2015

### CWatters

or even...

5. Nov 25, 2015

### azizlwl

Just remove the short-cicuited resistors(series resistors). No current flow through them.

Last edited: Nov 25, 2015
6. Nov 25, 2015

### Gbox

Why can I remove it? there is current there, here is what I got:

7. Nov 25, 2015

### azizlwl

There is no potential difference between both end of the conductor, thus no current flow across resistors parallel to it. Likewise no current flow up to the middle of the shorted resistors since no potential difference between them. If you say X volt on the right side, it will the same at the middle, which in turn equal to the right side of the lower resistor.
If you have a conductor carrying current, will there any difference to the circuit if you touch both ends of resistor to it.
It will be helpful if you could name all the resistors.

Last edited: Nov 25, 2015
8. Nov 25, 2015

### Gbox

From the current point of view (the arrows I added) this is what you mean? therefore where there is no current I can remove those resistors?

9. Nov 25, 2015

### haruspex

Are there any short-circuited resistors?
Numbering the resistors from 1, top to bottom, left to right within a horizontal: 3 , 4 and 5 are in parallel, but not short-circuited. Likewise 1 and 2. See CWatters' diagram.

10. Nov 25, 2015

### azizlwl

One correction. There is current flow in the lower right resistor.

11. Nov 25, 2015

### Gbox

So my diagram after azizlwl comment is ok?

12. Nov 25, 2015

### Staff: Mentor

Color or otherwise identify the relevant nodes in your circuit to begin with. Any components that connect to the same two nodes (colors) must be in parallel.

There's a parallel bonanza there!

13. Nov 25, 2015

### azizlwl

I think the color between the resistors should be red too, no current flow thus no p.d.

14. Nov 25, 2015

### Gbox

So which equivalent diagram is right this:

or this (although I am not sure as I removed short circuit cords):

15. Nov 25, 2015

### Staff: Mentor

I'm not liking either of them.

If you look at the colored diagram that I posted, all of horizontal resistors connect to the same pair of colors. That is, they all connect to the same two nodes. That is, they are ALL in parallel. So erase all but one and label it with the correct resistor value. Personally I'd choose to keep the one at the bottom right, just to clean up the diagram.

16. Nov 25, 2015

### SammyS

Staff Emeritus
@azizlwl

There is current flow through each and every resistor in this circuit.

(gneill has this covered very well..)

17. Nov 26, 2015

### Gbox

I came to this $I=\frac{21V}{2R}$

18. Nov 26, 2015

### SammyS

Staff Emeritus
19. Nov 26, 2015

### Gbox

Yes it should be 5, Thanks!

20. Nov 26, 2015

### SammyS

Staff Emeritus
By the way, even though the R/4 isn't correct, you should know that $\displaystyle \ \frac{R}{3}+ \frac{R}{4}\neq \frac{R}{7}\$.

21. Nov 26, 2015

### Gbox

I have wrote this?

22. Nov 26, 2015

### SammyS

Staff Emeritus
Maybe not explicitly, but you have R/3 and R/4 in series with a result of R/7 .

23. Nov 26, 2015

### Gbox

Ahh so it a mistake, I connected them in parallel

24. Nov 26, 2015

### Dopplershift

It's always best to simplify the circuit step by step. Take all the resistors in series and add them together and label it as one resistor. Then worry about the ones in parellel and use the inverse for each of them.