# Homework Help: Finding current and potential difference within a circuit

1. Oct 7, 2014

### mathguy831

1. For the circuit shown,
(a) Find the current in each resistor
(b) Find the potential difference between c and f

2. Relevant equations:
i2 + i1 = i3
i = V/R

3.
(a)
So solving for each resistor using the loop rule this is what I got,

(Resistor 3): E3 - IR3 = -E2 - E1
(Resistor 2): -E2 - IR2 = E3 - IR3
(Resistor 1): -E1 - IR1 = E3 - IR3

Then plugging in for the givens I got:
I3 = 52.5 kA
I2 = 1.67 kA
I1 = 2.49 kA

I'm trying to follow my notes the best I can, but I am not seeing any examples like this in my notes or in the text. I hope I'm at least on the right path.

(b)
So since the there are two different paths that could be taken to reach f from c I used the equation V=iR
to get:

(I3 - I2) R2 = the potential difference from c to f through battery 1
(I3 - I1) R1 = the potential difference from c to f through battery 2

I then subtracted the potential differences from each other to give me the overall potential difference at f.

Again, nothing in my notes or in the text are very helpful, so this is just based on the formulas I know.

Thank you in advance for the help.

Last edited by a moderator: Oct 7, 2014
2. Oct 7, 2014

### Staff: Mentor

You've defined three different currents above ( i2 + i1 = i3 ), but your equations are all using the same variable, I, for all the currents. That can't be good...

Equations should be labeled for individual loops, not for individual resistors.

Also, you should only need two loop equations to solve this problem. Notice that if you trace around the two obvious loops (the left one and the right one), you'll pass through every component at least once. All that is required is that the loops you choose cover all the components.
While you haven't shown how you arrived at those values, I can tell you that your currents should end up in the milliamp range, not kiloamp range. So definitely something amiss with your workings.
You only need to follow one path between points to establish the potential difference; all paths should yield the same result. It's like walking a terrain: The difference in elevation between your starting location and destination does not depend upon the path you take to get there.

3. Oct 7, 2014

### CWatters

What gneil said.

You also need to define the +ve direction of loop currents and mark them on the diagram. Far too easy to make a mistake with the signs later otherwise.

Is that meant to be a KVL equation? If so I can't make sense of it because E3, E2 and E1 aren't in the same loop.

Aside: KVL basically say that the voltages around a loop sum to zero so I prefer to write any such equations so they do indeed sum to zero... eg

Va + Vb + Vc + Vd = 0

rather than

Va + Vb = -Vc - Vd

I just find it easier to avoid mistakes with the sign.

4. Oct 7, 2014

### mathguy831

So I reworked part a) trying to follow what you guys recommended. I applied Cramer's rule as I managed to find something similar in my notes after all and this is what I got. It ended up in milliamp like gneill said so I hope this is the right answer. Can anyone confirm this please?

I am working on part B) right now so I'll upload my work when I finish with it as well.

Thank you.

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5. Oct 7, 2014

### mathguy831

So for part B) I found that the potential difference from c to f could be expressed using the equation:

Vc - E2 + i2R2 = Vf

Reworking the equation to get Vc - Vf on one side of the equation and after plugging in I found the potential difference to 54.4 Volts.

6. Oct 7, 2014

### Staff: Mentor

Sorry, I can't make out any of your work in the figure; it's too small and fuzzy on my screen. I can't see your work nor the results. Why not actually type them? Or at least the important parts!

Your value for the potential at c w.r.t. f doesn't look right. Your equation looks like it's assuming that i2 is flowing upwards through R2. Is that correct?

7. Oct 8, 2014

### CWatters

First step should be to mark up the diagram to define your currents.