# Circuits help

1. Feb 15, 2008

### jelsliger7

1. The problem statement, all variables and given/known data
The 10.00V battery in the figure is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is shown in the figure. Find a, the current in each branch and b, the potential difference Vab of point a relative to point b

2. Relevant equations
V = IR
1/Req=1/R1+1/R2+...+1/Rn
Req = R1 + R2 + R3
the sum of the current at each junction =0
the sum of the potential = 0 (loop rule)

3. The attempt at a solution
R in series
R' = 2+3 = 5
R''= 1+4 = 5

Now in series
1/Req = 1/R' +1/R'' +1/10
=1/5 + 1/5 + 1/10 = 5/10
thus Req = 2.0ohms

I=V/R
I=15/2 = 7.5A

i think i am just confused as to what to do if there are two different voltage supplies, how do i find Req?

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2. Feb 15, 2008

### ranger

The fact that they want you to find two branch currents calls for a much easier method. Just use mesh (or loop) analysis. You will get two equations and two unknowns (the branch currents).

To find Vab, just apply KVL around the loop containing points a and b. Better yet, you don't have to go around the entire circuit. Just start at point a, go to straight to b and work your way clockwise back to point a.

Given the topology of the circuit, I'm afraid that there isn't any easy method of finding Req wrt t0 points ab. Besides, you don't need req to solve this problem. Thinking about it only confuses you.

Last edited: Feb 15, 2008
3. Feb 28, 2008

### jelsliger7

hey i am still confused, how can i only have 2 unknowns when i have multiple resistors won't they all have different currents flowing through them? and then what exactly is the branch current?