(adsbygoogle = window.adsbygoogle || []).push({}); 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?

Please help thanks:)

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# Homework Help: Circuits help

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