# Resistors in Series, in Parallel

1. Feb 13, 2012

### rabcdred

1. The problem statement, all variables and given/known data

See the attached image.

2. Relevant equations

V=IR, Kirchoff's Law

3. The attempt at a solution

The voltage drop across resistors in parallel are equivalent due to Kirchoffs law (at least I think so?), so V=R(eq,top)I(top)=R(eq,bottom)I(bottom)--> I(top)(52)=I(bottom)(7+Rx)

For each resistor, an equation using Ohm's Law:
V1=I(top)(14)
V2=I(top)(38)
V3=I(bottom)(7)
V4=I(bottom)(Rx)

As the ammeter reads zero, I thought the voltage drop across resistor 1 and 3 were equal, so V1=V3. Rearranging the equation and substituting in the top equation-->
(V/14)(52)=(V/7)(7+Rx), which yields Rx=19.

I don't think this is right though. Please help! Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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• ###### Screen shot 2012-02-13 at 8.41.51 PM.png
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2. Feb 13, 2012

### PeterO

19Ω is a perfect answer - so probably your reasoning is correct too as answers like 19 don't usually appear by co-incidence.

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