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Posted by **Dav e** on Sunday, May 20, 2012 at 5:01pm.

a 120V battery, total of 12A, leads into a 8ohm resistor (R1), then breaks off into two parallel lines, R2 and R3. the resistance is unknown in R2, and in R3 the resistance is 3 ohms. I need to find the Amperage in R1,R2 and R3 as well as the resistance (ohms) on R2

I hope this explains it... HELP

- physics-series and parellel circuits -
**Dav e**, Sunday, May 20, 2012 at 5:02pmOOOPA and the voltage at R1, R2 and R3

- typo? -
**Damon**, Sunday, May 20, 2012 at 5:11pm12 amps through R1

volts across R1 = 8*12 = 72

That leaves 120 - 72 = 48 volts across R2 and R3

amps through R3 = 48/3 = 16 amps

You have a typo I think. Perhaps R3 is 30 and not 3 ohms?

- physics-series and parellel circuits -
**Dav e**, Sunday, May 20, 2012 at 5:38pmR3 is 3 ohms

- physics-series and parellel circuits -
**Damon**, Sunday, May 20, 2012 at 5:49pmWell, you have set up a physically impossible situation where R2 would have to be negative. You can not have 12 amps coming out of the battery and 16 amps through R3. Please check your problem statement carefully. Maybe you have 1.2 amps out of the battery?

- physics-series and parellel circuits -
**Dav e**, Sunday, May 20, 2012 at 5:50pmFor R1 voltage .. E=I*R

E=8*12=96 volts correct?

- physics-series and parellel circuits -
**Damon**, Sunday, May 20, 2012 at 5:58pmwhoops, right. I multiplied wrong

- physics-series and parellel circuits -
**Damon**, Sunday, May 20, 2012 at 6:00pm120 - 96 = 24 volts across R2 and R3

current through R3 = 24/3 = 8 amps

then 4 amps through R2

R2 = 24/4 = 6 ohms

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