physics-series and parellel circuits
posted by Dav e on .
I'll try to explain the schmatic drawing...
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
OOOPA and the voltage at R1, R2 and R3
12 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?
R3 is 3 ohms
Well, 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?
For R1 voltage .. E=I*R
E=8*12=96 volts correct?
whoops, right. I multiplied wrong
120 - 96 = 24 volts across R2 and R3
current through R3 = 24/3 = 8 amps
then 4 amps through R2
R2 = 24/4 = 6 ohms