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Homework Statement
The battery in the figure below has negligible internal resistance. Find
(a) the current in each resistor:
(b) the power delivered by the battery:
Homework Equations
V = IR
Resistor Connected in series
Req = R1 + R2 + R3
Resistor Connected in Parallel
1/Req = 1/R1 + 1/R1 + 1/R3
The Attempt at a Solution
I think that the resistors 3omega and vertical-2omega are connected in series, and the other 2 are connected in parallel.
So the ones in Parallel..
1/Req1 = 1/2 + 1/4
1/Req1 = 3/4
Req1 = 1.333
The ones in Series
Req2 = 3 + 2 = 5
so...
Req = Req1 +Req2 =1.333 + 5 = 6.333But to find the Current, I you just use Ohm's law
I = V/r
so the 3.0-omega resistor would be
I = 6/3 = 2
...but that's wrong and I am at a loss to know why?? thanks for the help!