(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Three resistors are connected across a battery as shown in the figure. (Let V = 141 V and R1 = R2 = R3 = 204 Ω.)

http://www.webassign.net/bauerphys1/25-p-057-alt.gif

(a) How much power is dissipated across the three resistors?

P1 =

P2 =

P3 =

2. Relevant equations

P=IV

3. The attempt at a solution

I solved for each of the voltages across each resistor. Each resistor is 68 ohms.

so the equivalent resistance of the circuit is 102 = 68 + 1/[2(1/68)]

Using that and the voltage 141 I get the current I = 1.38235 A

Using that I solve for V1 = 94 V and V2 and V3 = 46.999 V

So to get the power for R1 I would take the voltage 94 and multiply it by the current 1.38235 and get 129.9409 W but the answer is incorrect so I'm not sure what I'm doing wrong. The same goes for P2 and P3 when I do the same thing

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# Homework Help: Solving for power in circuits

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