Power Calculation for Three Resistors in a Circuit

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The discussion focuses on calculating the power dissipated across three resistors connected to a 141 V battery, with each resistor initially stated as 204 Ω. The user calculated the equivalent resistance as 102 Ω and derived a current of 1.38 A, leading to voltage values of 94 V for R1 and approximately 47 V for R2 and R3. However, the calculated power for R1, using the formula P=IV, yielded an incorrect result, prompting confusion about the resistor values. The user questions the accuracy of the resistance values provided, indicating a potential error in the problem setup. Clarification on the resistor values is necessary to resolve the discrepancies in power calculations.
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Homework Statement




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 =


Homework Equations



P=IV

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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I don't know, your original statement said the resistance was 204, do you have the right values?
 
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