Power Calculation for Three Resistors in a Circuit

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SUMMARY

The discussion centers on calculating the power dissipated across three resistors connected in a circuit with a voltage of 141 V and resistance values of R1 = R2 = R3 = 204 Ω. The user initially calculated the equivalent resistance incorrectly as 102 Ω instead of using the correct formula for resistors in series and parallel configurations. The correct current through the circuit is determined to be 1.38235 A, leading to voltage drops across the resistors of 94 V for R1 and approximately 47 V for R2 and R3. The user seeks clarification on the resistance values provided and the power calculations for each resistor.

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  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series and parallel resistor configurations
  • Familiarity with power calculations (P=IV)
  • Basic circuit analysis skills
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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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