Energy dissipated over time by resistors

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SUMMARY

The discussion centers on calculating the energy dissipated by resistors R1 (9.00 ohms) and R2 (23.0 ohms) connected to a 12.0 V ideal battery. The current through the circuit is 0.24 A. The power dissipated by the resistors was calculated using the formula P = i²R, resulting in a total power of 4.4928 J/s. Over a duration of 1 minute, the total energy dissipated amounts to 269.568 J. The user suspects an error in the calculation of the equivalent resistance (Rnet) and requests clarification on this aspect.

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scarne92
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Homework Statement



In the figure below, R1 = 9.00 , R2 = 23.0 and the ideal battery has an emf = 12.0 V.

hrw7_27-71.gif


(a) 0.24 A

(b) How much energy is dissipated by all four resistors in 1.00 min?

The Attempt at a Solution



Not sure where I went wrong.

P = inet2Rnet

P = i12R2 + i22R2 + i32R2 + i42R1

P = (0.24)2(23)+(0.24)2(23)+(0.24)2(23)+(0.24)2(9)

P = (0.24)2(23+23+23+9)

P = 4.4928 J/s

P = dW/dt

4.4928 = dW/60s

dW = (4.4928)(60)

W = 269.568 J
 
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I suspect you calculated Rnet incorrectly, and I get something different for inet than you.

If you show how you calculated Rnet, we might be able to spot the error.
 

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