Power Resistor Immersed In Water

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SUMMARY

The power dissipated through a 20-ohm resistor immersed in water, with an equivalent resistance of 40 ohms in the circuit and a current of 0.75A, is calculated using the formula P=I²*R. The correct calculation yields 11.25W for the 20-ohm resistor. The confusion arises from misinterpretation of the total power in the circuit, which is 22.5W, but this value represents the power dissipated across all resistors, not just the 20-ohm resistor. The insulating foam and water do not influence the power dissipation directly but affect the temperature of the resistor.

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  • Knowledge of thermal effects in electrical components
  • Basic algebra for manipulating equations
NEXT STEPS
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elements
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Homework Statement


For the circuit shown below the 20-ohm resistor is immersed in water surrounded by insulating foam; the equivalent resistance through the circuit is 40 ohms. The current through it is 0.75A. What is the power dissipated through the 20 ohm resistor?

Homework Equations



$$ P=I^2*R$$

The Attempt at a Solution


$$ P=0.75^2*20$$
$$ P=11.25W$$

The solution to this problem gives 22.5W, why is the power dissipated through the resistor doubled here? Does the water or insulating foam affect the power at all? or is power calculated for the entire circuit?
 
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The water and insulating foam do not affect the power dissipated, but they will certainly affect the temperature. The higher the insulation, the higher the steady-state temperature will be. Are there 2 resistors in the circuit? If 0.75 A is running through 40 ohms, then you can calculate the power from the equation you showed above. However, it should be P = I2 * R
 
Sorry forgot to give the picture
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elements said:
The solution to this problem gives 22.5W

It looks like the answer they gave is wrong.
 
That would be the power dissipated in the whole circuit (all the resistors). Perhaps check the wording of the question?
 

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