Power Resistor Immersed In Water

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Homework Help Overview

The discussion revolves around a circuit problem involving a 20-ohm resistor immersed in water and surrounded by insulating foam, with an equivalent resistance of 40 ohms and a current of 0.75A. Participants are exploring the power dissipated through the resistor and the implications of the surrounding materials on this power calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the calculation of power dissipated through the resistor, particularly why the provided solution suggests a different value than the original poster's calculation. Some are considering the role of the water and insulating foam in the power dissipation and whether the power calculation pertains to the entire circuit or just the resistor.

Discussion Status

There is an ongoing exploration of the problem, with participants offering insights into the relationship between current, resistance, and power. Some guidance has been provided regarding the interpretation of the circuit and the potential impact of the materials involved, but no consensus has been reached regarding the correct power dissipation value.

Contextual Notes

Participants note the possibility of multiple resistors in the circuit and question the wording of the original problem statement, indicating that there may be missing information or assumptions that need clarification.

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?
 
Last edited:
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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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