Power Absorbed In A Electrical Circuit

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SUMMARY

The power absorbed by the 4000 Ohm resistor in the electrical circuit is calculated to be 0.4 Watts. The solution involves applying Kirchhoff's Current Law (KCL) to derive the voltage across the resistor, which is determined to be 40 Volts. The calculation is finalized using the formula P=V²/R, confirming the power absorbed. The key to solving this problem lies in correctly identifying the direction of currents and applying KCL accurately.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with Ohm's Law
  • Ability to perform basic circuit analysis
  • Knowledge of power calculations in electrical circuits
NEXT STEPS
  • Study advanced applications of Kirchhoff's Laws in complex circuits
  • Learn about power calculations using P=V²/R in different circuit configurations
  • Explore techniques for identifying current directions in circuit analysis
  • Investigate common mistakes in circuit analysis and how to avoid them
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing electrical circuits will benefit from this discussion.

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Homework Statement


Find the power absorbed by element X, if it is 4000 Ohms.
(See attached picture for circuit)

Homework Equations


KCL: Sum of Currents into Node Must = 0

The Attempt at a Solution



I've already worked through a bunch of these type of problems, but I always mess up on which currents are positive and which ones are negative...which ultimately leads to the wrong answer. I know on this specific problem, the whole idea is to apply KCL in order to obtain a Voltage. That voltage will then be multiplied by 40 Ohms to give you the power absorbed by the resistive element.

I think my lack of understanding of KCL is keeping from getting these ideas. I guess as I run this problem through my head I use a reference location at the node above the 1000 Ohm resistor. And from my thinking I would come up with the following equation...

0=(.08)A - (v/1000[ohms])A - (.03)A - (v/4000[ohms])A

Then I would solve for V, but it ends up giving me the wrong answer. I'm assuming one of my values should be negative?
 

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Nevermind! Figured it out. The equation above is correct, it's when I was adding and subtracting to the opposing sides, I kept seeing .08A as -.08A.

It computes out to 40V on the Resistor. Then I use P=V2/R to obtain .4 Watts.
 

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