How Do You Calculate Voltage and Power in a 12-Ohm Resistor Circuit?

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Discussion Overview

The discussion revolves around calculating voltage and power in a circuit containing a 12-ohm resistor. Participants explore various methods for solving the problem, including mesh analysis and alternative approaches, while addressing discrepancies between their calculations and those presented in a textbook.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to analyze the circuit.
  • One participant suggests using mesh analysis and condensing the circuit to find equivalent resistance before writing mesh loops.
  • Another participant calculates the current in the circuit to be 0.4 A and derives a voltage of 4.8 V across the 12-ohm resistor, leading to a power calculation of 9.6 W.
  • However, this participant notes a discrepancy with the textbook's power value of 1.92 W, prompting questions about the correctness of their calculations or the book's information.
  • Another participant confirms the power calculation using the formula P = I^2R, arriving at the same power value of 1.92 W, suggesting a potential error in the earlier calculation.
  • One participant mentions that P can also be expressed as V^2/R, emphasizing that these equations are valid for calculating power absorbed by the resistor.
  • There is a suggestion that source transformation could simplify the problem, as mesh equations can become complex.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct power value, with some calculations yielding 9.6 W and others confirming 1.92 W. The discussion remains unresolved as to which calculation is correct.

Contextual Notes

Participants rely on various assumptions about circuit analysis techniques and the accuracy of textbook examples. The discussion reflects differing interpretations of the circuit's behavior and the application of relevant equations.

calcuseless
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Homework Statement



Find the voltage V and the power absorbed by the 12-ohm resistor.
http://i.imgur.com/xCTEI.png

Homework Equations



p = vi
v = ir
KVL
KCL

The Attempt at a Solution



Tried using KVL and KCL in various ways. This book gives horrid examples.
 
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calcuseless said:

Homework Statement



Find the voltage V and the power absorbed by the 12-ohm resistor.
http://i.imgur.com/xCTEI.png

Homework Equations



p = vi
v = ir
KVL
KCL

The Attempt at a Solution



Tried using KVL and KCL in various ways. This book gives horrid examples.

You can solve using mesh analysis if you wanted, but I'd condense that middle branch into an equivalent resistance first.

4 + (1/3 + 1/6)^-1 = 6. Next, write the mesh loops, but leave out that first loop with the current source (but include its effects in that the current going through the 2 ohm resistor is (I_1 - 6))
 
Last edited:
I found the current in the last loop to be .4 A. Using V = IR to solve to get the voltage, I get 4.8 V at the 12-ohm resistor. I then used P = VI to get a power of 9.6 W, but the book has 1.92 W. Did I do anything wrong or is the book wrong again?
 
calcuseless said:
I found the current in the last loop to be .4 A. Using V = IR to solve to get the voltage, I get 4.8 V at the 12-ohm resistor. I then used P = VI to get a power of 9.6 W, but the book has 1.92 W. Did I do anything wrong or is the book wrong again?

The current is [tex].4[/tex] and the voltage is [tex]4.8[/tex]. However, [tex].4 \times 4.8[/tex] is not [tex]9.6[/tex]. It's [tex]1.92[/tex](as the book says). Also, [tex]I^2 r = (.4)^2 12 = 1.92[/tex] as well if you wanted to avoid that extra calculation of voltage.

To see where the current-dependent equation comes from notice:
[tex]P=IV[/tex]
and
[tex]V=IR[/tex]
substituting [tex]IR[/tex] for voltage into the power equation brings us to:
[tex]P = I (IR) = I^2R[/tex]
 
Last edited:
Yes, like xcvxcvvc said: P is equal to I2*R as well as V2/R. These only work when you're finding the power absorbed by the resistor! Also, I think the easiest way to solve this (because mesh equations can get ridiculous and tedious after a while --> source transformation, if you have learned it!
 

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