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

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SUMMARY

The voltage across a 12-ohm resistor in a circuit can be calculated using Ohm's Law, where V = IR. In this discussion, the current was determined to be 0.4 A, resulting in a voltage of 4.8 V across the resistor. The power absorbed by the resistor is calculated using P = I²R, yielding a value of 1.92 W, which aligns with the textbook answer. The discussion emphasizes the importance of correctly applying formulas and understanding the relationships between voltage, current, and resistance.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with power calculations (P = IV and P = I²R)
  • Knowledge of Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL)
  • Basic skills in circuit analysis techniques, including mesh analysis
NEXT STEPS
  • Study mesh analysis techniques for circuit analysis
  • Learn about source transformation in circuit theory
  • Explore advanced power calculation methods in resistive circuits
  • Review examples of KVL and KCL applications in complex circuits
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone needing to understand voltage and power calculations in resistor circuits.

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

To see where the current-dependent equation comes from notice:
P=IV
and
V=IR
substituting IR for voltage into the power equation brings us to:
P = I (IR) = I^2R
 
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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