Simple Circuits for Stupid People

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  • Thread starter calcuseless
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  • #1
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.
 

Answers and Replies

  • #2
xcvxcvvc
394
0

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:
  • #3
calcuseless
9
0
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?
 
  • #4
xcvxcvvc
394
0
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:
  • #5
satchmo05
114
0
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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