Node Voltage Method: Find Voltage Across 6A Current Source

  • #1
Marcin H
306
6

Homework Statement


Find the voltage, V, across the 6A current source

Homework Equations


V=IR
Node Voltage Method

The Attempt at a Solution


Did I set this up correctly to find my voltage?
Screen Shot 2016-02-03 at 1.34.06 PM.png
 
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  • #2
Almost. Be careful of the signs of the terms.

Judging by the sign you gave to the 6A supply term you're summing currents leaving the node. So why are your last two terms negative?
 
  • #3
gneill said:
Almost. Be careful of the signs of the terms.

Judging by the sign you gave to the 6A supply term you're summing currents leaving the node. So why are your last two terms negative?
Oh, woops. Read the polarities incorrectly. So if I change the last 2 to positive, it's good?
 
  • #4
Yes, that would be a correct expression. Make it an equation by setting it equal to zero :smile:
 
  • #5
gneill said:
Yes, that would be a correct expression. Make it an equation by setting it equal to zero :smile:
Thanks!
 
  • #6
gneill said:
Yes, that would be a correct expression. Make it an equation by setting it equal to zero :smile:
Got stuck again! :nb) The next question asks "What is the power associated with the 10V source?" I can find it using P=IV=I^2R=V^2/r, but how do I find the current through the 10V source?
 
  • #7
Each term in the node equation represents a branch current. One of the terms corresponds to that current. Pick it out and use it now that you've found the node voltage.
 
  • #8
gneill said:
Each term in the node equation represents a branch current. One of the terms corresponds to that current. Pick it out and use it now that you've found the node voltage.
Ohh. So I use (Vx-10)/20= (36-10)/10 = 2.6A And then using P=IV I can do (2.6A)(10V)= 26W***?
 
Last edited:
  • #9
That would be the idea, yes.
 
  • #10
gneill said:
That would be the idea, yes.
Ok thanks! 13W***
 
  • #11
Check your calculation for the current. (36 - 10)/10 = ?
 
  • #12
Marcin H said:
Ohh. So I use (Vx-10)/20= (36-10)/10 = 1.3A And then using P=IV I can do (1.3A)(10V)= 13W***?
gneill said:
Check your calculation for the current. (36 - 10)/10 = ?
Woops! 26 watts. :confused:
 
  • #13
Marcin H said:
Woops! 26 watts. :confused:
Try again. Show your calculations.
 
  • #14
gneill said:
Try again. Show your calculations.
What? Where did I go wrong? I = (36-10)/10 = 26/10 = 2.6A. Using that current and P=IV I can find the power in the 10V source. P = (2.6A)(10V) = 26 Watts
 
  • #15
Marcin H said:
What? Where did I go wrong? I = (36-10)/10 = 26/10 = 2.6A. Using that current and P=IV I can find the power in the 10V source. P = (2.6A)(10V) = 26 Watts

ORIGINAL QUESTION: what is the power associated with the 10V source?
 
  • #16
Marcin H said:
ORIGINAL QUESTION: what is the power associated with the 10V source?
Ah. My mistake. Sorry. I was thinking of the power associated with the branch as a whole. You have the correct result.
 
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