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EECS: Find i_0 in a circuit with 4 resistors and 1 indep. current source

  • Engineering
  • Thread starter VinnyCee
  • Start date
  • #1
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(a) Using nodal analysis, determine [itex]i_0[/itex] in the circuit.

(b) Now use linearity to find [itex]i_0[/itex].


ch4prob4.jpg



Work so far:

[tex]i_0\,=\,\frac{v_1\,-\,0}{6\,\ohm}\,=\,\frac{v_1}{6\,\ohm}[/tex]

KCL@[itex]v_1[/itex]: [tex]\frac{v_1\,-\,0}{3\,\ohm}\,+\,\frac{v_1\,-\,0}{6\,\ohm}\,+\,\frac{v_1\,-\,v_2}{2\,\ohm}\,=\,0[/tex]

KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]

[tex]v_1\,-\,\frac{1}{2}\,v_2\,=\,0[/tex]

[tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,-9[/tex]

Whe I put these two equaitons into a 3 X 2 coefficient matrix, I get

[tex]v_1\,=\,-\,9V[/tex]
[tex]v_2\,=\,-\,18V[/tex]

Then I plug into the first equation:

[tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(-\,9V)}{6\,\ohm}\,=\,-\,\frac{3}{2}A[/tex]

Is this the correct value for [itex]i_0[/itex] or have I got the sign wrong?
 

Answers and Replies

  • #2
161
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VinnyCee said:
KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]
You are repeating the same mistake here as you did in the previous exercise.
 
  • #3
489
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What is that, can you explain?
 
  • #4
berkeman
Mentor
56,455
6,369
VinnyCee said:
What is that, can you explain?
The sign on the 9A is wrong. The other two terms in that KCL are for currents *out* of the node. So since the 9A is into the node, it has to be -9A in that equation. Makes sense?
 
  • #5
489
0
OIC! There has to be a minus in there somewhere if all of the terms are on one side and it's equal to zero. Whoops!

Using your correction:

KCL @ [itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,-\,9A\,=\,0[/tex]

[tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,9[/tex]

Which switches the signs on the [itex]v_1[/itex] and [itex]v_2[/itex]!

[tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(9V)}{6\,\ohm}\,=\,\frac{3}{2}A[/tex]
 
  • #6
161
0
Yes, that's much better. And be careful when you formulate the KCLs next time.
 

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