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CIRCUIT ANALYSIS: 4 resistor, 2 Voltage Source, V.C.V.S., C.C.C.S. - Find Io & Vo

  • Engineering
  • Thread starter VinnyCee
  • Start date
489
0
1. Homework Statement

Using nodal analysis, find [itex]v_0[/itex] and [itex]I_0[/itex] in the circuit below.

http://img248.imageshack.us/img248/7325/chapter3problem301dy.jpg [Broken]


2. Homework Equations

KVL, KCL, V = i R, Super-node


3. The Attempt at a Solution

So I added 3 current variables, 3 node markers ([itex]V_1\,-\,V_3[/itex]), a super node, a ground node, and marked a KVL loop.

http://img404.imageshack.us/img404/6940/chapter3problem30part26cn.jpg [Broken]

[tex]V_0\,=\,V_3[/tex] <----- Right?

Now I express the currents:

[tex]I_0\,=\,\frac{V_1\,-\,V_2}{40\Omega}[/tex]

[tex]I_1\,=\,\frac{100\,-\,V_1}{10\Omega}[/tex]

[tex]I_2\,=\,\frac{4\,V_0\,-\,V_1}{20\Omega}[/tex]

[tex]I_3\,=\,\frac{V_0}{80\Omega}[/tex]

KCL at [itex]V_1[/itex]:

[tex]I_0\,=\,I_1\,+\,I_2[/tex]

[tex]\left(\frac{V_1\,-\,V_2}{40}\right)\,=\,\left(\frac{100\,-\,V_1}{10}\right)\,+\,\left(\frac{4\,V_3\,-\,V_2}{20}\right)[/tex]

[tex]7\,V_1\,-\,V_2\,-\,8\,V_3\,=\,400[/tex]


KCL at super-node:

[tex]I_0\,+\,2\,I_0\,=\,I_3[/tex]

[tex]3\,\left(\frac{V_1\,-\,V_2}{40}\right)\,-\,\left(\frac{V_0}{80}\right)\,=\,0[/tex]

[tex]6\,V_1\,-\,6\,V_2\,-\,V_3\,=\,0[/tex]


KVL inside super-node:

[tex]V_3\,-\,V_2\,=\,120[/tex]


Now I put those 3 equations into a matrix and rref to get [itex]V_1\,-\,V_3[/itex].

[tex]\left[\begin{array}{cccc}0&-1&1&120\\7&-1&-8&400\\6&-6&-1&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{cccc}1&0&0&-1688\\0&1&0& -1464\\0&0&1&-1344\end{array}\right][/tex]

[tex]V_3\,=\,V_0\,=\,-1344\,V[/tex]

But -1344 Volts seems too high (or low) doesn't it? Should I have also expressed the currents that I did not mark at the short wires between ground and [itex]V_1[/itex] and [itex]V_3[/itex]? Maybe I should use KVL 1 loop instead of the super-node KVL expression?
 
Last edited by a moderator:

Answers and Replies

mjsd
Homework Helper
725
3
your diagram as it stand contains two wires that short out V1 and V3 making them trivial with respect to ground chosen..should they not be there?
 
Last edited:
489
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Nope, they are there.
 
mjsd
Homework Helper
725
3
in that case your (effective) diagram changes dramatically, i m not even sure whether it is consistent?...anyway, if it is consistent: your circuit theory then tells you: I3=0, V_0=0, I2=0, V1=V3=0 so can join them up and form a loop containing the 120V source and 40 ohm resistor..and I_0=3A

it is actually ok i think... consistency wise speaking
 
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489
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So how would I go about getting a system of equations for [itex]V_1[/itex] thorugh [itex]V_3[/itex]?
 
mjsd
Homework Helper
725
3
you wanna find V_0 and I_0, why worry about v1, v2 and v3? they are introduced for node voltage analysis purposes only... but if you really want a "system" of equations: you have
V1=0, V3=0, V2 = -120

all w.r.t. the ground chosen
 
489
0
OIC, [itex]V_1[/itex] and [itex]V_3[/itex] are 0 becasue they are both connected to ground through the wires you were asking about.

So for [itex]I_0[/itex] we have

[tex]I_0\,=\,\frac{V_1\,-\,V_2}{40\Omega}\,=\,\frac{(0)\,-\,(-120\,V)}{40\Omega}\,=\,3\,A[/tex]

Right?
 
mjsd
Homework Helper
725
3
yes. NB: but for this question, introducing V1-V3 are not necessary, just observe that those wires short out several components leading to a simpler circuit and solve by inspection or doing a simple KVL loop (in this case).
 
489
0
So, to get [itex]V_0[/itex], I do a KVL loop around the [itex]80\Omega[/itex] and [itex]40\Omega[/itex] and the 120 V source?

[tex]V_0\,+\,40\,I_0\,-\,120\,V\,=\,0\,\,\longrightarrow\,\,V_0\,=\,120\,-\,40(3)\,=\,0\,V[/tex]

So [itex]V_0[/itex] is really zero?
 
mjsd
Homework Helper
725
3
by the way, I got V0=0 by inspection. but what you have done is also correct. i guess showing that the circuit is consistent after all
 

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