# Question on Circuit Analysis

Discussion in 'Homework Help' started by Digit0001, Mar 10, 2011.

1. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
Hi
I have two problem both dealing with finding voltage using nodal anaylsis. When you have two dependant or independant voltage sources, Is the supernode equation determine identifying the node points separately for each voltage source or together as one?

For instance in the first question (refer to attachment doc)
View attachment tut2qns.doc
First SuperNode independent voltage source 3V : i1 = i3

Second SuperNode dependent voltage source 4Vo : i1 + i2

KVL for 4Vo - V = 0

rest of the equations

i1= (Vo-V)/3
i2= -3+V
i3= 4Vo-V

P.S

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Well simply you start with equations for the current that enter the V node

I1 = (Vo - V)/3Ω
I2 = (0 - (V - 3))/1Ω
I3 = (4Vo - V)/5Ω

(Vo - V)/3Ω - (0V - (V - 3V))/1Ω - (4Vo - V)/5Ω = 0

Now we write equation for Vo node

(0V - Vo)/2 - ( Vo - V)/3 = 0

And now we can solve for Vo and V
Vo = 10/9 ; V = 25/9

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3. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
One thing i don't understand when using nodal analysis with voltage source aren't you suppose to create a supernode?

4. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
according to your equations i have tried it a number of times and i cannot get the answer that you have got.
This is what i got from your equations:

1) -7Vo+13V = 45
2) -5Vo+2V=0

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
In your circuit we have only one supernode.

$I1 = \frac{V_B - V_A}{3\Omega}$

$I2 = \frac{0V - V_D}{1\Omega}$

$V_D = V_A - 3V$ (supernode)

$I3 = \frac{V_C - V_A}{5\Omega}$

So for node A we can write from KCL.

$I1 + I2 + I3 = 0$

All current entry the A node, no current (0A) leaves the node.

Well, yes because I made the mistake in the first equation when I rewriting the equation.

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Last edited: Mar 11, 2011