# Nodal analysis

Discussion in 'Homework Help' started by nyasha, Apr 14, 2009.

1. ### nyasha Thread Starter Active Member

Mar 23, 2009
90
2
Guys l have a final exam tomorrow and l desperately like to like to know how does the the voltage at node D become :

Vd=-(Vc-Va)=Va-Vc

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Calculate ix = (VC-VA)/2. You see why this is, I hope. Ask if you don't.

Then the dependent source feeding VD has a voltage VD =-2*ix. Substituting what we calculated for ix above, we have VD = -2*ix = -2*(VC-VA)/2 = VA-VC.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,711
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The OP hasn't anything more to say, but I know that tnk and HGMJR want to try out their skills by solving this one.

What is the voltage at each node?

And, what is the impedance (incremental resistance) to ground at each node?

I've given it some effort, and I will say that attention to detail is necessary!

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The sweat is pouring from my brow! Calculator is smoking!

I pity the poor student ["nyasha"] if that's a typical exam question.

OK "electrician" I just did the voltages - you're a hard man asking for the impedances!

I hope ...

VA = 3.243 V
VB = 5.270 V
VC = 0.811 V
VD = 2.432 V
VE = VX = 0.270 V

And I didn't use matrices to solve the simultaneous equations

BTW - what are the impedances?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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PS - Is there an exploding head "smiley" face?

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,711
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"I pity the poor student ["nyasha"] if that's a typical exam question."

I was thinking the same thing! The question he actually asked seemed to me rather trivial compared to the full solution of the network.

"VA = 3.243 V
VB = 5.270 V
VC = 0.811 V
VD = 2.432 V
VE = VX = 0.270 V"

Yep. That's what I get, too. I was hoping HGMJR would show us how to do it using Millman's theorem.

As far as impedances, probably a good way to find them is to attach a 1 amp current source to each node, one at a time, and re-solve the network. Then subtract the voltage at that particular node (without the extra current source) from the voltage there with the 1 amp injected. That difference voltage (divided by 1 amp, of course) should be the impedance.

I get:

Z@VA = .5946
Z@VB = .1622
Z@VC = .1892
Z@VD = .4054
Z@VE = .1622

You'd think the impedance to ground on top of a voltage source (such as node VC and VD) would be zero, but for a controlled source it's not necessarily so!

7. ### kiranag Member

Apr 14, 2009
12
0
From the circuit, the current i(x) is flowing from V(c) to v(a) through the resistor 2ohms. So it can be written as the difference between the voltages divided by the resistor. Hence i(x)=[V(c)-V(a)]/2. Then from the circuit again V(d) is nothing but the negative of voltage of the dependent source 2i(x). but we know what is i(x). Hence V(d) is -[V(c)-V(a)]=V(a)-V(c). Hope you understood this.

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,711
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Well, what do you think, tnk?

Do those impedances make any sense? Have you tried to calculate them?