# Nodal Analysis questions

Discussion in 'Homework Help' started by Digit0001, Mar 13, 2011.

1. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
Hi

The two questions i am having trouble getting the correct equations.
View attachment tut2qns.doc
Can someone tell me where is my mistake?

Question 1
SuperNode
5 = i2 + i3 + i4

Node 1
i1 = i2 + i6 + i5

KVL
6 + V3 - V1 = 0

Question 2

For SuperNode
i1 + i3 + i2 + i - i3 = 0

For Node V3
i4 + i3 = i

For KVL
2i + V2 + V1 =0

P.S

Last edited: Mar 13, 2011
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
In question 1 the 3A source to the left side is translated in the circuit to the right as a 6V equivalent source. The equivalent circuit shown on the right is incorrect, as the 6V source should have a series resistance of 2Ω. This is not a supernode example.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It would have been "nice" if you had acknowledged the fact that you had made a mistake in the schematic attached with your original post, rather than just correcting the problem and leaving my post in no-man's land.

In any case there are several errors in your equations as well.

This is the set ...

i2+i5+i6=5

i6-i3-i7=0

i4+i5+i7=0

i2=V1/4

i3=V2/2

i4=(12-V3)/8

i5=(V1-V3-6)/2

i6=(V1-V2)/8

i7=(V2-V3)/4

4. ### Digit0001 Thread Starter Member

Mar 28, 2010
89
0
i have done Question 2 again and these are the equations i get but the answer is incorrect.

supernode V1-V2
i1 + i2 + i3 + i4 = 0

Node V3
i3 + i4 = i

KVL Supernode
2i + V1 - V2 = 0

5. ### peter_morley Member

Mar 12, 2011
179
0
Can you make v2 and v3 also a supernode? I am also learning nodal analysis and number 2 is a problem I have never come across...quite a mess.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
This was my attempt at a solution for the second problem. Interesting challenge although I didn't specifically use supernode approach.

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7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
To solve problem 1, it's better not to convert the 3A current source in parallel with the 2 Ω resistor into a voltage source. Rather, convert the 12 volt source in series with the 8 Ω resistor into a 1.5A current source in parallel with an 8 Ω resistor. Then there is no supernode and a nodal solution is straightforward.

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A supernode solution for problem 2 goes like this:

All three nodes comprise the supernode, so the three currents from the supernode to ground must add to zero: I1 + I2 + i = 0

That becomes a nodal equation:

Eq 1: V1/4 + V2/1 + V3/4 = 0

The next 2 equations are constraint equations for the two sources:

Eq 2: V1 - V3 = 12

Given that 2*i = 2*(V3/4) = V3/2, we have:

Eq 3: V1 - V2 = V3/2

Solving these three equations gives us:

V1 = -3
V2 = 9/2
V3 = -15

Digit0001 and t_n_k like this.