# Thévenin's Equivalent Circuit

1. Apr 11, 2012

### An1MuS

As seen from a and b.

My teacher did the norton's equivalent in the class, and now for studying purposes i was trying to get to the Thévenin's one. However it seems my equations are somehow wrong, and i can't figure out why.

The Eq resistance is 8||2, which is 1.6 ohms.

I tried to apply Nodal analysis first

so,
for node V1: $$\frac{20-V_1}{3}=\frac{V_1-V_3}{2}$$
for node V2: $$\frac{V_1-V_3}{2}=6+\frac{V_2-0}{5}=$$
The relationship between two voltages is also known, which is $$V_2=V_3+10$$

which gives V1=2 ; V2=0 ; V3=-10.

V1-V2 should indeed be 2, because from my teachers resolution, Vab = 2V. I don't get is how can V2 be 0? Then there can't be another voltage drop when it reaches the ground (0V)

Also, then i tried mesh analysis:

For mesh 1$$20+10=3I_1+2I_1+5(I_1+I_2)$$
For mesh 2 $$0=5(I_2-I_1)+v$$
$$I_2=6$$

Where v is the voltage drop across the current source.

It gives I1 = 6 and v (which is equal to the V2 of the nodal analysis) = 0 again.

2. Apr 11, 2012

### Jony130

My answer is that :
V2 is 0V because 6A current source short 5Ω resistor.
Also notice that (20V + 10V) / ( 3Ω + 2Ω ) = 6A

So from superposition point of view we have this situation

V2' = 30*R2/(R1 +R2) = 15V

And

V2'' = - (6A * 5/(10)) * 5 = - 15V

and V2 = V2' + V2'' = 15 + (- 15V) = 0V

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3. Apr 11, 2012

### An1MuS

Thanks :) it means the current source is just discharging current. Like i had a current source between two earth connections.

Another question:

In nodal analysis, when calculating a current entering / leaving a node we do $$\frac{V_x-V_y}{R}$$ where $$V_x > V_y$$ My question is: what if we don't know which of the two nodes has higher voltage when writting the equations? Or should it always be possible to know ?

4. Apr 12, 2012

### Jony130

You don't need to know which of the two nodes has higher voltage.
All you need to do is assume that one of a voltage nodes has a higher voltage than the other node. And you can treat all nodes individually and assume that the node your currently examining has higher voltage.

For example for this circuit

We can assume Vin > Va > Vout

And we can write nodal equation

$$\frac{(Vin-Va)}{R1}=\frac{Va}{R3}+\frac{(Va-Vout)}{R2}$$

$$\frac{(Va-Vout)}{R2}=\frac{Vout}{R4}$$

Or we can assume when we examining Va node that Va has the highest voltage.
And this means that all current flow-out (leave) from the Va node.

$$\frac{(Va-Vin)}{R1}+\frac{Va}{R3}+\frac{(Va-Vout)}{R2} = 0$$

And the same think we can do for Vout

$$\frac{(Vout-Va)}{R2}+\frac{Vout}{R4} = 0$$

Simply I assume that all the current leaving the node. If so this means that the voltage at this node has to be be the highest.

http://www.wolframalpha.com/input/?...+++(A+-+B)/100+=+0+,+B/100+++(B+-+A)/100+=+0+

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