# Quick Thevinin Question

#### Digit0001

Joined Mar 28, 2010
100
Hi
Can someone tell me how would i do this question

I have tried to use a nodal analysis but still incorrect answer

P.S

Last edited:

#### Jony130

Joined Feb 17, 2009
5,457
I don't see any picture ??

#### Jony130

Joined Feb 17, 2009
5,457
First you need to find voltage across RL when the load is disconnect.
The open circuit voltage Vab = Vth
The second step is to find the current flowing from a to b when a & b are shorted together.
And then Rth = Vth/Isc

So know sow as your solution

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#### Digit0001

Joined Mar 28, 2010
100
ok for Vth i get

(9-Va)/2 = 3Vx-Va
2(2Va) = 9-Va
4Va = 9-Va
5Va = 9
Va = Vth = 9/5

Then when i find In = (33/6)/4

so R = Vth/In = (9/5)/(In) = 4.4 ohms

What is wrong?

#### jegues

Joined Sep 13, 2010
733
ok for Vth i get

(9-Va)/2 = 3Vx-Va
2(2Va) = 9-Va
4Va = 9-Va
5Va = 9
Va = Vth = 9/5

Then when i find In = (33/6)/4

so R = Vth/In = (9/5)/(In) = 4.4 ohms

What is wrong?
You wrote your KCL(or nodal analysis) incorrectly,

You wrote a "current entering the node = the other current entering the node."

Just write a simple KVL around the loop and express Vx in terms of the current flowing, solve Vab.

#### Jony130

Joined Feb 17, 2009
5,457
Simply nodal equation
(9-Va)/2 + 3Vx-Va = 0

Or we can write one equation like this

I = (9V - 3Vx) / ( 2Ω + 1Ω) = ( 9V - 3Vx)/3Ω

I = (9V - 3(2Ω*I))/3 = (9V - 6I)/3

I = (3A - 2I)

3I = 3A

I = 3A/3 = 1A

And Vx = 3*(1A*2Ω) = 6V

Vth = Vx + 1A*1Ω = 7V