# Thevenin's theorem with current and dependent voltage source.

Discussion in 'Homework Help' started by rdj, Mar 26, 2010.

1. ### rdj Thread Starter New Member

Mar 25, 2010
5
0
I like to find Eth, Rth and I at 6 ohm resistor...see attachment.

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This is a homework help forum. You need to show some effort of your own before you get any help.

3. ### rdj Thread Starter New Member

Mar 25, 2010
5
0
6 +(V1) -Eth +3(V1) = 0

6 +4V1 -Eth = 0

since 6 ohm resistor open circuited 3A will go V1, hence

V1 = 3

Eth = 6 + 4*3 = 18 volts

solving for I short circuit:

Isc = 3 - V1

6 + V1 +3V1 = 0
V1 = 1.5

Isc = 1.5 A

Rth = Eth/Isc = 18/1.5 = 12 ohms

solving for I at 6 ohms:

I = Eth/(Rth+6) = 18/(12+6) = 1 A

6 + (V1) - 6*I + 3(V1) = 0

6 + 2 - 6*2 + 3*2 = 2,

Can someone help me to correctly solve this by Thevenin's Theorem?

Thanks!

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,702
493
Right here is where you made your error. You have to add in the 3 amp current source. This will make Isc = 4.5 amps. I think the rest of it will work out if you do that.

5. ### rdj Thread Starter New Member

Mar 25, 2010
5
0
The Electrician,

Thanks!

solving for I short circuit:

Isc = 3 - V1

6 + V1 +3V1 = 0, 4V1 = -6

V1 = -1.5

Isc = 3 - (-1.5) = 4.5

Rth = Eth/Isc = 18/4.5 = 4 ohms

solving for I at 6 ohms:

I = Eth/(Rth+6) = 18/(4+6) = 1.8 A