I like to find Eth, Rth and I at 6 ohm resistor...see attachment.

 

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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

Checking the answer:

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!

 

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

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.

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

solving for I at 6 ohms:

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

Checking the answer:

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!

 

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

Checking the answer:

V1 = 3 - I = 3 - 1.8 = 1.2 volts

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


6 +(1.2) - (6*1.8) + (3)*(1.2) = 0,