Thevenin equivalent by source transformation

[Castello]

Homework Statement

Find the Thevenin equivalent at terminals a-b

V1 = 70V
V2 (dependent) = 4Vo
R1 = 10Kohm
R2 = 20Kohm

Homework Equations

$V=RI$
$1/req = 1/r1 + 1/r2 + ... +1/rn$

The Attempt at a Solution

Found the current i = 1mA by KVL using Vo = 10ki then did source transformation for both tension sources: 7mA and 2mA, parallel with 10kohm and 20Kohm respectively. Then, calculated the equivalent resistor (20/3)kohm and subtracted the 7mA by the 2mA current sources. Transforming again, It gives me Vth = 33,3V and Rth = 6,6kohm but the answer is 60V and 2.9kohm.

I've tried many times this exercise with source transformation (as some others), had different results, and sometimes looks like that the source transformation doesn't work so generally that i thought (or I'm applying it in a wrong way).

 

[ehild]

Homework Statement

Find the Thevenin equivalent at terminals a-b

V1 = 70V
V2 (dependent) = 4Vo
R1 = 10Kohm
R2 = 20Kohm

View attachment 236913

Homework Equations

$V=RI$
$1/req = 1/r1 + 1/r2 + ... +1/rn$

The Attempt at a Solution

Found the current i = 1mA by KVL using Vo = 10ki then did source transformation for both tension sources: 7mA and 2mA, parallel with 10kohm and 20Kohm respectively. Then, calculated the equivalent resistor (20/3)kohm and subtracted the 7mA by the 2mA current sources. Transforming again, It gives me Vth = 33,3V and Rth = 6,6kohm but the answer is 60V and 2.9kohm.

I've tried many times this exercise with source transformation (as some others), had different results, and sometimes looks like that the source transformation doesn't work so generally that i thought (or I'm applying it in a wrong way).

Your method is not correct.
Determine the Thevenin equivalent emf and resistance from the open circuit voltage Uo and short-circuit current Is between a and b: the emf is equal to Uo and Rth=Uo/Is.

 

[Castello]

Your method is not correct.
Determine the Thevenin equivalent emf and resistance from the open circuit voltage Uo and short-circuit current Is between a and b: the emf is equal to Uo and Rth=Uo/Is.

Sorry, i didn't get how to do this. I mean, what i previously understood from source transformation with thevenin theorem is that if i could transform a circuit just in a tension source in series with a resistor, than it would be the Vth and the Rth. Isn't it? This kind of resolution worked in many exercises, and some others not.

 

[ehild]

Sorry, i didn't get how to do this. I mean, what i previously understood from source transformation with thevenin theorem is that if i could transform a circuit just in a tension source in series with a resistor, than it would be the Vth and the Rth. Isn't it? This kind of resolution worked in many exercises, and some others not.

Yes, according to Thevenin theorem, a circuit between two points can be replaced by a single voltage source and series resistor. The question is, how to do this. In general, the Thevenin source voltage is equal to the voltage between a and b if nothing is connected between these points - the open-circuit voltage (Uo).
To get the Thevenin-equivalent resistance, one needs the short--circuit current (Is) between a and b, that is, the current that would flow between a and b if those points were connected by a single wire with zero resistance. The Thevenin-equivalent resistance is Rth = Uo/Is. This method is general, applicable for all circuits. For circuits with independent sources only, there are other methods to get Rth, but these methods might fail for circuits with dependent sources. Your circuit contains a dependent source.

 

[gneill]

This circuit does not appear to be amenable to source transformation (current source to voltage source or vice versa) as a way to reduce the circuit. If you convert the independent voltage source and its series 10 k resistance to an equivalent current source then you would "lose" the sense resistor for Vo. Converting the dependent source and its series resistor to a current source leaves you no further ahead either. So you will need to use another strategy to solve this circuit.

 

[ehild]

When connecting different loading resistors to the output of a linear circuit, the relation between the output voltage and current is represented by a straight line, like in the figure. The idea of Thevenin's Theorem is to replace the circuit with a single voltage source and series resistance producing the same U-I characteristics.As a straight line is determined by two points, it is enough if the open circuit voltages (I=0) and short circuit currents (U=0) are the same for the circuit and its Thevenin-equivalent.

 

[Castello]

Ok, now i see that it's not as general as i was thinking. Thank's! @ehild @gneill