I am a bit confused with this circuit.I uploaded the circuit as you can see I already removed the resistance where I need to calculate the thevenin resistance. I know how to find the Thevenin voltage. Our mission is to find the
Thevenin equivalent circuit at R1 resistance.
simulate this circuit – Schematic created using CircuitLab
I know how to find the Thevenin voltage but if I don't understand how to find the Thevenin resistance.
Let is remove the voltage sources:
As you can see I already removed the R1 resistance and named the nodes A and B.
I know that the R7 and R3 in series connection.But I don't understand what to do with R6 R4 and R2 they are in star connection.
My quenstion it is useful to use star-delta transformation ? If yes how to do it properly?
Edit: I want this circuit to be reduced to this:
As I mentioned above I know how to find voltage I need the Rth and star-delta transformation is little confusing.
Best Answer
There are two ways to find Thevenin's resistence. The first one is how are you doing, trying to find the equivalente circuit saw by R1 resistance (just short circuit all independent voltage source and open circuit all independen current source).
In your case, it is not easy to see the equivalent circuit saw by A and B terminal, so let us try the second way.
The second way is, from your original circuit, make a short circuit between terminals A and B (i.e. a short circuit between the terminal over the R1 resistor), and calculate the short circuit current (Isc) that pass between those terminals. The Thevenin's resistance will be:
$$ R_{th} = \frac{V_{th}}{I_{sc}} $$
EDITED:
I really did not understand what you have commented, so I'll try to explain a bit better what I have said.
To calculate Thevenin equivalence from this circuit:
simulate this circuit – Schematic created using CircuitLab
You calculate Vth and Isc
$$ V_{th} = 40 + I_2 * 40 + I_1 * 20 $$ where $$ I_1 = \frac{ \begin{vmatrix} 60 & -20 \\ -20 & 80 \end{vmatrix}}{ \begin{vmatrix} 120 & -20 \\ -20 & 80 \end{vmatrix}} = 0.4783 ~A $$ $$ I_2 = \frac{ \begin{vmatrix} 120 & 60 \\ -20 & -20 \end{vmatrix}}{ \begin{vmatrix} 120 & -20 \\ -20 & 80 \end{vmatrix}} = -0.1304 ~A $$ So, $$ V_{th} = 44.35 ~V $$
To calculate Isc, just short circuit terminals A and B, and calculate Isc
simulate this circuit
$$ I_{sc} = \frac{ \begin{vmatrix} 120 & -20 & 60 \\ -20 & 80 & -20 \\ -20 & -40 & 40 \end{vmatrix}}{ \begin{vmatrix} 120 & -20 & -20 \\ -20 & 80 & -40 \\ -20 & -40 & 60 \end{vmatrix}} = 1.3784 ~A $$
In this way: $$ R_{th} = \frac{44.35}{1.3784} = 32.1755 ~ohm $$
Sorry to solve your exercise, but I didn't find another way to explain this.