Thévenin equivalent circuit

[Nadun Palmada]

Hi guys can someone please help me with the following problem:

I have figured out the Rth by turning the current sources to open and voltage sources to closed loop and i got this to be 50ohm but not sure how to do the Vth.
Thanks

 

[(*steve*)]

you need any two of:

- open circuit voltage
- short circuit current
- equivalent resistance

from any 2 you can get the third, and that leads to a fairly simple Thevenin or Norton equivalent.

 

[Nadun Palmada]

I am not sure how to work out the open circuit voltage! I have come to these assumptions: No current will flow through either the 4ohm or the 6ohm resistor. Therefore would i jst take the voltage drop across the 40ohm to find the Vopen circuit?

 

[hevans1944]

I am not sure how to work out the open circuit voltage! I have come to these assumptions: No current will flow through either the 4ohm or the 6ohm resistor. Therefore would i jst take the voltage drop across the 40ohm to find the Vopen circuit?

Don't forget to include the 520 V that is dropped across the 260 Ω resistor. This voltage is in series with the voltage dropped across 40 Ω resistor to determine the open-circuit voltage between nodes A and B.

 

[Nadun Palmada]

So would Vth just be 520+40 = 560v?

 

[GPG]

what polarity is the voltage across 40Ω?

 

[Nadun Palmada]

what polarity is the voltage across 40Ω?

Oh i see! since the current is going from bottom to top, the bottom will be + and the top will be -, thus it will be 520-40= 480v?

 

[GPG]

Tick