# Help needed in a particular example of Thevenin case

1. Dec 10, 2004

### abhinay

I tried every which way, but am getting a wrong answer for the Thevenin equivalent voltage in the following circuit. I tried Norton too, but got a different answer and neither of them matches the correct answer from the book. Please help me ........ I'm too frustrated :(

Here's the circuit : ..... All resistance values are in Ohms. Small info : equivalent resistance of the circuit is 10 ohms

http://img131.exs.cx/img131/9840/a140.jpg

And btw....... u don't need to calculate the values and stuff....... If anyone can just tell me algebraically what's happening :) But please help asap :) I got only one day b4 my exam

2. Dec 10, 2004

### learningphysics

I got 0.704V for the open circuit voltage. Is this correct?

What I did was, create open circuit where the R is... Now you can ignore the 2.4 ohm resistor Then calculate what V is at the node where the 100ohm and 40ohm resistors intersect. Use KCL here:

To make it easier combine the 40,20,20 and 10ohm resistors into one:

I get 9.23 ohms.

So the KCL equation is:

V/9.23 + (V-50)/100 + 0.50 -0.25= 0

I get V=2.11 volts

Use voltage divider to get the open circuit voltage (voltage across the 10ohm resistor)
OC voltage= 2.11 (10/(10+20))=0.704V

Let me know if the answer is right.

Hope this helps. Do you see the way to get equivalent resistance? You can either find OC voltage, and short circuit current then divide... or a quicker way... looking into the circuit create a short circuit for the voltage sources and open circuit for current sources, and calculate the resistance you see looking into the circuit... I get 10 ohms.

3. Dec 11, 2004

### rayjohn01

I get 2.11 volts and 10.85 ohms

4. Jan 24, 2005

### heman

i could not grasp the concept of open circuit voltage here ...just after calcutating the resistance..plss help in this.

5. Jan 28, 2005

### learningphysics

Sorry for not answering this earlier. I created an open circuit where R is... For thevinin equivalence you need the open circuit voltage here. So remove R, and calculate what the voltage is across the two points where R used to be. The bottom point is ground 0 volts. We need to voltage at the upper point.

So after I calculate V, which is the node where the 100ohm and 20ohm resistors meet... I use voltage dividier to get the open circuit voltage. Remember that 0 current goes through the 2.4 ohm resistor (open circuit here). So the open circuit voltage is 10/(10+20) * V. Or you can get the current going through the 10 ohm resistor first... this is (V-0)/30. Voltage across the resistor is then (V/30)*10.

I hope this helps. Let me know if you have more questions.

6. Jan 28, 2005

### dduardo

Staff Emeritus
Source transformation works very nicely. I get Rth=10 Ohm and Vth=1.9 V.

1) Convert the 50V and 100 Ohm in series to 1/2 A Up and 100 ohm in parallel.
2) The 1/2 A Up cancels with the 1/2 Down.
3) 100 Ohm || 40 Ohm || 20 Ohm = 200/17 Ohm
4) Convert the 1/4 A Up and 200/17 Ohm in parallel to 50/17 V and 200/17 Ohm in series
5) 200/17+20 = 540/17
6) Convert 50/17 V and and 540/17 Ohm in series to 1/4 A Up and 540/17 Ohm in parallel
7) 540/17 || 10 = 540/71 Ohm
8) Convert 1/4 A Up and 540/71 Ohm in parallel to 135/71 V and 540/71 Ohm in series
9) Rth = 540/71 +2.4 = 10 Ohm
10) Vth = 135/71 = 1.9 V

7. Jan 28, 2005

### learningphysics

You made a mistake here in step 6. It's not 1/4 A. It's 5/54 A.