# Thevenin Equivalent Problem

1. Jan 22, 2010

### mmmboh

1. The problem statement, all variables and given/known data

I am being asked to draw the Thevenin equivalent for

and to calculate the voltage that would appear across RL if it were connected across the 2 output terminals.

Using Kirchoff's law I found that the voltage across RL is -12.5 V. However I am really suppose to do it using Thevenin but I can't figure out how to find the equivalent.

I found the voltage at the node between the two resistors to be 25 V because I1+I2=0, so (50-V1)/5+ (-V1)/5=0, which means V1 = 25V, and I1=5A. Then I found the voltage across the output terminals to be 50V-5Ax5ohm-5Ax10=-25V.
This would mean that Veq=25V...which I believe is wrong, and I also found Req to be 5ohm...which is probably also wrong...

Thanks :)

2. Jan 22, 2010

### ideasrule

An easy way to do this is to notice that when nothing is connected between the ou tput terminals, the only closed circuit consists of the 50V battery and the two resistors. The voltage drop across the first is logically 25V, so the top terminal must be 15V.

As for the equivalent resistance, when all batteries are shorted out, the two resistors are arranged in parallel.

3. Jan 22, 2010

### mathman44

You mean 25V for the top terminal right? And, the two 5 resistors are in parallel when the battery is shorted?

What is the effect of reducing the circuit to one loop on the controlled voltage source, which is affected by I1? I can't make sense of this...

4. Jan 22, 2010

### mmmboh

Isn't there another voltage drop when you pass the current controlled voltage source, so the voltage at the top terminal would be 25V-10I1? I calculated I1 to be 5A, so that means the voltage would be -25V at the top terminal....I think I am doing something wrong :S...and the equivalent resistance I calculate to be 2.5 ohm, I think that is right, but I am confused about how to go about the other stuff.