Thevenin's Equivalent Help

[sjgallagher2]

For some reason I find immense difficulty in figuring out these circuits. I must be missing something somewhere Here's my issue. I have this circuit:

And I need to find the Thevenins equivalent circuit. When I first got the problem I confidently threw some voltage divider equations around and got the thevenin voltage as 0.75V. But that's wrong. It's actually 0.5V. Why? Thats where I need help! If you could explain that to me I would be very thankful.

More info on what I did to get my answer.
First I used a voltage divider equation on the 200ohm resistors. That resulted in 1/2(3V) = 1.5V. Then I put that voltage in the next voltage divider equation with the 100ohm resistors, which came out as 1/2(1.5) = 0.75. It seemed so simple, but I went wrong. Where?

Someone suggested putting the two 100ohm resistors in series and solving from there but I'm not sure that's the right thing to do

 

[keith1200rs]

You started at the wrong end, I think.

Start at the right hand side. You have two 100 ohms in series making 200 ohms. They are across 200 ohms making 100 ohms. That is in series with 200 ohms which is across 3V. So the 3V is multiplied by 100/(100+200) = 1V. The output potential divider halves that so 0.5V.

Keith

 

[sjgallagher2]

But where do the contacts go for that inbetween step when you combine the two resistors? If you sum the two hundred ohm resistances doesn't that put one of the contacts in the middle of the resistor now? Here's a picture of what I think you're saying the second step is:

I didn't put the A and B contacts because I'm not sure where they go now that we've combined those two resistors :S

 

[keith1200rs]

Yes, you diagram is correct and from that you can hopefully see there is 1V at the junction where all 200 ohms resistors meet( 3* 100/300)?

So, then you split the 200 ohms back to 100+100 so the voltage is halved = 0.5V.

Keith

 

[SanjKrish]

**broken link removed**

Thevenin voltage is the voltage felt at the terminals of A and B. When u connect a voltmeter across A B the measured voltage will be thevinin voltage..

In order to find it analytically first combine all the resistors from load to voltage source such that u end up with one voltage source and one resistor.. now expand the resistors one at a time and find the voltage across each. Thus u can get the voltage across the load terminals.. Check the diagram for better explanation

 

[Marvel_tronix]

But where do the contacts go for that inbetween step when you combine the two resistors? If you sum the two hundred ohm resistances doesn't that put one of the contacts in the middle of the resistor now?

At this in between step(perfectly said), you don't consider about the terminals A & B. Taking the 100Ω, the one across AB, in series with the other one is the consequence of this. The calculated resistance of 100Ω[=((200Ω)||(200Ω))], that comes in series with 200Ω, is actually the equivalent resistance that comes into effect for the voltage divider rule. Remember that voltage divider can be applied only to resistances connected in series with no other branch originating from any of internal node(s) connecting to some another element or network. So, we have to find the effective resistance b/w those two points A & B.
Also, the voltage calculated now using the voltage divider is the voltage that is present at the node connecting the 2 200Ω & 1 100Ω resistances.
This method would be continued even if some more such ladder n/ws(resistance connected in ladder form) are presented.

 

[pebe]

Point A is effectively fed with a voltage of 0.5V, relative to B, through a 66.67ohm resistor.