# Thevenin Equivalent Circuits

#### werd2501

Joined Aug 28, 2014
4
How do I solve this?

Not sure how I begin solving this. I'm supposed to draw the thevenin equivalent circuit for this (as well as Norton). I'm used to the source being a voltage, but I'm not sure how to deal with this one.

Do I first find the open-circuit voltage?

Someone has told me that the 0.5*30 = 15V. Then they used the voltage divider rule on the 40 ohms resistor multiplied by 15 to get the answer of 6V.

I don't understand why this is it. Could someone explain this to me?

Isn't the 0.5A source in parallel with a 30 ohms and 70 ohms resistors? So how is the voltage divider law done as though the resistors are in series..?

Thanks.

#### anhnha

Joined Apr 19, 2012
880
Do I first find the open-circuit voltage?
Yes. You need to find open-circuit voltage (Thevenin voltage) and Thevenin resistance.
Someone has told me that the 0.5*30 = 15V. Then they used the voltage divider rule on the 40 ohms resistor multiplied by 15 to get the answer of 6V.

I don't understand why this is it. Could someone explain this to me?
The person is using Thevenin-Norton equivalencies.
And from that you convert a current source in parallel with a resistor into a voltage source in series with that resistor.
You can read it more here:

#### werd2501

Joined Aug 28, 2014
4
I figured maybe I should go back to basics... so...

I'm looking at this textbook with a basic exercise and I just feel like I'm missing something.

What even?

The open-circuit voltage is the voltage at R1? Is that V=IR = 10*5=50? Short circuit current is still 5A? R_T = V_OC / I_SC = 50 / 5 = 10? I don't know. o.o

#### t_n_k

Joined Mar 6, 2009
5,455
You correctly stated that the open circuit voltage is 50V - the 5A flowing in R1 produces 50V. No current in R2 means the same voltage appears across the open terminals.

The short circuit current is not 5A. The 5A will split between R1 & R2 now effectively in parallel, with the short circuit current being the same as that which would flow in R2. Use the current divider rule.

With R1 & R2 in parallel

$$I_{R_2}=I_{short \ circuit}=I_{source} \times \frac{R_1}{(R_1+R_2)}$$

#### werd2501

Joined Aug 28, 2014
4
Okay, so that's 1A. V/I=R

50/1 = 50 ohms.

-----

I kinda ended up working backwards in the first problem and finding the thevenin resistance as (20+30+10)//40 = 24 Ohms.

I still am confused on how to work out the open circuit voltage in the first problem though.

#### t_n_k

Joined Mar 6, 2009
5,455
OK - re the first problem and finding the Thevenin voltage.

As in the last problem consider how the source current would divide among the various circuit elements.

You'll note the 20 ohm, 40 ohm and 10 ohm are effectively in series with the output terminals open. That makes 70 ohm total. This 70 ohm is effectively in parallel with the 30 ohm. So the 0.5A source current splits between the two branches with

$$I_{30}=\frac{70}{70+30} \times 0.5 = 0.35 \ A$$

$$I_{70}=\frac{30}{70+30} \times 0.5 = 0.15 \ A$$

So there is 0.15A flowing in the 20 ohm, the 40 ohm and the 10 ohm. What's the voltage drop across the 40 ohm resistor with 0.15A current? This is the Thevenin voltage.