# Thevenin Equivalent With Two Voltage Sources

I've been working through a few examples of Thevenin Equivalent circuits, and came across this one.

Find the Thevenin Equivalent circuit as seen by a load resistance between points A and B.

Finding the equiv. resistance wasn't too hard (75ohms), but I'm having trouble calculating the equiv. voltage.

I've attempted the questions, but am not sure I got the correct answer. Here's my working.

The total voltage of the circuit is 12V because of two aiding voltage sources being in series. From this, you can use a voltage divider between the two 150ohm resistors, getting a value of 6V. Another voltage divider can be used to find the voltage at node B, that being 2.4V.

Not sure I've done the next thing correctly -

I've assumed that Node A is 3V, and since the Thevenin voltage is the difference between Node A and B, Thevenin Voltage = 3-2.4 = 0.6V. Have I done this question correctly? Anywho, any help is greatly appreciated.

• You seem to be trying to solve this as one problem. I would split it in 2 simple ones: 1) 9V + 150 ohm + 150 ohm and 2) 3 V + 100 ohm. Now make Thevenin equivalent for each. What do you get ? for 1 you get 300 ohms in parallel with a current source and for 2 something similar. Now redraw your schematic and note how only node A and B are remaining making the problem trivial to solve. Dec 15 '15 at 15:46
• There is not 12 V across the two 150 ohm resistors. Dec 15 '15 at 15:58
• @Gregd'Eon The source voltage is 12V, I don't think I mentioned there was 12V across the two 150 ohm resistors. Dec 15 '15 at 16:01
• That means you are misunderstanding voltage dividers. The voltage between the two 150 ohms is not 6 V. Dec 15 '15 at 16:03
• Oh, my bad! The voltage divider should be 12 * ((150+100)/150 + 250) - giving 7.5V? Dec 15 '15 at 16:08

I think there was some confusion in the comments, so I'll write up an answer.

The circuit we're talking about is: simulate this circuit – Schematic created using CircuitLab

Note that I've left out the load - it has nothing to do with your Thevenin equivalent.

Your goal is to make a circuit that looks like: simulate this circuit

so you need to find two things: the equivalent resistance and the Thevenin voltage. You already found $R_{eq} = 75 \Omega$, so I won't go over that.

The Thevenin voltage of a circuit is the same as the open circuit voltage: when you leave the load disconnected, $V_{th} = V_{ab}$. That means that all you need to do is find $V_a$ and $V_b$ with no load.

$V_a$ is easy - it's just 3 V - so the harder part is $$V_b = (3V + 9V) \frac{100 \Omega}{150\Omega + 150\Omega + 100\Omega} = 3 V$$ so $$V_{th} = V_a - V_b = 0 V$$

That means the Thevenin equivalent of this circuit is just a 75 ohm resistor - there's no voltage across the terminals.

• This the correct method approaching a Thevenin/Norton problem, in every sense. The comments below the question made the problem more confusing for the asker than it should be. Dec 15 '15 at 17:29
• I agree - there is no need to use the Thevenin transformations. It's usually much simpler to find any 2 of V_oc, I_sc, and R_eq. Dec 15 '15 at 17:30
• @Gregd'Eon Thanks for clearing things up - may I ask though, what are Thevenin transformations? Has that something to do with the Norton Equivalent? Dec 15 '15 at 17:44
• Exactly. You can switch back and forth between Thevenin and Norton equivalents without affecting the rest of the circuit. This lets you do some things more easily -- sometimes doing these transformations puts components conveniently in series/parallel -- but it's not really a bulletproof method. Dec 15 '15 at 17:50

here we can use superposition theorem.when 9v is active and 3v is shorted,we should use voltage devider rule to find the voltage of 100 ohm.this is 9/4(B+ & A-).now,3v is active and 9v is shorted,we again use voltage devider rule to find the voltage of (150+150)ohm.this is 9/4(B- & A+).so totally v_th=9/4-9/4=0