# A circuit question about Thevenin's Theorem

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#### v9260019

##### Member level 2 thevenin theorem calculate voltage 2 loop

Hello All The circuit Only have dependent supplies ;
If i want to find the Thevenin's equivalent circuit from a,b.That is, finding Roc and Vth
my question is: why "Vth' must =0

Thanks a lot

#### v_c viva questions on thevenin theorem

Well, Vth is defined as the open circuit voltage that you see at terminals a-b.
OK, just try to calculate this voltage. The voltage across the 10Ω resistor is 10I. This voltage also equals Vab=Vth. Now, if you can calculate the value of I, you just multiply by 10 and that is your Vth, right? OK. Let's try to find the current I. Due to the open circuit, the current flowing in the 5Ω resistor (from left to right) is also I. So the voltage across the 5Ω resistor is 5I (+ 5I -), from left to right. So let's now write a loop voltage equation using Kirchoff's voltage law. Using a clockwide loop, we can say that sum of voltage rises = sum of voltage drops and this gives
voltage rise in dependent source = voltage drop in 5Ω + voltag drop in 10Ω
2I = 5I + 10I or
0 = 13I so I=0 and Vth=10I = 0

The moral of this story is that if all you have are dependent sources, eventually you can write all the voltages and current in terms of the dependent variable. And when you write any KVL or KCL equations, you will get one like the above where the only value of the variable satisfying the equations is making the varialbe equal to zero.

Best regards,
v_c

#### fireflies

##### Newbie level 5 choosing thevenin circuit

To calculate Rth, you need to set all voltage and current sources to zero ( short circuit and open circuit, respectively) and then you can find the resulting resistance between a and b.

Simply, Vth = 0 will give you a short circuit which is needed to perform Thevenin's theorem.

#### v_c use test source to solve thevenin resistor

fireflies, you cannot do that directly in this circuit. You are right,
to find Rth, you have to set the independent sources to zero, but note that this circuit has no independent sources. Do not short circuit the 2I source!

to get Rth, you have to attach at test source at the output terminals a-b. So put a test source of Vtest=10V (with positive at a and negative at b). Then Rth=Vtest/Itest, where Itest is the current coming out of the positive end of Vtest. Why did I choose 10V? Actually, you can choose anything, but with Vtest=10V, the voltage across the 10Ω resistor is 10V so I=1A. With I=1A, the 2I dependent voltage source becomes a voltage source of 2V. Now the voltage across the 5Ω resistor is 8V (- 8V +), with more positive end at the right end of the resistor. So the current flowing in the 5Ω resistor is 8V/5Ω= 8/5 A (this current is flowing from right to left). Now you can calculate Itest (the current that is leaving the + end of Vtest). Itest = current in the 10Ω (from top to bottom) + current in the 5Ω (from right to left). So Itest = 1+8/5 = 13/5A. Now Rth=Vtest/Itest=10V/(13/5)=(50/13)Ω≈3.85Ω

So the thevenin equivalent circuit is just a resistance since Vth=0 (short circuit).

Hope this helps.
Best regards,
v_c

#### ziyas

##### Member level 1 thevenin dependent source

your circuit has not an independent voltage or current sourse and thevenin equivalent shows us the basic circuit of complex circuit.

#### v_c solved sample thevenin circuits

ziyas,
Are you making a comment or a statement? I already said there are no independent sources in the circuit and already solved the problem in the previous post.

#### v9260019

##### Member level 2 thevenin theorem

Hi v_c In above figure, if let the current source be a dependent current source 5Ix.
Using KVL or KCL, i will get :5Ix=5Ix or 4Ix=4Ix. In this case any value of Ix will work~~~right??? And i cannot get Vth=0
What's the problem
Thanks

#### v_c solved circuit of thevenin theorem

Now, if you wanted to work on this circuit, you would need to write two nodal equations. For this you would make the bottom node (node b) reference (0V) and write node equations (Kirchoff's current law) at node a and node to the right, call it node c. You will get two equations and two unknowns (va and vc). Solve for voltage va and that is your voltage Vab or Vth. But there is an easier way...

OK, take a look at the figure that I uploaded. The circuit at left is your original circuit. To make the circuit simpler, I did a little trick. I took the circled area (which looks like a Norton equivalent circuit) and converted it to a Thevenin equivalent circuit. When you do that, you get the second circuit where the value of the dependent voltage source is (5Ix)*(1$\Omega$)=5Ix (which is in volts). Now if you combine the two resistors in series and move the source, you will see the simplified circuit which is the one at the right. Now this exactly looks like the circuit you had originally asked about -- only the values are different. Can you do it now? I will let you write the equations and prove to yourself that Vab=Vth is indeed 0V. Now put the test source as before and find Rth. The value of Rth might surprise you!

Best regards,
v_c

Answer: $R_{th}=\infty, V_{th}=0$. So the Thevenin equivalent circuit looking into a-b is just an open circuit!

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