Open-circuit voltage in Thévenin equivalent

In summary: This raises me 2 questions:- Did I do something wrong? Was there any inconsistency? Because I don't get why I got the wrong answer. If the textbook answer was $$V_{OC}=V_{AB}=-5V$$ I would get it but this way. no... - And this bring me to my 2nd question: how do I choose the direction of the open circuit voltage? I arbitrarily chose the BA direction but I don't know if I'm allowed to chose this arbitrarily. I don't see anything meaning that I can't choose the AB direction. How can I choose?It would be easier if you used source transformation. Convert the current sources into voltage sources and reduce the circuit into
  • #1
Granger
168
7

Homework Statement


Determine the Thévenin equivalent of the following circuit with respect to terminals A-B

fLF0lgo.png


Homework Equations


$$U=RI$$
The sum of the currents in a node is zero. The sum of voltages in a mesh is zero.
I'm always considering the currents in the resistors directed downwards. The open circuit voltage in directed upwards (BA direction).

The Attempt at a Solution


So because A-B is opened, there is no current in the 1.5k resistor, so we can erase it.

Than we apply the superposition principle:

1 - We turn off the voltage source with a short circuit and the 15mA current source with a open circuit.

Therefore we have all the resistors and the current source in parallel.
The 2k and 2k resistors in the right are in parallel so we can change it to a 1k resistor.
Now using KVL and KCL

$$I1 + I2 + 30mA=0$$
$$1k\Omega I2-1k\Omega I1=0$$

Solving it we get to $$I1=-15mA$$
$$V_{OC1}=V_{BA}=-(-15mA)1k\Omega=15V$$

2 - We turn off the current sources with open circuits.

Applying KVL and KCL
$$I1 + I2 + I3=0$$
$$2k\Omega I2-1k\Omega I1=0$$
$$10V + 2k\Omega I3-2k\Omega I2=0$$

Solving this system we have
$$I2=1.25mA$$
$$I1=2.5mA$$

$$V_{OC2}=V_{BA}=-(2.5mA)1k\Omega=-2.5V$$

3 - We turn off the voltage source with a short circuit and the 30mA current source with a open circuit.

The 3 middle resistors are in parallel so we can change it to a single resistor of 0.5 k.

Therefore the current across this resistor is exactly 15 mA.

$$V_{OC3}=V_{BA}=-(15mA)0.5k\Omega=-7.5V$$

Combining this
$$V_{OC}=V_{OC1}+V_{OC2}+V_{OC3}=5V$$

However my book gives us the answer $$V_{OC}=V_{BA}=-5V$$

So I have a signal wrong...
This raises me 2 questions:
- Did I do something wrong? Was there any inconsistency? Because I don't get why I got the wrong answer. If the textbook answer was $$V_{OC}=V_{AB}=-5V$$ I would get it but this way. no...

- And this bring me to my 2nd question: how do I choose the direction of the open circuit voltage? I arbitrarily chose the BA direction but I don't know if I'm allowed to chose this arbitrarily. I don't see anything meaning that I can't choose the AB direction. How can I choose?
 
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  • #2
It would be easier if you used source transformation. Convert the current sources into voltage sources and reduce the circuit into a single loop.
 
  • #3
I get the same answer you did. Is the book's convention that ##V_\text{BA} = V_\text{B}- V_\text{A}##? If so, it's probably just a typo in the book's answer.
 
  • #4
Let's say the voltage across any resistor will be [typical]=VFB,VDB,VCB.
At first diagram -supplied from VCD=10 V VFB1,VDB1 etc.
from second diagram supplied form current source of 30 mA VFB2,VDB2 etc.
and at the third VFB3,VDB3 etc.
The sign[+/-] will be according the current direction: entering the node + leaving -.
For current source the current will split into 3 branches -1kohm,2 kohm,2kohm according to an equivalent resistance of 3 parallel Req=1/(1/1+1/2+1/2) multiplied by Isource[15 or 30 mA] and divided by the branch resistance value.
The voltage source [10 V] current will be V/Req [this Req=2+2*1/3=2.667] ohm.
The result is:
At first diagram -supplied from VCD=10 V VFB1=VDB1=2.5 VCB1=-7.5 V
from second diagram supplied form current source of 30 mA:
VFB2=VDB2=VCB2=15 V
At third diagram VFB3=VDB3=VCB3=-7.5 V
The total equivalent VFB=VFB1+VFB2+VFB3=10
VDB=VDB1+VDB2+VDB3=10
VCB=VCB1+VCB2+VCB3=0
Viewed from A-B Rth=1.5+1/(1/1+1/2+1/2)
I=IAF=0 if A-B is open and V=VFB=10V
Rth=2.667 Vth=10
I=VFBnew/1.5 if VAB=0 [A-B short-circuit - in this case you have to introduce a fourth branch of 1.5 kohm parallel and recalculate at all from the beginning.
upload_2017-8-21_11-49-14.png

upload_2017-8-21_11-49-49.png

upload_2017-8-21_11-50-11.png
 
  • #5
Sorry. The conventional current direction of a battery as a source it is from "+" to "-". I took it as a receiver [in charging process]. Actually VFB1=-2.5 and :sorry:then VAB[=VFB]=5 V indeed.
 

Related to Open-circuit voltage in Thévenin equivalent

1. What is open-circuit voltage in Thévenin equivalent?

The open-circuit voltage in Thévenin equivalent is the voltage measured across the output terminals of a circuit when there is no load or current flowing through it. It is the maximum voltage that a circuit can provide to a load.

2. How is open-circuit voltage calculated in Thévenin equivalent?

The open-circuit voltage in Thévenin equivalent can be calculated by first finding the Thévenin voltage (VTH) and the internal resistance (RTH) of the circuit. The open-circuit voltage is then equal to the Thévenin voltage.

3. What is the significance of open-circuit voltage in Thévenin equivalent?

The open-circuit voltage in Thévenin equivalent is important because it represents the maximum voltage that a circuit can provide to a load. It is also used to simplify complex circuits into a single voltage source and a single resistor, making it easier to analyze the circuit.

4. How does open-circuit voltage affect the behavior of a circuit?

The open-circuit voltage in Thévenin equivalent does not affect the behavior of a circuit as long as the load connected to the circuit has a higher resistance than the Thévenin resistance. However, if the load resistance is lower, it can cause the circuit to behave differently and the open-circuit voltage will have a significant impact on the output voltage and current.

5. Can open-circuit voltage be measured in a real circuit?

Yes, open-circuit voltage can be measured in a real circuit by using a voltmeter. The voltmeter should be connected across the output terminals of the circuit with no load connected. This will give the open-circuit voltage value of the circuit.

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