Engineering Open-circuit voltage in Thévenin equivalent

[Granger]

Homework Statement

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

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?

 

[cnh1995]

It would be easier if you used source transformation. Convert the current sources into voltage sources and reduce the circuit into a single loop.

 

[vela]

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.

 

[Babadag]

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.

 

[Babadag]

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 then VAB[=VFB]=5 V indeed.