OK new circuit.



First I created my equations:
V1 = I1*5.4kΩ + I2*1.2kΩ = 25V therefore 5.4x + 1.2y = 25
V2 = I1*1.2kΩ + I2*2.2kΩ = -75V therefore 1.2x + 2.2y = -75

Using dual simultaneous equation solver:
The goal here is to solve
5.4x+1.2y=25 and
1.2x+2.2y=-75 for the variables x and y.

The solutions to your equations are:
x= 13.8889 and y= -41.6667
(so I1 = 13.89mA, I2 = -41.67mA, I3 = I1+I2 = -28mA which are wrong according to the answer key)

And from LTS:
--- Operating Point ---
V(n001): 25 voltage
V(n003): 75 voltage
V(n002): 39 voltage
(at this point I assume V(n001) = V1, V(n003) = V2, V(n002) =??? what is this node?)

I(R3): -0.0325 device_current
I(R2): 0.036 device_current
I(R1): 0.0035 device_current

I(V2): 0.036 device_current
I(V1): 0.0035 device_current
(this tells me V1/I1 = R1+R3, so 25V/0.0035A = 7142Ω which should be 5200Ω?)
(and V2/I2 = R2+R3, so 75V/0.036A = 2083Ω which should be 2200Ω?)

And from the answer key:
VR1 = 58V
VR2 = 42V
VR3 = 33V
I1 = 14.5mA
I2 = 42mA
I3 = 27.5mA

So I am totally lost here...

 

As a side issue... there is no ASCI code for the "therefore" symbol but there is a UNICODE U+2211. I tried the ALT(holding down) +2211(on the num pad) and get ú instead of "therefore". I get the same thing even without entering the + on the num pad. ú

Any pointers here?

 

Below is my simulation results.
I labeled the nodes, so they can be readily identified.

Why did you invert V2 and then give it a negative voltage (which is the same as it being non-inverted with a positive voltage)?
That may be why your equations don't work.
But I'm not much into doing math, so you'll have to figure that out for yourself.

 

Show us the exact circuit from the book without your integration.
Because as it was already pointed out by crutschow the V2 is now a positive voltage source (with respect to ground ).

 

I suspect that you are trying to analyze this circuit:



As you can see I drawn the loops (mesh) currents.

The red is I1 and the blue is I2.

So I'm starting to writing the KVL by starting at point A

I1*R1 + (I1 + I2)*R3 - V1 = 0 (1)

I1*4kΩ + (I1 + I2)*1.2kΩ - 25V = 0 (1)

And the KVL for I2 but this time I'm starting at B (I can start from any point I want).

(I2 + I1)*R3 + V2 +I2*R2 = 0 (2)

(I2 + I1)*1.2kΩ + 75V +I2*1kΩ = 0 (2)

Solving this I obtain

https://www.wolframalpha.com/input/?i=x*4+++(x+++y)*1.2+-+25+=+0+;+(y+++x)*1.2+++75+++y*1+=+0

I1 = 14.5mA ; I2 = - 42mA

And the negative sign in the answer means that, in reality, I2 current is flowing in the opposite direction to the one I have drawn.

And I3 = I1 + I2 = 14.5mA + (-42mA) = - 27.5mA

Which again means that I3 is flowing in the opposite direction to the one I drew.

The real situation is shown here

 

Fig. 9-33

 

OK for clarity lets take one solution at a time. Simultaneous equations first.

I1*R1 + (I1 + I2)*R3 - V1 = 0 (1)

I1*4kΩ + (I1 + I2)*1.2kΩ - 25V = 0 (1)

which is: I1*5.2kΩ + I2*1.2kΩ = 25v so 5.2x +1.2y = 25 which is what I had

(I2 + I1)*R3 + V2 +I2*R2 = 0 (2)

(I2 + I1)*1.2kΩ + 75V +I2*1kΩ = 0 (2)

which is: I1*1.2kΩ + I2*1kΩ=-75V so 1.2x + 2.2y = -75 also what I had

Using the wolframalpha calculator:

which is correct

using the webmath calculator:
The goal here is to solve
5.2x +1.2y=25 and !!! found the error I put in 5.4x+1.2y=25 previously
1.2x + 2.2y=-75 for the variables x and y.

The solutions to your equations are:
x= 14.5 and y= -42 now the correct answer

Now the rest will calculate correctly!!! This is why I was using the calculator to reduce manual calculation errors but if I don't key it in correctly it will do exactly what I told it to do and garbage in = garbage out!

 

BTW Jony I much prefer your calculator. Much easier data entry and I particularly liked the graphing of the resultants. Now a short break and onto LTS to get it also working the solution. On your diagram I have become accustomed to the electron flow method instead of conventional. Took a while to remap my synapses but used to it now.

 

On your diagram I have become accustomed to the electron flow method instead of conventional. Took a while to remap my synapses but used to it now.

During circuit analysis like this one, I do not think about conventional/electron flow. Because we can pick the current direction arbitrarily. You can choose any direction you like it doesn't matter at all.

. Now a short break and onto LTS to get it also working the solution

Just remove the minus sign from V2 and you done.

 

Corrected the V polarity for V2...


Now:


OK that looks good except that I(R1) should be +0.0145?
With Op Pt I cannot point and click with the probe tool for V and I?

Having a hard time getting my head around opposing polarity nodes. But that will come with practice and experience gained.

 

During circuit analysis like this one, I do not think about conventional/electron flow.

As I have already learned what I assume is not always right about current flow until the math supports it. Particularly with these opposing voltage nodes. Why am I thinking opposing? They are additive? Except for the crossover on R3.

 

except that I(R1) should be +0.0145?

In LTspice resistor will have a "polarity", flipping the resistor will fix this.
https://forum.allaboutcircuits.com/threads/direction-of-current-flow-in-ltspice.154700/

With Op Pt I cannot point and click with the probe tool for V and I?

For sure you can that. Just click on the node to see the voltage.

 

In LTspice resistor will have a "polarity", flipping the resistor

Ah yes forgot to check that. Argghh once again struggling with flow direction but that will come with more practice.

For sure you can that. Just click on the node to see the voltage.

Ahhh... I was expecting to see the probe instead of the crosshairs. It doesn't give me the I for the device but with the nodeV/R it's there easily enough.

Yes I watched that video some time ago but it was a good refresher.

OK time to put branch currents down as done and now onto Node-Voltage Analysis. Again many thanks to you and Crut for the help. I really do appreciate your time and assistance.

Sam