Ltspice helpp

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arthurwinchester

Joined Jun 1, 2023
5
First use the pencil icon to draw all the wires like this:
View attachment 295518
Then use the resistor symbol to place the resistors. When you place a component on top of a wire it removes the short across the terminals of the component. When you grab a symbol to place, the Ctrl-R key combination will rotate the symbol and Ctrl-E will flip the symbol side-to-side.
View attachment 295520
Place a voltage source on the left-hand side using the letter V or select it from the components menu, which is between the diode symbol and the hand symbol. Also place the current source on the right-hand side.
View attachment 295523
Lastly you want to add a behavioral voltage source whose name in the component directory is "bv"
View attachment 295524
Once you get this far then we have to show you how to edit the values, construct a simulation command, and run it.
Hi Papabravo!! thank you a lot for helping me with that circuit. I have another circuit right now that I thought I did it right, but I'm having some trouble getting the same answer as the manual version. Could you take a look plz? 1686539720499.png
this is the problem, I need to find v1,v2 and v3 with nodal analysis. Here is my way of doing it 1686539773799.png
when I run, it gives me this: 1686539828067.png
but the answer should be: V1=-3V, V2=4.5V and V3=-15V. any ideias?
 

WBahn

Joined Mar 31, 2012
32,840
but the answer should be: V1=-3V, V2=4.5V and V3=-15V. any ideias?
Always ask if the answer makes sense. You say that the answers SHOULD be V1=-3V, V2=4.5V and V3=-15V. But have you checked to see whether those answers are consistent with the problem?

One of the nice things about most engineering problems is that the correctness of the answers to a problem can usually be verified from the problem itself.

Assume that the given answers are correct. From those answers, calculate the currents in all of the resistors. Then check to see if the constraints imposed by the voltage sources is correct.

As an example, the voltage output by the CCVS is

V_ccvs = (-3 V) - (4.5 V) = -7.5 V

The controlling current is i, which is

i = -15 V / 4 Ω = -3.75 A

The relationship between them is given as

V_ccvs = 2 Ω · i

Is that satisfied? If not, the answer is wrong. If it is, then it still has a chance of being right, but you need to verify that all of the voltage and all of the currents satisfy KVL and KCL. This only takes a minute or so and is something you need to get in the habit of doing religiously.
 

WBahn

Joined Mar 31, 2012
32,840
View attachment 296251
this is the problem, I need to find v1,v2 and v3 with nodal analysis. Here is my way of doing it
It appears that "your way of doing it" is NOT doing nodal analysis, as specified by the problem, but rather to through the problem as a simulator and hope that it gives you the answer.

View attachment 296252



when I run, it gives me this:
View attachment 296253
but the answer should be: V1=-3V, V2=4.5V and V3=-15V. any ideias?
If I(R3) = -1.5 A, what should the voltage difference between V1 and V2 be, according to the problem schematic?

Does that jive with the simulation results?

Here's a big hint:

The control equation for your controlled source is not properly specified in the problem -- it should be

(2 Ω) · i

Even if you had used the correct kind of source, how do you know whether I(R3) is the proper control current for the source, as opposed to -I(R3)?

Do you know what the polarity rules are for device currents in the simulator?
 

crutschow

Joined Mar 14, 2008
38,508
So using a current-controlled current source looks correct?
Nope.
Missed that it should be a current-controlled voltage source. :oops:
With that change the calculated values appear correct.
Do you know what the polarity rules are for device currents in the simulator?
Although it is assigned, resistor voltage polarity vs. current direction is not apparent in LTspice since it is not shown on the schematic resistor symbol.
I modified the resistor .asy symbol file with a dot to indicate that current going into the dot side gives a positive voltage at the dot to resolve that ambiguity.

1686579730241.png
 

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