Hi,

Can anyone help me simulate this exercise on Ltspice? I've learned to use it but can't work with it very well yet.

I would like to calculate Va(t) with the simulation.



With this I got an empty simulation
Is this correct: 60cos(50t + 20°) = 60sin(50t - 70°)?
50t = 50 / 3.14 = 78.53Hz
60cos(50t + 20°) = 60sin(78.53Hz - 70°)?

Thanks in advance

 

hi 442,
It helps if you post your LTS asc file if you have a sim query.
E
post your LTS asc file

 

hi 442,
It helps if you post your LTS asc file if you have a sim query.
E
post your LTS asc file

Ok, here it is

 

hi 442,
As you are still learning LTS.
Create a Voltage source simulation circuit for the Answer to the question, simulate and post the plot result for:

Va(t) = 95.83 cos(50t + 24.1)

I can then see if you understand the method.

E

 

Srry but I don't understand it so well when I simulate I get this

 

hi arthur,
As this is Homework I can only give hints, not answers.

Look at this sim of the question I asked, note the V source type is a BV

E,
BTW: when you do the final part, note the Gnd ref is at the bottom of the right side of the circuit.

 

Just for reference, here is a cleaned up drawing, the first schematic shown in this thread with an apparent "answer". We are in the process of looking over that answer also with an update on the way

 

Hello again,

I meant to get back here sooner but i guess the OP hasnt come back yet anyway.

Here is a simulation of the circuit showing the desired output without showing any numerical data yet.
The simulation is the first pic, the second pic is another simpler way to analyze the circuit by rearranging some of the components a little which then makes it a ground referenced voltage shown as "Va" in the second schematic. By ground referencing the two sources, it turns the analysis into a two source voltage divider problem where we can use superposition for the two sources. It's a lot simpler then. It is also informative to do it the long way too though with the original schematic.
Ignore the two extra sources on the left and right, and ignore R3 which was used to provide a DC path to ground for the node it is connected to.

Just to note, the exact numerical results are very close to the answers given for the original problem. The answers given are close to exact but there is a small almost unimportant difference probably not a big deal.