If i2(t) = 5cos(100t - 300) in the circuit below, find vg(t).

 

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If i2(t) = 5cos(100t - 300) in the circuit below, find vg(t).

Hi,

Also, i guess the first transformer is ideal?

 

Hi,

Also, i guess the first transformer is ideal?

Yes the first transformer is ideal, I can't understand of how the 2nd transformer is of any use if I need to find the voltage of the source using current of the 2nd loop??

 

Hello again,

We might call the load the 100 ohm resistor in series with the 400uf capacitor, and so the second transformer couples the left part of the circuit to the load. That means that the current i2 is dependent on the load and the properties of the second transformer (as well as the left side of the circuit with the ideal transformer). If the coupling of the second transformer where to change, the current i2 would change too. Thus, the second transformer is very much part of the circuit and so very much part of the problem.

For many circuits we dont do the analysis in this manner. Most of the time we just want to know what the output is with a given input. In that case we would come up with an equation like:
Vout/Vin=F1(s)

or similar. But note that in that more typical solution we would find it easy to solve for either Vout *or* Vin.
More typically we see:
Vout=Vin*F1(s)

but what is to prevent solving for Vin instead:
Vin=Vout/F1(s)

So here we see we got Vin knowing Vout and F1(s) and it was not that hard to do.
What this means is that we could find Vin knowing other things too if we had to, such as in ths problem i2.,
So we could first find i2 and then try to work it into a solution for Vin instead of i2, which probably isnt that hard to do once we have the solution for i2.
More strictly speaking, we could find (as in many other problems):
i2/Vin=F(s)

and then solve for Vin instead of i2.

We could talk about ways to handle that second transformer if you have not encountered that yet. One way for example is to use a "T" equivalent.

So you think you can proceed now or no?
You should at least try one time to do this and see what you can get out of it.