How Do You Solve a DC+AC Transformer Circuit Problem?

  • Thread starter Thread starter ZetaOfThree
  • Start date Start date
  • Tags Tags
    Transformer
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
ZetaOfThree
Gold Member
Messages
109
Reaction score
23

Homework Statement


Find ##V_{out}## for the following circuit:
m7dhn7.jpg

We are given that the transformers are ideal and ##\frac{n_1}{n_2}=a>1## where ##n_1## is the number of windings on the coil on the left and ##n_2## is the number of windings on the left.

Homework Equations


KVL
Transformer equation: ##\frac{n_1}{n_2}=\frac{V_1}{V_2}## where the V's are the voltages in the transformers.

The Attempt at a Solution


Just use superposition of the voltage sources. I've got the AC part down, but I'm not so sure how to effectively handle the DC part. We can use KVL to write down a system of differential equations that relate the currents that flow through the three loops in the circuit. I was able to solve it using Maple, but the solution is pretty messy and Maple was unable to evaluate the solutions in the limit as ##t \rightarrow \infty##. Is there any way to easily see a steady state solution for the DC part of this problem, without using software as a crutch?

Thanks in advance!
 
Physics news on Phys.org
Transformers do not couple DC across windings; at steady state there's no change in flux from DC current in an inductor. No change in flux means no changing flux through the other windings to induce a current there.
 
  • Like
Likes   Reactions: ZetaOfThree
Thank you for the response!
The only problem I have with that is what's going on in the rightmost loop in the circuit on the left (the one with the capacitor and the inductor). KVL gives ##V_C + V_L = 0##, but if there is not change in current, then ##V_L=0##, so ##V_C=0##. How can it be that there is no charge buildup on the capacitor?
 
At DC steady state an ideal inductor looks like a short circuit, so zero volts potential across it. The capacitor is effectively shorted out and "disappears" from the circuit.