# DC+AC Transformer Problem

1. Oct 1, 2014

### ZetaOfThree

1. The problem statement, all variables and given/known data
Find $V_{out}$ for the following circuit:

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.

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

3. 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?

2. Oct 1, 2014

### Staff: Mentor

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.

3. Oct 1, 2014

### 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?

4. Oct 1, 2014

### Staff: Mentor

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.