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It sounds like you're trying to understand an AC circuit using DC behavior. …

The key thing to understand in your circuit is that Vs is an ideal voltage source, which means it doesn't control the current at all. The inductor isn't "opposing" current forced from Vs, it's changin …

Phase is just a way of describing time differences in periodic signals and events. Remember that, from a mathematical standpoint, a true periodic signal is eternal: $$f(t + T) = f(t), \ \ \ -\infty …

Any real voltage source must be built and turned on, which is a transient event (AC). Real circuits also receive noise from background radiation and other sources. …

Phasor analysis does not represent a sine as a sum of two complex exponentials. It's doing something more subtle -- taking advantage of linearity. Let's say you have a circuit with a voltage source \$ …

The capacitor might be rated for exposure to AC, but you can't "store" AC power in a capacitor. Even if you somehow could, 1.14uF is orders of magnitude too small. …

The usual way to find a Thevenin equivalent for a circuit with dependent sources is to connect a 1 amp test source to the input. Find the voltage across the terminals, then divide by 1 amp to get the …

The resistor and capacitor voltages are out of phase. Their peak values don't happen at the same time. You can see this in the CircuitLab simulation: Your equation for the reactance of the capacito …

Now that we have a practical definition for DC, let's look at AC. We're only going to consider sinusoids here, since you can use Fourier transforms to express any signal in terms of sinusoids. …

I'm not an electrician, but I can help with the circuit analysis. The formula for resistance is: $$R = \rho \frac L A$$ where \$\rho\$ is the resistivity of the steel, \$L\$ is the length of the wire, …

First off, in all cases that I know of, AC is converted to DC using a rectifier of some sort. The AC isn't connected to the switching element. There's a family of SMPS called off-line converters. …

Adding a diode in the AC path will prevent reverse conduction through the body diodes. If you want to draw current during both half-cycles, use a bridge rectifier. …

In the AC analysis, both the 10V DC source and the 0.7V drop in the diode disappear, and you're left with only the 8 \$\Omega\$ equivalent resistance. … Calculate the effect of a small input variation (AC) using the linear approximation. Add the DC and AC results together. …

The problem is sloppy terminology. You're using "DC" to mean two different things: A circuit in which none of the voltages or currents change over time. This is a mathematical ideal. A voltage or cu …