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The 22.5ms label is wrong, but the rest of the graph is right. There's equal horizontal spacing between 0ms, 5ms, 10ms, and 15ms, with about half as much space between 15ms and the final +5τ' tick. Th …

It should tell you how to construct a part number for a specific capacitor. …

You need an initial state for the 1 F capacitor. And you can't just short together two capacitors at different voltages. That would imply infinite current and violate conservation of energy. …

The capacitor acts like a short to ground, so the current is \$V_i/R\$. After a long time, the voltage has stabilized and we effectively have a DC circuit. … The capacitor acts like an open circuit, so \$V = V_i\$ and no current flows. The transition between these two states in an exponential decay. …

To see this more clearly, imagine removing the load and leaving only the capacitor and the diode rectifier. As the AC voltage rises, it charges the capacitor through the diodes. … At the peak of the AC half-cycle, the AC voltage becomes greater than the capacitor voltage. The diodes turn on and the AC source charges the capacitor back to its maximum value. …

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. …

You're on the right track, but your formula is wrong. R1 and R2 are in parallel, so you need to use the formula for combining parallel resistors to get the Thevenin resistance.

D1 protects the circuit if the voltage supply is connected backwards. C3 reduces voltage ripple from the supply; giving a quieter DC voltage source for the rest of the circuit. Diodes do not prevent o …

It's common to put a 0.1uF capacitor close to the chip to provide a low-inductance power supply. In this role, the capacitors are called "bypass capacitors". …

If the capacitor and voltage source are modeled as ideal, you probably shouldn't have a switch in the circuit. Discontinuities in capacitor voltage are not allowed. …

It sounds like you want to do three things: Connect each generator coil to one capacitor. Connect the two capacitors to each other in series. Not connect the generator coils to each other. …

To the rest of the circuit, it looks like there's a current through the capacitor, since electrons go into one end and come out the other. … If you replace the battery with an open circuit, no current flows and the capacitor remains charged. …

To get the voltage to start rising, you need to be pumping charge onto the plates of the capacitor. I think you're trying to apply DC thinking to an AC circuit. … It's what really matters to the capacitor. …

But from the outside, we see charge moving into one side of the capacitor and moving out of the other side. So in circuit theory we pretend that current is flowing "through" the capacitor. …

The resistor and capacitor voltages are out of phase. Their peak values don't happen at the same time. … For a 1 microfarad capacitor at 55 Hz, this gives: $$X_C = \frac 1 {j 2 \pi (55\ \mathrm{Hz} \cdot 1\ \mathrm {\mu F})} = -j2.89\ \mathrm{k\Omega}$$ To find the voltage across the capacitor, use your …