Q. Why do we have a AC power distribution system vs. DC?
A. Because with the development of the transformer (cir. 1885), AC could travel better than DC and did not require the use of large, costly, distribution wires.^{(1)}

However AC power must be stepped down and rectified to produce a stable DC voltage before it can be used locally. Consider the transformer circuit below to step down
$$\
Vin = 120V \ AC \ at \ 60Hz \ to \ Vout = 12V
$$
We want to drive the output load, RL. The inductance ratio is the square of the turns ratio, so we pick a 100:1 inductor ratio to step down the voltage by a factor of 10. In LTSpice, the spice directive K1 L1 L2 1 (no transformer leakage) ties the two inductors. The phasing dot beside the indicator indicates the winding direction. Current enters L1 at its phasing dot and exits L2 at its phasing dot.

$$\
Turns \ ratio = 10:1 \\
Vin = 120V \ AC \ at \ 60Hz \ step \ down \ to \ 12V
$$

The next step is to rectify the AC input, i.e. get rid of the negative voltage. Let's add a diode rectifier.

The diode conducts only in the positive half-cycle, so a half-wave rectifier, meaning we're losing the negative half of the AC power cycle when the diode does not conduct.

We can remedy this situation by the following bridge arrangement. Let's consider the full-wave rectifier below:

Positive cycle flow: current flows through diodes D1 and D4 (orange arrows)

Negative cycle flow: current flows through diodes D2 and D3 (purple arrows)

Now to get a DC voltage, we simply add a capacitor on the output, as below.

The capacitor charges up when the input voltage is increasing, and the next cycle picks up before the capacitor completely discharges. Notice, in (the zoomed-in) fig. 1(b), we have a ripple on the output. We also need to allow a start up time of ~5ms for the capacitor to charge during the first cycle. We can increase the capacitor to minimize the output voltage ripple, but this may result in a longer startup time.