Electrical – How much power am I consuming from using an AC-DC converter

Also, its quite unreasonable that my charger, that uses 1.5A AC, would consume 345W and only output 19V * 4.74A = 90.06W

That would be unreasonable but that doesn't happen. The 1.5 A of current is at the lowest AC supply voltage that your charger can tolerate and still work. From what you are saying that sounds like 110V RMS. So now your input power will be 165 watts but that's still sounding a bit high so maybe your charger can operate at the "generally accepted" "world-standard" of 85V to 265V. At 85 volts RMS and 1.5 amps, the input power is 127.5 watts and this is a power efficiency of: -

Power efficiency = 100% x output power / inputted power = 71%.

Also, it is rated above 80% efficiency, which doesn't meet the previous calculation.

If it is rated at above 80% efficiency this may be at the 230V range with efficiency dropping to 71% at the lower end of the range. Or you may have measured the value incorrectly.

Whether DC or AC, instantaneous power is instantaneous voltage x instantanous current. You then find that if you average the instantanous power over time, it becomes real power of which your can be billed for. Here are various scenarios: -

How can I calculate an electric device's power usage given its current and AC voltage?

If you want to do it accurately you have to use a wattmeter and this gives you true average power - at the heart of any wattmeter is a device that multiplies the waveforms of voltage and current.

These days power supplies like yours are usually what is known as "power factor corrected" but if yours is an older type supply, the current and voltage waveforms may both look very different and not be truly in-phase. That makes simple power estimations problematic.

You're right in thinking that current and voltage varies according to time in an alternating current circuit. However, when analysing circuits with alternating current and simple resistive loads, it is much easier to work with average power instead of instantaneous power. You can use RMS voltages and RMS currents to simplify your analysis.

The 120V voltage rating you get out of a wall outlet is given in RMS. You can multiply this number by \$\sqrt2\$ to obtain the amplitude of the sine wave.

Your circuit analysis does seem right on the most part. Assuming your input voltage remains constant, adding resistances in parallel decreases overall resistance and implies increased power. Adding resistances in series increases overall resistances and implies decreased overall power.

If you rearrange for the voltage, you can see how the resistance \$R\$ of the circuit affects the power consumption of the circuit.

\$P = \frac{V_{rms}^2}{R} = \frac{120^2}{R} = \frac{1.44\times10^4}{R}W\$

Lastly, there's a few reasons why you might not observe the lightbulb to flicker when current changes direction. One of which is that a light bulb doesn't dim instantly when power goes to zero (try switching off the lamp and you'll see the filament fades away slowly) so it still produces light as the instantaneous power momentarily crosses over zero. Furthermore, 50/60Hz is generally fast enough that the flickering is indiscernible by humans.