The smaller capacitors will not pass as much current at 60Hz, so the bulb will not light (or glow as brightly)
We can calculate the expected wattage:
For 1uF:
1 / (2 * pi * 60 * 1e-6) = 2652 Ohms.
If we add this to the bulb resistance (say 50 ohms as it will be somewhere in between cold and hot) we get 2700 ohms (we can be rough here as the bulb resistance makes little difference)
For a 115V line, that's 115 / 2700 = ~43mA.
So the apparent wattage will be around 115V * 43mA = ~5W
The real wattage dissipated by the bulb will be much less, around 100mW.
We calculate this by I^2 * R. So with a 50 ohm filament resistance we get 0.043^2 * 50 = 92mW.
Not enough to light the bulb up. For the 2.2uF the result would be around 900mW (may glow a bit?)
If we do the same for the 22uF capacitor, we get:
1 / (2 * pi * 60 * 22e-6) = 120 Ohms.
sqrt(120^2 + 285^2) = 309 Ohms.
115V / 309 = ~372mA
115V * 372mA = ~43W (~39W dissipated by filament)
So you see the larger cap makes a big difference.
Similarly, how to calculate the reactance for a capacitor when a
square wave is passing through it. What is the formula?
There isn't a capacitive reactance associated with a square wave. The very concept of reactance depends on the context of sinusoidal excitation.
When we solve AC circuits in the phasor domain, it is taken for granted that the circuit is in sinusoidal steady state, i.e., all sources are sinusoidal of the same frequency and all transients have decayed away.
That fact is this: one cannot meaningfully sum phasors or reactances for sinusoids of different frequencies.
Now, that doesn't mean that you can't apply the concept of reactance to find the capacitor voltage across for a square wave current through.
Since (ideal) capacitors are linear, we can decompose the square wave into sinusoidal components, find the associated sinusoidal voltage for each component, and then sum to voltage components to find the total voltage.
Recall the fundamental phasor domain relationship for capacitor voltage and current:
$$\vec V_c = \dfrac{1}{j \omega C}\vec I_c $$
where \$\omega\$ is the angular frequency of the associated sinusoid.
Now, let
$$i_C(t) = a_1 \cos(\omega t + \phi_1) + a_2 \cos(2\omega t + \phi_2) + a_3 \cos(3\omega t + \phi_3) + ...$$
For each sinusoidal component, there is an associated phasor. For example, for the first component, the associated phasor is
$$\vec I_{c_1} = a_1 e^{j\phi_1}$$
Thus
$$\vec V_{c_1} = \dfrac{a_1 e^{j\phi_1}}{j \omega C}$$
so that
$$v_{C_1}(t) = \dfrac{a_1}{\omega C}\cos(\omega t +\phi_1 - \frac{\pi}{2}) $$
Repeat for each term in the series and then sum to find the total capacitor voltage.
Note that we have not defined a reactance to the entire current waveform nor can we define such a thing. Instead, we
(1) found the reactance to each sinusoidal component
(2) converted each resulting phasor voltage back into the time domain
(3) summed the individual time domain voltage components
Best Answer
If pin 3 is high (+5V) and C4 is discharged, both sides of C4 are high (+5V), The right hand side of C4 is held high by R10. The left hand side is held high by pin 3. There is 0V across C4.
If pin 3 goes low (-5V), C4 cannot instantaneously charge from 0V to 10V so taking it's left hand side down to -5V takes the right hand side to -5V. This -5V is seen at pin4 on IC6.
R10 then slowly charges C4 bringing the voltage at it's right hand side towards +5V.
The circuit would function the same if I arbitrarily changed the labels +5V and -5V to, for example, +10V and 0V (or GND). In which case you might or might not want to replace the word "negative" with something else.
No, See the "Differentiator" example in this tutorial
No current needs to flow through or across the capacitor because the voltage across it is initially unchanged. If the voltage with respect to a fixed reference point (ground) at the left hand is changed, the same change will be seen at the right hand side. No current need flow. Subsequently there are changes but these involve external flows of current.