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enter image description here

Here's what the marking scheme says: "Capacitor gets charged first and acts as an insulator/blocks current".

However, electrons flow from negative to positive, so shouldn't the electrons pass the bulb first, therefore causing it to light?

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    $\begingroup$ It is so unfortunate that we give students textbook problems like this without first giving them a battery, a bulb, and an assortment of capacitors to play with. $\endgroup$
    – John Doty
    Mar 18, 2023 at 15:23

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Let's modify your circuit:

enter image description here

Now suppose that SW1 is closed, and SW2 is open. The voltage across the capacitor, therefore, is zero.

Next, let's open SW1, and then close SW2. What will happen when SW2 makes contact is, a brief pulse of current will flow through SW2 and the lamp as the capacitor is "charged up" to the supply voltage. The current, as a function of time, is

$I(t)=\frac{V}{R}e^{-\frac{t}{RC}}$

Where $V$ is the supply voltage, $R$ is the resistance of the lamp, $t$ is the time elapsed since SW2 was closed. At $t=0$, the capacitor acts like a short circuit, and the current through the lamp is just $V/R$ (Ohm's law.) But, as the voltage across the capacitor rises toward the supply voltage, the current will approach zero.

https://en.wikipedia.org/wiki/Capacitor#DC_circuits

The quantity $RC$ is the so-called time constant of the circuit. For larger values of $R$ or larger values of $C$ the capacitor will take longer to charge up.

If the capacitor is huge (e.g., like the big electrolytic capacitors used in linear power supplies) and if the amount of current needed to light the lamp is small, then you may see the lamp instantly light up, and then fade to dark over a period of several seconds.

If the capacitor is small (e.g., like capacitors used in audio circuits and digital logic circuits) then the current pulse might be too short and too weak for the lamp to emit any light at all.

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Assuming that the capacitor is initially uncharged, when the circuit is constructed, an initial flow of current is created, causing the capacitor to start charging. This results in a brief period of time where the light bulb illuminates. As the capacitor reaches its maximum charge capacity, the flow of current in the circuit decreases, causing the light bulb to gradually dim until it eventually turns off.

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  • $\begingroup$ Have you tried this with the given component values? $\endgroup$
    – John Doty
    Mar 18, 2023 at 17:14
  • $\begingroup$ Using simulations, the light bulb lights for ~1.2ms. $\endgroup$
    – Ozzy
    Mar 18, 2023 at 17:27
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    $\begingroup$ Can you make a 1.2 ms pulse of light with an ordinary incandescent bulb? $\endgroup$
    – John Doty
    Mar 18, 2023 at 17:29
  • $\begingroup$ Is 1.2 ms "gradual"? $\endgroup$
    – John Doty
    Mar 18, 2023 at 17:30
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It’s a poorly-formed question. You’re right that there is a transient current when the circuit is connected. A future chapter in your textbook (or a chapter in a future textbook) will talk about “resistor-capacitor circuits,” and will teach you to calculate that the transient current here will vanish after a few microseconds. Whether the bulb blinks during this brief interval depends on its construction. An incandescent bulb doesn’t get hot enough to glow in a few microseconds. An LED, on the other hand, can blink that rapidly (though it’d be tough to observe naked-eye).

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    $\begingroup$ RE: "(though it’d be tough to observe naked-eye)", it doesn't matter how short the blink is, if it produces enough photons for the eye to detect, you'll see it...you just can't resolve two separate blinks that are less than ~50 ms apart. $\endgroup$
    – The Photon
    Mar 18, 2023 at 15:21
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    $\begingroup$ … which gets into the question of how much power goes through the hypothetical LED in those few microseconds, the intensity of the room lights, and a bunch of other hypothetical details. Many years ago I debugged a photomultiplier detector by sending smaller and briefer pulses to an LED until the photomultiplier was detecting ten-ish photons in each pulse. $\endgroup$
    – rob
    Mar 18, 2023 at 16:55

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