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Let's say we have a normal circuit with a light bulb, with wires and a battery.

When one places a capacitor in this circuit, how is the light bulb able to light up, even when the capacitor prevents the flow of charge? Also, why does it dim and then go out eventually?

Then when the battery is removed from this circuit, how is the light bulb still able to light up? And what is happening when the light bulb dims and goes out in this situation as well?

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3 Answers

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First, note that the light bulb is essentially just a glorified resistor. As current flows through the filament, Joule heating causes the filament to get hot and emit light.

When one places a capacitor in a circuit containing a light bulb and a battery, the capacitor will initially charge up, and as this charging up is happening, there will be a nonzero current in the circuit, so the light bulb will light up. However, the capacitor will eventually be fully charged at which point the potential between its plates will match the voltage of the battery, and the current in the circuit will drop to zero. This is when the light bulb will dim and then fizzle out.

When the battery is removed from the circuit, there is nothing to maintain the potential difference between the plates, and the capacitor will discharge. As this happens, there will once again be a nonzero current flowing through the circuit, and the bulb will light up. However, the current will steadily decrease as the capacitor discharges and will eventually drop to zero at which point the bulb will go off.

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This question is straight out of chapter 20 of the third edition of Matter & Interactions by Chabay and Sherwood (Wiley, 2010). The deep, underlying issue here is that for the first time ever that I'm aware of, an introductory physics textbook correctly teaches that a capacitor's fringe field must not be neglected! If it is neglected, then capacitors simply cannot work they way most introductory physics textbooks claim they work. Understanding the charging of the capacitor requires a thorough understanding of chapter 19 and the feedback mechanism described therein. The feedback mechanism drives the surface charge distribution and the circuit behaves accordingly. You're quite correct that the physical gap between the capacitor's plates creates a conundrum, at least at first. Then, you realize that there is an electric field between the place and, more importantly, a fringe field in the wire just outside the plates.

So, go back and make sure you thoroughly understand chapter 19 and then apply those ideas to chapter 20. The text does a quite excellent job of illustrating all this, so be sure to follow along, actually READ the text, do the exercises AS YOU COME TO THEM IN THE TEXT, complete the accompanying diagrams yourself without referring to the text. Here's a VERY important hint: you need neither the concept of resistance nor the concept of capacitance to answer your question. All you need is the concept of electric field and the concept of surface charge distribution.

M&I is the ONLY, yes ONLY, introductory textbook that gets capacitors right so you're actually being exposed to some groundbreaking foundational material here. Take advantage of it!

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Why is the fringe field required to understand charging a capacitor? Or maybe, what do you mean by a "fringe field in the wire just outside the plates? It seems to me that considering the "standard" E-field between the plates is sufficient to comprehend the displacement current. Thanks. –  Art Brown Mar 11 '13 at 0:08
    
The fringe field is the field outside the plates in the wires attached to the capacitor. The fringe field plays a very important role in charging and discharging, and this role is easiest to see in discharging. The fringe field starts the process of discharging. –  user11266 Mar 11 '13 at 1:05
    
Sorry, I'm not getting it. My mental model is: I connect a resistor to a charged capacitor, with the wires and cap plates modeled as perfect conductors. Current steps up in the resistor, wire, and cap (ohm's law and kcl), and that current is exactly balanced by the displacement current between the cap plates. Where does the fringe field enter? Thx again. –  Art Brown Mar 11 '13 at 1:23
    
See chapter 20 of Matter & Interactions by Chabay and Sherwood (Wiley, 2010). You're thinking macroscopic, I'm talking microscopic. Resistance, capacitance, and Ohm's law are irrelevant. –  user11266 Mar 11 '13 at 2:46
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A Short answer is here,

Firstly the capacitor gets charged.Though it takes infinite time to reach the battery potential the current is reduced to considerably low values. So obviously bulb won't glow.

Note:As capacitor gains more and more charge its potential increases finally reaching the value same as the battery.So potential across bulb tends to zero.

Second case the capacitor is at the same potential as the battery was. So now the potential across the bulb is some value so current flows. but the brightness of the bulb varies exponentially.

I have not used maths to explain because I don't know if you know Differential equations.

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