Question

A lamp is connected in Series with a Capacitor. Predict your observation for dc and ac connections. What happens in each case if the Capacitance of the Capacitor is reduced ?

Answer 1

What do you think will happy with the DC circuit?

Answer 2

I'm sorry but I seriously dont know ,can you give me a hint ?

Answer 3

What will happen is in short order the plates of the capacitor will charge up and the current will stop flowing.

Answer 4

In other words, very quickly, the potential difference across the plates of the capacitor will equal the potential difference across the power source. And then there is no potential difference in the wires, no electric field, no electron flow, no current.

When that happens the lamp will not emit any light either.

Answer 5

Hence with a DC circuit, the lamp will flash briefly as the capacitor plates charge, but then it will dim asymptotically to zero.

Answer 6

Make sense?

Answer 7

How does the capacitor get charged ?

Answer 8

Let me refer you to a video that will show you. One sec.

Answer 9

Okay :)

Answer 10

Watch this, beginning at 24:00
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-7-capacitance-and-field-energy/

Answer 11

Thanks a lot ,It makes sense
So even if the capacitor is reduced there will not be any charge ?

Answer 12

And What happens in case of ac ?

Answer 13

No, just less charge. By definition capacitance C = Q/V, where Q is charge and V is potential difference.

Answer 14

Hence with constant V, C down => Q down

Answer 15

Ah Okay

Answer 16

And in case of ac ,What's the difference?

Answer 17

Well, imagined you charged the capacitor, and then reversed the polarity of the power source: negative to positive, positive to negative. What would happen?

Answer 18

Current flows ?

Answer 19

yes and ...?

Answer 20

lamp will emit light

Answer 21

Yes and ...?

Answer 22

Does the system now tend to equilibrium?

Answer 23

Hm No?

Answer 24

i didnt understand

Answer 25

If I attach a capacitor and lamp with a DC source, does the system go to equilibrium?

Answer 26

1min let me think

Answer 27

Scroll up, I described the situation and you should have just seen it on the video.

Answer 28

I guess yes ,since everything comes down to zero ?

Answer 29

Do you live in the US btw?

Answer 30

No

Answer 31

Even so. See you can buy a simple electronics kit with some components: capacitors, resistors, lamps, power sources. If you do these simple experiments yourself, you'll be able to see.

Answer 32

Anyway, in this situation, we attach the power source and the capacitor plates charge until the V across the plates equals the V of the power source. Hence the system does reach equilibrium.

Answer 33

No potential difference then from the plates of the capacitor to the power source
=> No electric field between the capacitor plates and the power source
=> No force on electrons
=> No electrons move
=> Current = 0

I.e., the system is in equilibrium.

Answer 34

Make sense? That's the DC case. You need to understand this to understand the AC case.

Answer 35

Yes ,Understood

Answer 36

Ok. Now, think again what happens if we take the wires from the power source and flip them. Negative -> Positive, Positive -> Negative.

Now what happens?

Answer 37

Power source is still DC.

Answer 38

Current starts to flow .

Answer 39

Yes, until what?

Answer 40

Until the plates are charged again, this time with positive and negative reversed. This happens like before, very quickly.

Answer 41

Oh okay

Answer 42

The lamp will flash as current flows through it.

Answer 43

But if we don't do anything, the system reaches equilibrium again.

Answer 44

No current will flow.

Now, what happens if you reverse the wires in the power source again?

Answer 45

Current will again flow

Answer 46

Until?

Answer 47

It gets charged again

Answer 48

the capacitor. yes.

Answer 49

Yes capacitor

Answer 50

So the lamp will flash for a moment as the charge flows through it.

Answer 51

Suppose the lamp flashes for 0.1 of a second. What would happen if I changed the power source 10 times a second?

Answer 52

Positive -> Negative, Negative -> Positive
I do that 10 times in one second.

Answer 53

it will light for a second

Answer 54

?

Answer 55

Yes. And what if I change the polarity 50 or 100 times a second?

Answer 56

It will light for 5 or 10 seconds

Answer 57

No. I'm changing the polarity 50 times a second for just one second. How long will the lamp shine in that one second?

Answer 58

its just one second isn't it ?

Answer 59

Yes.

Answer 60

Change it once and all the electrons go racing around the circuit until the voltage difference V over the capacitor plates equals the potential difference V over the power source.

Then change it again, and they all go racing back the other way to try and form equilibrium again.

But then before they have time to do that, we change the polarity again, and they go racing back the other way to get back to the first attempted equilibrium.

And then we change the polarity again, and they immediately start racing back the other way.

Answer 61

Current is constantly flowing through the circuit. Current is constantly flowing through the lamp, so the lamp will shine the entire time.

Answer 62

Now, does this situation remind you of an AC power source?

Answer 63

like we keep changing the polarity ?

Answer 64

Yes. In an AC power source, polarity is changing 50, 60, 100 times a second, or whatever the frequency of the AC power source is.

Answer 65

Oh okayy

Answer 66

Hence provided the AC source has a sufficiently high frequency, electrons are going to be constantly moving => current => lamp emits light.

Answer 67

Make sense?

Answer 68

Yes

Answer 69

So When C is reduced , less charge so it'll shine less brightly ?

Answer 70

Yes, C down => Q down => less electrons racing around => lower current => less light emitted

Answer 71

Okay James ,Thank You sooooo Much!! for this great Explaination :)