4
\$\begingroup\$

I connected an LED in series with the capacitor (any type of capacitor. I used one of 10 μF) and connected them with the 9V DC battery. I connected the positive pin of the capacitor directly to the positive side of the battery and attached the positive pin of the LED with the negative pin of the capacitor (as the LED and the capacitor are in series combination) and connected the negative pin of the LED to the negative side of the battery. So, The bulb should gradually turn dim and then turn off and it did. I followed a diagram of the circuit but I cant attach it with the question due to the lack of population.

enter image description here

My question is, why does this happen? Why does the light turn dim when attached to a capacitor in series?

I am quite new in electronics. Therefore I am not quite familiar with the complex detailed descriptions but still I am trying to give as much details as possible. Pardon me if there has been any mistakes in my question.

\$\endgroup\$
5
\$\begingroup\$

Short answer: Current stops when the capacitor gets charged up to the battery voltage.

When current flows through the circuit, the bulb lights up. In this case you can consider the bulb as a 'current detector'. Current in this case is flow of charge, and the charge carriers are electrons. They are pushed around because the battery pushes them with an electric field.

Now look at the capacitor symbol. It indicates that there is a gap between the plates. Electrons normally don't jump this gap. But you know there is current because the bulb initially turns on. So what is happening? Electrons are effectively being accumulated at one of the plates of the capacitor, while the opposite plate has electrons being removed, "charging it". But they can't accumulate indefinitely, because the more they accumulate, the more the repel each other. It basically takes energy/work to do this, and you can consider that the capacitor is storing this energy, and can effectively be released later on. As the capacitor gets charged, the voltage accross it augments, until the battery cannot push more electrons. At this point the capacitor voltage has equalized the battery voltage. No more electrons flow, the bulb finally turns off.

To release the energy stored in the capacitor, remove the battery from the circuit and connect the wires together. You should see the same effect (bulb turning on, then dimming until off), because the battery no longer keeps those electrons pushed and they return to neutrality via the bulb.

\$\endgroup\$
  • \$\begingroup\$ So that means the capacitors work as a container of electrons that captures the electrons flowing from the -ve side of battery to +ve side and contain them for a certain limit, if the limit is exceeded, the electrons don't flow from the battery towards the capacitor. Isn't it? \$\endgroup\$ – shiladitya basu May 5 '13 at 9:35
  • 5
    \$\begingroup\$ @shiladityabasu: Capacitors store energy, not electrons. A charged capacitor contains no more electrons than an uncharged capacitor, it's just that they have been moved around to a different place inside. At this new position they have more electric potential energy. Current flows while the electrons are being relocated. \$\endgroup\$ – RedGrittyBrick May 5 '13 at 12:02
  • 1
    \$\begingroup\$ On capacitors: amasci.com/emotor/cap1.html \$\endgroup\$ – Phil Frost May 5 '13 at 17:01
  • 2
    \$\begingroup\$ @shila Correct. The only thing to clarify is that the capacitor as a whole does not have more electrons, but one of its plates gets an accumulation of electrons, while the other lacks roughly the same amount. Its ability to have this done to it with with ease is its capacitance, so for a given voltage it has more charge accumulated, which means more current (or for longer) flowed to reach this state, which means it can store more energy per (square) volt. \$\endgroup\$ – apalopohapa May 5 '13 at 17:08
1
\$\begingroup\$

If you had two batteries; one charging the other, the current stops when the more discharged battery reaches the same level of charge as the other battery.

The capacitor isn't a battery but it behaves like one in this circuit - it progessively becomes charged with an increasing voltage which progressively opposes the electrons flowing into it.

A state of equilibrium is reached after a certain length of time and no more current flows.

Smaller value caps charge quicker - if you had a 10nF cap instead of 10uF, you wouldn't notice the LED illuminate at all - it would happen so quickly. If you had a supercap it might take minutes or hours until the LED got dim.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.