Make Electronics-Exp 9b

CAPACITOR & DISPLACEMENT CURRENT: Play around with the 2 SPDT switches below the power source. The one on the left is to open / close the circuit, the one on the right to discharge the capacitor. Simulation speed = 1s/s. Close the left SPDT and the LED lights up and fades away. Open & close again the left SPDT and the LED does nothing this time (very brief and dim light in the actual experiment). The LED only lit-up the first time we closed the circuit when the capacitor's starting state was uncharged. If you close the circuit again with the capacitor already in a charged state then the LED will not light-up (or very briefly and dimly in the actual experiment). Open the circuit with the left SPDT again and this time close the right SPDT. It will discharge the capacitor. Do not forget to re-open the right SPDT and now close the left SPDT again. The LED will light-up again and fade away. MY UNDERSTANDING: No current is actually going through the plates of the capacitor. Negative charges get attracted to the negative plate of the capacitor and push away the negative charges that were on the other side on or near the other plate. Meanwhile that leaves holes on the + plate of the capacitor and that plate becomes more positively charged as more negative charges are pushed away from the + side of the capacitor and towards the + side of the power source. This flow of electron between the + plate of the capacitor and the + side of the power source is the displacement current. Those electrons were displaced by the charge on the - plate of the capacitor. The catch here is that unlike other circuits where it seems there is an unlimited supply of current (as long as the power supply is on), here we are limited by the amount of charges that can move from the + side of the capacitor to the + side of the power source. It happens once and that's it. Once the plates of the capacitor are fully charged, there is no more displacement current. That is why the LED lights up a little bit and then fades away when we close the circuit starting with an uncharged capacitor but (almost) nothing happens with the LED when we do it again starting with a charged capacitor. When we start with a charged capacitor, no more charges can be added to the plates and no new displacement current can be generated. Only discharging the capacitor (using the SPDT on the right) prior to closing the circuit will allow us to light-up the LED for a little bit.

published 7 years ago