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In the picture below we see 3 LEDS in a parallel circuit. When the switch is closed only LED 2 and LED 3 are on. LED 1 isn't on because it's a diode in the opposite direction. You observe what happens when the switch is closed and opened by opening the switch first and then closing it. Explain:

  • Which LED(S) are on for a short while after opening the switch;
  • Which LED(S) are not on for a short while after closing the switch.

p.s. - I probably made some mistakes in translation. Here is the image of the circuit:

enter image description here

My Problems:

I can't answer any of these questions. I know for sure they have something to do with electromagnetic induction. But I'm having trouble grasping what will happen in these situations. Can someone help? I know that if a switch is open and closed there will be induction for a short time.

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Isn't this a picture from a textbook, and the explanations to help you answer the questions can be found nearby? – JYelton Jan 17 at 16:53
@JYelton So you assume that every single individual can make every question flawlessly if he just has a bunch of formulas? No, of course not. Also, no. – ZafarS Jan 17 at 16:54
@JYelton Not all textbooks work that way. In fact most of the text books I've used don't have detailed explanations nearby. – AndrejaKo Jan 17 at 16:55
Also the pictured circuit of how to kill an LED. – AndrejaKo Jan 17 at 16:57
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@AndrejaKo as long as the switch isn't closed too long, there is nothing here that will kill the LED. – Phil Frost Jan 17 at 19:37
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1 Answer

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Inductors resist changes in current. To change the current in an inductor, you must apply a voltage. If you remove the voltage, the inductor will make its own voltage so that the current can keep flowing. Of course, making this voltage requires energy, and as that energy moves out of the inductor and into something else, the current will eventually reach zero, when there's no energy left in the inductor.

So, when the switch is closed for a while, there is a current in the inductor. When you open the switch, that current must keep going. Where can it go in this circuit, if its prevented from going through the battery by the open switch?

switch closed switch open

A diode allows current to flow only in one direction. It's no mistake that their schematic symbol looks like an arrow. If the switch is opened, there is only one place for the same current to keep flowing, and that's through LED1 and LED2. It will keep flowing here, even without help from the battery, until all the stored energy in the inductor has been transferred to the LEDs.

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Well if I'm correct, the current can flow to either the switch or to LED 1 or LED 3. LED 3 won't turn on because of the diode, but LED 1 will turn on. Is this correct? – ZafarS Jan 17 at 20:03
@ZafarS not really correct. Current can't flow "to the switch" if the switch is open. There must be a closed loop for any current to flow. Imagine the electricity flowing out the + end of the battery, through the inductor, through the switch, and back into the - of the battery. Which way is the electricity flowing in the inductor? If the switch is now open, how can electricity continue flowing through the inductor, considering that it can flow only one way through a diode, and you need a complete loop for it to flow? – Phil Frost Jan 17 at 20:57
It first goes from the left to the right.. but for it not to go to the switch it must go back from right to left? – ZafarS Jan 17 at 21:12
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@ZafarS: don't think about current going to things. Think about current going through things. There must be a closed loop for current to flow. You are right about the current flowing left to right through the inductor. There is one other closed loop around which current can flow that way, which doesn't go through the switch. Where is it? – Phil Frost Jan 17 at 21:39
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@ZafarS everyone had this problem at some point. I've added two pictures, which should be good for at least two thousand words. – Phil Frost Jan 17 at 21:53
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