# Electromagnetic induction circuit

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:

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.

• Isn't this a picture from a textbook, and the explanations to help you answer the questions can be found nearby? – JYelton Jan 17 '13 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. – JohnPhteven Jan 17 '13 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 '13 at 16:55
• Also the pictured circuit of how to kill an LED. – AndrejaKo Jan 17 '13 at 16:57
• @AndrejaKo as long as the switch isn't closed too long, there is nothing here that will kill the LED. – Phil Frost Jan 17 '13 at 19:37