Hi I am a Korean student studying physics. Recently I was solving a physics problem about coil emf and I have a question.

As you see in this circuit and the graph, when you close the switch the emf of the coil reaches '-E'(voltage of the battery). However when you open the switch the emf reaches far more than 'E' Why does this happen? I already asked my physics teacher and he said that in theory when you open the switch the emf should be 'E'

How does the voltage reach far more than E?

• electronics.stackexchange.com/questions/309616/…
– G36
Commented Jul 23, 2017 at 16:45
• Do not ever excuse yourself for speaking another language that is not your native tongue. Commented Jul 23, 2017 at 21:19

Theoretically, the inductor follows V = L* di/dt. If the switch is opened instantaneously, then di/dt is infinite, so V is also infinite. In reality, there are small stray capacitances which will cause the Voltage spike to be less than infinite, both in amplitude and in rise time. Also, as noted elsewhere, an arc will likely form across the switch. This will also affect voltage spike.

I have no idea why your professor thought the emf should be equal to V when the switch is open. Theoretically, when the switch is open, the battery current is zero, and the battery does not really enter into the analysis.

• Thanks for your answer! I think I understand now why the emf is larger than E for this circuit when the circuit is open. However I noticed that the when I asked my teacher about this, the circuit was not actually picture above! It's a little bit different. Could you take a look at this picture and tell me if what my teacher said is true for this circuit?i.imgur.com/Ou1SvFB.png Commented Jul 23, 2017 at 17:36
• @NohJeong-Yeon what is the symbol at the top? Commented Jul 23, 2017 at 18:16
• @HarrySvensson It says it's a neon lamp Commented Jul 23, 2017 at 18:28
• @NohJeong-Yeon Link You'll still get something above E. Commented Jul 23, 2017 at 18:39
• Ah! Now I'm with you 100% I thought you talked about the link I made. Commented Jul 23, 2017 at 21:17

The inductor stores energy and the energy stored is used by the inductor to try and keep the current flow through it constant. So the instant the switch opens, there is a current flow that the inductor tries to maintain but, the only circuit path includes the fractionally open contacts. Therefore, to maintain current flow it will generate a voltage that can form a small arc. This can be as high as several hundreds or thousands of volts.

When the switch is opened, all the energy gets transferred into the parasitic capacitance across the switch contacts. Given Energy = 0. 5 * C * V * V, a very small cap requires the V to increase as squareroot.