In this experiment: experiment

Where neon lamp doesn’t light when the switch is closed because the voltage is much lower than that required to ionize the gas, but it only lights up momentarily when the switch is opened.

The only explanation for this is that the rate of decay of current is much greater than its rate of growth so this equation could be satisfied:

eind = −L•dϕ/dt

So why is that?


The rate of change of the current is proportional to the voltage across it.

When the switch is closed, that voltage is limited to V.

But when the switch opens, the voltage across it can be very high as the current drops quickly. It rises all the way to the voltage that causes the bulb to flash.

  • $\begingroup$ But sir, differentiating both equations of growth and decay give the same equation with different signs, doesn’t that mean the rate is the same? I can send you the equations if you want $\endgroup$ – user3407319 Jan 11 at 14:16
  • $\begingroup$ What happens is determined by equations and the specific conditions. These appear as the externally-determined values in those equations. Yes, V=Ldi/dt works for both the increase and decrease in the current. But "V" means "the voltage across the inductor", which doesn't have the same value during the increase and the decrease. $\endgroup$ – Bob Jacobsen Jan 11 at 14:38
  • $\begingroup$ Is there any equation to determine it? And by the way this wasn’t the equation i was speaking of. $\endgroup$ – user3407319 Jan 11 at 14:40
  • $\begingroup$ I am talking about the differential of both of the equations in this photo goo.gl/images/hrDmC4 with respect to time $\endgroup$ – user3407319 Jan 11 at 14:43
  • $\begingroup$ The right side of the equations in that link have V in them. If V is i.e. 9 volts due to a battery, the result is different than if V is 90 volts to flash a lamp. $\endgroup$ – Bob Jacobsen Jan 12 at 15:18

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