in volume I, chapter 15, section 2, the example circuit involving the neon light is confusing. why would closing the switch produce very little voltage as the current increases from zero to "some magnitude" but opening the switch produces enough voltage to light the bulb as the current drops from "some magnitude" back to zero. if the voltage is proportional to the rate of change of the current, wouldn't opening and closing the switch produce changes in current (and therefore voltage) that were identical, but of opposite polarity? if there is another factor that lowers the rate of current increase on switch closure, it is not made clear in the example.

thanks for the great course so far,
russ h.

 

No,

when you close the switch the inductor tries to maintain the initial steady stae current through it which is zero in this case. Thus the inductor just induces a voltage which is enough to oppose this change in current but it is small because the resistance of the circuit is small.

When you open the switch the inductor wants to maintain the current through it. But because the resistance of the neon lamp (initially) is high it produces a high voltage as to keep the current (flowed through the inductor) the same. However, due to resistive losses the current decrreases exponentially.