Di/dt for Growth and Decay in LR Circuit

[aim1732]

Why is the emf induced in an inductor much larger when the switch connecting it to a battery is closed than when the switch is opened? Somewhere in a text I had read that di/dt is very large when an LR circuit breaks but then its value for growth and decay in an LR circuit is same---(i₀R/L)e^-tR/L with a corresponding plus and minus sign. Any ideas?

 

[yungman]

For one thing, when you only the switch, the current got from I to 0 in very very short time. so

$$\frac{dI}{dt}\rightarrow -\infty$$

That's why you can see arc in a lot of case on the contact of the switch. It can reach KV range.

Look at the equation, when the switch open, R is approach infinity, this mean decade is very fast even the voltage shoot up very high. I did design to optimize the decade time balance with limit overshoot by putting a resistor in parallel with the electro magnet ( inductor ) to achief low enough overshoot voltage but still fast enough decay time. If you parallel too small a resistor, the overshoot will be smaller, but the decay time would increase. When your switch go from open to close, the parallel resistor is not in the picture because it is parallel to the coil. The resistor in the equation is really the internal impedance of the voltage that drive the coil. So it is totally different.

 

[Bob S]

Here in thumbnail is a simulation of a pre-1970 automobile ignition circuit. The switch opens at 2ms and 6 ms. It closes at 0, 4, and 8 ms. Note the 25-kV pulses at 2 ms and 6 ms, and the barely visible pulses at 0, 4, and 8 ms. Typically, the switch ("points") and capacitor ("condenser") are inside the distributor. The L/R time constant applies only when the points are closed. The LC time constant applies when the points open.
Working on classic automobile ignition circuits like this one is very educational, especially if you forget to turn off the ignition switch when you pull off the distributor cap and adjust the point gap.

Bob S