What happens to the energy in an inductor or capacitor when opened?

AI Thread Summary
When an inductor is disconnected from a DC source, the magnetic energy can cause a high voltage spike, potentially resulting in a spark, which is a practical application in spark plugs. The energy in a charged capacitor remains until it discharges through internal leakage, posing risks if not handled carefully, especially in high voltage systems. The voltage generated by an inductor when disconnected can be captured and increased using a diode and capacitor, leading to a step-up converter effect. This phenomenon allows for higher voltages to be achieved than the original supply, demonstrated by experiments with coils and batteries. Overall, conservation of energy is maintained in both cases, illustrating the principles of electrical energy behavior in inductors and capacitors.
RohanKulkarni
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Hi guys..i have got a question...suppose i connect an inductor to a dc source until it is fully charged..as soon as it gets full of magnetic energy,i disconnect it from the circuit..remember that it has magnetic potential energy..suppose an idiot like me opens up the inductor (as inductor is just a coil) Where would the trapped energy go?
Similarly what would happen if i open a charged capacitor..where will its energy go?
 
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Never fear, conservation of energy stands up to all challengers.

When you try to reduce the current in an inductor it causes a very high ##L\frac{dI}{dt}## voltage. That might cause a spark to appear across the switch and the energy goes into the spark. That is exactly how we make the spark plugs spark in a gasoline engine.

We can think of scenarios where it is hard to see where the energy goes, but when we look close enough, energy will always be conserved.
 
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And the energy in a charged capacitor stays there until the capacitor bleeds down from internal leakage. This is why you have to be very careful when working on high voltage power supplies - the power can be shut off, but the internal capacitors are still fully charged. Then somebody sticks their finger in there, and KAPOW!
 
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The effect anorlunda describes is very handy. The high voltage generated by the inductor when it's disconnected from the DC source can be captured by feeding it through a diode into a capacitor. If you keep connecting and disconnecting the inductor you can end up with a higher voltage on the capacitor than you started with in the DC supply. This is called a step up converter because the voltage is stepped up, and such circuits are quite common.
 
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Back in high school, I had a 10,000 turn air core coil as a result of failed attempt to make my own transformer to build a bug zapper (I shorted the primary coil). When I connected the coil to a 1.5 volt D cell battery, I got jolt when disconnecting the battery. So I took the coil to school, and connected it to an oscilloscope. We measured 1600 volts peak from a 1.5 volt battery.

Probably a good thing I had removed the steel transformer core and/or did not think of using a more powerful power supply.
 
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I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...
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