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Why are they dangerous connection?

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humblelearner

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I would appreciate that someone could explain to me why are they dangerous connection and how to improve or amend them.
 

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LEFT:
THe capacitor initially is uncharged so there is 0V across it. What else has 0V across it? A piece of wire (aka short-circuit). That initial current pulse to charge the cap might be too high if the capacitance is too large.

RIGHT
Current in and inductor cannot change instantly, it only changes gradually (sort of like how capacitor voltage). When you suddenly disconnect current from an inductor the inductor tries to keep the current flowing at the same value (only allowing it to decrease gradually). How does it do this? It generates a huge voltage spike that keeps the current flowing (current will flow anywhere if the voltage is high enough). This voltage spike is damaging.
 
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May I know what will the damage be on the left diagram which means if I really connect it that way, which part of the circuit would be damaged?
 
"E" could be damaged due to inrush current if "C" is large since "C" is effectively a short circuit when the switch is initially closed.

If "E" has a built in fuse, then it's fuse might eventually blow due with repeated switch closures, again due to the high inrush current if "C" is large.

Also, if the "K" switch contacts are underrated the contacts might "eventually" weld shut due to inrush current.

Sometimes a small value series resistor is placed between "E" and "K" to limit the inrush current.

creakndale
 
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