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As one can see there is a circuit in which consists of a high voltage source, high value Inductor, high value capacitor and a switch. Suppose switch is closed for some time and current flows through the circuit .Now, both Inductor and capacitor are charged and energy is stored in them.

Now, I want to ask few questions

  1. When one attempts to open the switch ,why the switch burns out due to Charged capacitor?

  2. When one attempts to close the switch ,why the switch burns out due to Charged Inductor?

enter image description here

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closed as off-topic by Jon Custer, ACuriousMind, user36790, Dan, Norbert Schuch Sep 16 '16 at 7:58

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  • $\begingroup$ Usually the problem with inductive loads is opening the switch... $\endgroup$ – Jon Custer Sep 14 '16 at 16:06
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    $\begingroup$ "The switch closed for a long time" condition of your circuit is that the capacitor has the same voltage across it as that of the voltage supply and no current is flowing in the circuit. I therefore think that the switch is not in danger of burning out when it is opened. $\endgroup$ – Farcher Sep 14 '16 at 16:10
  • $\begingroup$ @Farcher suppose voltage source is of 100 V ,and voltage across Inductor / capacitor is 50 V. $\endgroup$ – user3559780 Sep 14 '16 at 16:23
  • $\begingroup$ @Jon Custer but why? $\endgroup$ – user3559780 Sep 14 '16 at 16:28
  • $\begingroup$ As @JonCuster has pointed out when you open the switch the current in the circuit will collapse very rapidly and so there will be a large voltage induced across the inductor which may well cause an arc to form across the switch contacts. $\endgroup$ – Farcher Sep 14 '16 at 16:31
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The switch will not burn in the schematic you have provided*.

  • When the switch is closed (assuming 0 initial conditions), the charge current of the capacitor will be limited by the inductor. (*) Of course, if the current rating of the switch is smaller than even that limited current value, it can still be damaged.

  • After some time (assuming there are parasitic resistance somewhere in the circuit), the capacitor will be charged to the value of V, and the inductor will carry 0 current. Note that no energy will be stored in the inductor.

  • When you later open the switch, it will again not be damaged, since no current would flow through it anyway.

If we assume that all components are ideal, then the circuit will oscillate forever. In that case, if you open the switch the moment the current though it reaches its maximum, the switch may get damaged as the ideal inductor will generate potentially infinite voltage when the current though it stops. Whether an ideal switch is able to withstand infinite voltage is another question.

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  • $\begingroup$ please assume that as the switch was closed for some time and charge is stored in Inductor and capacitor. Let V=100, And voltage across Inductor / capacitor is 50V. And then you observe the above cases given in question part $\endgroup$ – user3559780 Sep 14 '16 at 17:23
  • $\begingroup$ Inductor cannot store any charge, perhaps you mean magnetic flux? If the voltage across C and L is 50V, that would mean there is a current flowing through the inductor and the switch will take damage when opened (assuming it takes damage from electric arcs). $\endgroup$ – Dmitry Grigoryev Sep 14 '16 at 17:28
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    $\begingroup$ Hint: stating all conditions in your question gets you shorted and more precise answers. $\endgroup$ – Dmitry Grigoryev Sep 14 '16 at 17:31

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