# Dissipation of energy in inductors

Please explain the dissipation of energy in a circuit containing a resistance and inductor after the switched in the circuit is opened.I mean how do you explain that the energy that is across inductor gets dissipated without any further closed circuit?

• Please include a schematic diagram of your circuit so we can understand what is the effect of opening the switch? – The Photon Feb 4 at 18:20
• Here is an article on the effect of inductive loads on switches. Hope it helps.artisancontrols.com/UtilityImages/ProductSupportPdf/… – Bob D Feb 4 at 21:24

A schematic is needed to give a complete answer, but consider the following:

The voltage across an ideal inductor is given by

$$V(t)=L\frac{dI(t)}{dt}$$

The inductor will therefore generate a high voltage to resist a large change in current which would be the case in attempting to open the circuit with a switch. If the switch is an air gap type, the inductor will create a high enough voltage to create an arc across the switch contacts, ionizing the air in the gap and allowing current to flow through the ionized air. The energy stored in the inductors magnetic field is thus dissipated in the combination of the arc and the series resistor.

Hope this helps.

• @JustyLaurent Answer was acceptable? – Bob D Feb 4 at 20:07
• What if the arc does not happen and the current does not continue through the gap? – Ján Lalinský Feb 4 at 21:15
• @JánLalinský But it does happen. Switches that control inductive loads have to have special inductive load ratings since arcing due to inductive loads can damage the contacts. – Bob D Feb 4 at 21:19
• It happens if the electric field is strong enough. If not, the air behaves as isolator and there is no arc. – Ján Lalinský Feb 4 at 21:23
• @JánLalinský Here is an article on the effects of inductive loads on switches.artisancontrols.com/UtilityImages/ProductSupportPdf/… – Bob D Feb 4 at 21:23