How does large over currents passes through a device during turn-on

Thread Starter

khatus

Joined Jul 2, 2018
95
I understood how does large inductive over voltage spikes(the voltage across inductor summed up with the source voltage)
generated during turn off of a device
but now i want to know How does large over currents passes through a device during turn-on??If possible please explain with picture
 

WBahn

Joined Mar 31, 2012
29,976
You have a very nonspecific question, so the best we can do is provide examples of how it can happen.

If this is homework, it would help if you posted the actual question.
 

Reloadron

Joined Jan 15, 2015
7,501
The inrush current is dissipated by the device at turn on. This is figured into the design of the device. Knowing what the inrush current will be and the duration of the inrush current figures into the equation. I have a 120 Volt 100 Watt incandescent lamp laying here. The cold DC resistance of the lamp is right about 9.5 Ohms. So if I look at this 120 V / 9.5 Ohms = 12.63 Amps. That would be the instantaneous current but at turn on the resistive filament quickly heats and the current drops to about 0.833 Amps. However, for a short period of time the bulb will draw well over ten times its normal operating current. The filament is designed to handle the inrush current so the bulb dissipates the inrush current for the time necessary. Fuses are also designed around inrush currents.

When necessary the inrush current can be controlled. This is inrush current limiting. For example bringing an AC motor slowly up to speed. So it depends on the case and design and knowing the duration of inrush current.

Ron
 

MaxHeadRoom

Joined Jul 18, 2013
28,617
As per Ron's post, a AC motor at turn on is essentially a transformer with a shorted turn secondary, a high current will flow due to the DC resistance of the winding , inrush, and until rotation takes place, decreasing the 'secondary' current.
Max.
 

ian field

Joined Oct 27, 2012
6,536
I understood how does large inductive over voltage spikes(the voltage across inductor summed up with the source voltage)
generated during turn off of a device
but now i want to know How does large over currents passes through a device during turn-on??If possible please explain with picture
Turn on surge is likely to be a full wave rectifier charging a reservoir capacitor from fully discharged. Some parts like lamps have a positive temperature coefficient - they settle at normal current once up to running temperature.

An inductor as part of the load has lines of magnetic flux around it - interrupting the current lets the magnetic field collapse and produce a back emf.
 
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