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How does large over currents passes through a device during turn-on
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khatus:
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
fsr:
Well, let's go with a classical one: transformer, rectifier and capacitor PSU. When you turn the device on, the capacitor is completely discharged, so as the current "thru" a capacitor is I = C * (dv / dt) and the voltage potentially goes from 0 to the peak AC voltage of the transformer very quickly, the current is going to be really large. Diodes don't limit current, so it's going to pass thru them into the capacitor.
Discharged capacitors are probably the #1 cause of inrush current in circuits, i believe. Incandescent bulbs also have very low resistance when cold, but who uses them anymore? :D
Dave:
If we are talking about transformers, residual flux and having the transformer connected to the line at the wrong (read: less than ideal) moment, will cause the core to saturate and the current will go sky high.
khatus:
IDEngineer:
I'm presuming your question is "Why does 'inrush' current occur?" where inrush means a surge of current at turn-on. If so, let's consider the opposite question: Why doesn't the steady-state current remain high after turn-on? An easy, non-math answer is that something changes after the inrush current. If you were dealing with a pure resistance, the current wouldn't change - there would be a single, constant value based on Ohm's Law. So what can we think of that changes with current flow? How about reactance?

Let's start with capacitive reactance. As others have answered, current flowing into the total system capacitance (which we presume to be discharged) can, for easy analysis, be thought of as an almost dead short, so the inrush current is limited primarily by the current source's ability to provide it. As the current begins to charge the total system capacitance, its voltage increases, and the difference between the capacitive charge voltage and the charging voltage is reduced, which reduces the current flowing into that capacitance. Eventually the two voltages achieve parity and capacitive inrush current drops to zero, leaving you only with the steady-state current required by the rest of the system. (Again, I must emphasize that this is a simplified explanation.)

Next we have inductive reactance. For an easy thought experiment, an inductor can be considered a dead short at DC. So when you first turn on the current, the inductor presents basically zero impedance to the current flow and you have very high inrush current. But the effect of current in an inductor is to create a magnetic field, which in turn (pun intended!) increases the reactance of the inductor and thus its "resistance" to further changes in current. Just as with capacitive reactance above, you thus have an effect that initially permits near-infinite current to flow but then begins resisting that current as its magnetic field builds up.

Thus you have an easy two answers to the question "What can we think of that changes with current flow?" If your load were a pure resistance, inrush current wouldn't occur. But since all real-world systems have reactance too, inrush current is very real.

Does that help?

EDIT: Look up "Cuk Converter" and check out the first inductor. When the converter's switch is closed, that inductor presents a dead short to the input current source. How can that possibly work? The answer is the same as above... inductive reactance limits the current flow once the magnetic field is established. Otherwise that first inductor would act as a very expensive fuse.
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