Question

When a DC current passes through the coil, it will produce a magnetic flux around the coil. Is this the reason why the coil get heated? Because the magnetic flux will be stable after all as DC current does not oscillate?

Answer 1

For a coil V=L di/dt is the voltage equation.since there is no change of current in DC ,di/dt=0 except during the transient period.Then V=0.This means short circuit.So high currents will flow and burn out the winding. it can also be explained in another way.

for a coil Z=R+jωL. ω= 2*pi*f,For DC f=0,so ω=0 ,Z=R.For a coil L will be more and R will be less.So Z will be very less and very high currents will flow and damage the circuit.

If we neglect the proper impedance of the wire, the theoretical impedance of the coil is proportional to the frequency : Z_coil = j.L.ω with ω = 2.π.f and L the inductance of the coil.

If the current is DC then f = 0 and consequently Z_coil = 0, the coil acts only as a wire.

Obviously the wire has a little impedance that should be taken in account if necessary, according to your needs. It can cause some Joule losses for example, depending on the cross section of the wire and the current value.

Supposes a coil is there. Initially the switch of the circuit is off. So, no current passing through it. Now at some time "t", you trigger the switch ON. So from current=0A the coil now faces a sudden change of current through it. Thus, for a moment flux is generated through the coil and it opposes the current initially. But as this is DC, flux generated remain just for a fraction of a second(called Transient period) and then the coil allows the passage of DC Current.

So, in short COIL ACTS AS AN OPEN CIRCUIT FOR TRANSIENT PERIOD AND THEN AS SHORT CIRCUIT WHEN A DC FLOWS THROUGH IT.