Electrical Engineering - Inductors - Discussion

@All.

I'm guessing that the frequency here is infinite because it's an instantaneous change, giving you a theoretically infinite resistance in the inductor (2*π*f*L), so approximately all of the voltage drop is in the inductor at t=0.

To break it down, a non-changing voltage level (constant DC voltage) has a frequency of 0, while an instantaneously changing voltage (such as a sudden flick of a switch) results to an infinite frequency (or extremely high frequency).

Then with impedances, we can infer that inductors resist high frequencies based on the formula while capacitors resist low frequencies more.

So, theoretically infinite frequency, with an inductor component in series, would result in a formula for the voltage drop of

V(inductor)=V(source)*2*π*frequency*inductance/(2*π*frequency*inductance+resistance)

With the voltage drop in the resistance to virtually zero.

Yes.

In case of AC. Above answer is right as current cannot change instantaneously due to impedance.

But here battery is DC. Here inductor will act as a conductor (just like resistor). No ideal inductor exists. Ideal condition can be taken only for theoretical calculation as well as simulation purpose.

In real time. Inductor will offer some finite resistance to drop the voltage like resistor.

Since we have no idea about finite resistance in inductance. Hence we assume ideal condition 0V.

Under transient condition (Here:-closing of switch) inductor behaves like open circuit so voltage across inductor is 6.

After than current in the circuit gradually increase and after finite time inductor behaves short circuit so that voltage across it 0.

Here we know from transient analysis that when switch is on (t=0) then initially inductor behave like a open circuit. So no current flow through the circuit because of open circuit therefore no we drop across are and input we appear across L.

Though it is DC source. At 0 instant the voltage will bot be 6 V.

The voltage has to raise from 0 to 6 volts, which is change in voltage. This change is not acceptable by the inductor. Hence initially the inductor acts as a open circuit.

Here Voltage does not change.

R=1000ohm,
V=6V,
Then, I= V/R=6/1000=0.006A.

And again R will not change because switch is off means R does not equal to zero,
Hence R=1000 ohm.

Then V =I * R = 0.006 * 1000,
= 6V.

As the nature of inductor is to oppose the change of current through it, it initially opposes the current flow through it. Thereby it acts as an open circuit. & total drop appears at inductor.

At instant, there is no change in current of inducter, so the current will for some time .

V=IR
I=6/1=6 mA

After switch off current will remain same so the voltage will be
v=6*1=6 V.

I agree with @Soumya sen. Here we're considering transient condition. Thus inductor is open. Hence voltage across inductor is 6V. Current=0 ==>voltage drop across resistor is 0.

As according to AC formula, V=Vm(1-e^-rt\L) for RL circuit the value of voltage at inductor will be zero, for the instant when switch is closed dc will behave same as AC.