# Question about the inductor plz

Discussion in 'General Electronics Chat' started by fankoushat, Mar 8, 2010.

1. ### fankoushat Thread Starter New Member

Mar 8, 2010
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0
Hi guys ,

All articles and lessons say " when connecting an inductor with a dc source, the first value on the inductor will be almost the same as the dc source value, then dropes gradually". How, we know that the induced voltage equals to the value of L multiplied by di/dt, so they should say that the value on the inductor at the first point, will be L di/dt, not " almost the source value".

If we consider the induced voltage to be almost the same as the source value, so any inductor with any value of L will produce the same value of Vind at the first moment!!!!!!

I think we can produced higher values of the induced voltage ( also may be higher than the source itself) if the number of turns were high enough.

I also read the lesson " Inductors and calculus ", the instructor considered a virtual experiement by connecting a neon lamp in parallel with the inductor, he said ' after closing the switch , the voltage on the coil will be the same as the source, where the source was 6 volts, so the neon lamp will not be lite, but after opening the switch (note that the inductor has very big number of turns) the neon lamp will be lite, why not at the begining?

plz give me more explaination, thanks alot

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Consider the case of an ideal 10V DC voltage source connected via an ideal [non-arcing] switch to an ideal 10H inductor in parallel with an ideal 1000Ω resistor.

At time t=0 the switch is closed and current begins to flow in the circuit.
For the parallel resistor the current jumps immediately to a constant 10mA.
For the ideal inductor the current rises linearly at the rate of 1A/sec - based on Ldi/dt.

Suppose at time t=1second the switch is opened - no arcing of the switch occurs.

What do we observe after the switch is opened?

The current in the inductor at t=1 sec is 1A and will now be flowing via the resistor rather than from the DC source. If the 1000Ω resistor has a current of 1A flowing then the voltage across it (and the inductor) will be 1000V - rather than the original 10V from the source. Additionally, the voltage across both the resistor and inductor will be of the opposite polarity to that observed when the 10V source was connected via the switch.

What led to this condition? When the inductor was connected to the 10V source it was accumulating energy in the magnetic field set up within it by the current flow - the inductor was being "charged" by the current drawn from the 10V source in the time interval from 0 to 1 second.

As time progresses from 1 second onwards, this stored energy will be progressively dissipated as heat in the resistor. At some point in time all the stored magnetic energy will be lost as heat and current will no longer flow. This is where the concept of the circuit time constant also arises. The decay of current in the circuit is governed by a "natural" exponential relationship whose time constant is determined by the values [physical properties] of L and R.

3. ### fankoushat Thread Starter New Member

Mar 8, 2010
17
0
OK, Thanks for this answer alot, but what is the case if we connect a 10V dc battery with 1000 ohm pure resistor and 10 H pure inductor in series??

Then, consider we replaced this inductor with another one 10 mH pure inductor.

My question is, Do both inductors produce almost 10 v at t=0 ( the source voltage)?

so this equation (Vind = L di/dt) is wrong.

4. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
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That equation predicts the voltage drop across an inductor for a given rate of change in current.

At t=0, the current is zero; it takes time to build up.

Alternatively, if you think current starts to flow in zero time, then di/dt (change in current / time) is infinite, therefore Vind is also infinite, so limited only by the supply voltage.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
I agree with rjenkins.

It can be shown mathematically for the series RL case that the voltage across the inductor is a decaying exponential function where, for a DC source voltage Vs

$e_{L}(t)=V_{s}{\epsilon^{-\frac{Rt}{L}}$

At t=0

$e_{L}(0)=V_{s}$

Which is what one expects ...

At t=L/R (one time constant)

$e_{L}(L/R)=V_{s}\epsilon^{-1}=0.368V_{s}$

At t=∞

$e_{L}=0$

It doesn't matter whether the inductor is 10H or 10mH the principle is the same - the only difference being that as the inductance gets smaller (for the same resistance value) then the changes happen progressively more quickly - i.e. the time constant gets shorter.

The point also should be made that throughout the time interval discussed above [t=0 to infinity], the inductor voltage is always governed by the relationship e=Ldi/dt where i is the inductor current.

If Vs=10V and L=10H at t=0, i=0 but di/dt=1A/sec

If Vs= 10V and L=10mH at t=0, i=0 but di/dt=1000A/sec

Last edited: Mar 8, 2010
6. ### fankoushat Thread Starter New Member

Mar 8, 2010
17
0
That means , if L increase, the time interval increases, so di/dt at the beginning becomes less so that Vind = L di/dt = Vsource. And if L decreases, the time interval decreases also, so di/dt at the beginning becomes more higher , so that Vind = L di/dt = also Vsource.

Yes, that is the right answer, Thanks alot I appereciate your assistance very much

7. ### fankoushat Thread Starter New Member

Mar 8, 2010
17
0
Excuse me, to make really sure of this answer, I have another question.

If we connect a pure inductor directly with ac source, the phase difference between the induced voltage on the inductor and the current passing through the circuit will be 90 degree. But what happen if we increase or decrease L, according to the above answer, changing L ( which means changing XL ) wiill affect only the current signal. More XL less amplitude for the current signal, Less XL , More amplitude for the current signal, but the induced voltage signal will never chang, and will always be the same as the source voltage ( where the circuit is pure inductive ) , right?