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When an inductor is connected to a DC source, current starts passing through inductor. This generates magnetic field so flux associated with inductor changes. So a back emf is generated. As a result net emf decreases and so does current and magnetic field intensity. So magnetic flux change is there. Won't again an emf be induced in such a way to oppose the back emf? Thinking that all these happens in small time interval......

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2 Answers 2

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Yes. That's why you need a differential equation $$ L \frac{dI}{dt}= V, $$ whose solution (with $I=0$ at $t=0$) is $I=Vt/L$.

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  • $\begingroup$ What that differential equation mean or signify? $\endgroup$ Commented Aug 26, 2020 at 14:00
  • $\begingroup$ It tells you what the current in amps is at a time $t$ after you connect the DC voltage $V$ across the inductor. $L$ is the inductance in henrys. $\endgroup$
    – mike stone
    Commented Aug 26, 2020 at 16:12
  • $\begingroup$ What is the relation of this to the question? $\endgroup$ Commented Aug 26, 2020 at 16:46
  • $\begingroup$ It's the answer to what you were asking. At each point the back EMF must be is equal to the applied EMF (Kirchhoff Voltage law) and so $I$ increases to keep it so. If it is not the answer, then your question is unclear. $\endgroup$
    – mike stone
    Commented Aug 26, 2020 at 17:01
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As a result net emf decreases and so does current . . . .

This is not true, rather the rate of change of current decreases whilst the current continues to increase to an asymptotic value equal to the emf of the power supply divided by the resistance of the circuit.

In theory, the rate of change of current stays the same and the current just keeps increasing if there is no resistance in the circuit.

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