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When an inductor is connected with a voltage source we get equal and opposite voltage on inductor against the source voltage. That equal and opposite voltage gradually decreases with time which allows the current(caused by the source voltage) to gradually rise.

My question is what makes the equal and opposite voltage in an inductor to fall gradually ?

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  • $\begingroup$ The voltage across an ideal inductor connected to an ideal constant voltage source does not fall gradually since, by KVL, it must be identical to the voltage across the voltage source. So, are you in fact thinking about a series RL circuit where, as the voltage decreases across the inductor, the voltage increases across the resistor? $\endgroup$ – Alfred Centauri May 19 '18 at 20:28
  • $\begingroup$ even its a RL circuit current will rise in the circuit when equal and opposite voltage across the inductor will fall. @AlfredCentauri $\endgroup$ – Alex May 19 '18 at 20:35
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You can't just connect a pure inductor to a voltage source. Even in a thought experiment there has to be some resistance. Even if your inductor is a superconductor the voltage source has an internal resistance.

OK, so to begin with the current is zero, and the inductor has a potential $L{dI \over dt}$ which is equal to the applied voltage $V$. That means ${dI\over dt}=V/L$ and after some small time $t$ a current $Vt/L$ flows. That produces a voltage drop across the resistor, meaning that much less across the inductor. As time goes by the current continues to increase, the voltage across the resistance increases (tending eventually to $V$) and the voltage across the inductor falls (tensing eventually to $0$).

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  • $\begingroup$ thanks alot for the answer. what is wrote about is theoratical. theoratically its possible to attach the inductor with a voltage source. $\endgroup$ – Alex May 19 '18 at 20:15
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    $\begingroup$ "You can't just connect a pure inductor to a voltage source. Even in a thought experiment." - Why? Assuming the ideal circuit theory context, there's no problem with say, a constant voltage source connected to an inductor with a switch that is closed at $t =0$. In that case, the voltage across the inductor is a step function $v_L(t) = V_S\,u(t)$ and the current through is a ramp $i(t) = \frac{V_S}{L}t\, u(t)$ $\endgroup$ – Alfred Centauri May 19 '18 at 20:21
  • $\begingroup$ In linear circuit theory with idealized elements one may never connect an ideal current source in series with an inductor nor an ideal voltage source in parallel with a capacitor, but a voltage source in series with an inductor or a current source in parallel with a capacitor are allowed. $\endgroup$ – hyportnex May 19 '18 at 20:40
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theoratically its possible to attach the inductor with a voltage source

Yes, in the context of ideal circuit theory, it is possible to do so without contradiction.

Let the voltage source have constant voltage across $V_S \gt 0$ and the inductor have inductance $L$. The inductor is connected to the voltage source at time $t = 0$. By KVL, the voltage across the inductor is given by

$$v_L(t) = V_S\, u(t)$$

where $u(t)$ is the unit step function.

The circuit current is described by a very simple differential equation:

$$V_S\,u(t) = L \frac{di}{dt}$$

with solution

$$i(t) = \frac{V_S}{L}\, t\, u(t)$$

In other words, the current is zero for $t \le 0$ and increases at a constant rate for $t \ge 0$.

Note that the current is unbounded (does not reach a limiting value) as $t \rightarrow \infty$ which is clearly unphysical. For a physical circuit, the internal resistance of the voltage source and/or the inductor will limit the current.

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Assuming the DC voltage source is ideal, the voltage on its terminals is constant. If the real (resistive) coil is connected to those terminals, the voltage across the coil will be the same, constant.

What you probably mean is that as the current increases, the induced EMF on the coil decreases in magnitude; in the end the current is constant and induced EMF is zero.

It is important to understand the difference between the voltage and EMF. They have the same units, but they are due to different agents: the voltage is due to the voltage source but the induced EMF is due coil wire carrying time-varying current. When coil is connected to a voltage source, the EMF works against the voltage, so the current is low and only gradually increases to its final value.

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  • $\begingroup$ so why the back emf decreases with time? thats the main question. when back emf will decrease then the current will be able to increase. $\endgroup$ – Alex May 20 '18 at 20:04
  • $\begingroup$ Because induced EMF is proportional to rate of change of current. As time goes, current approaches to its maximum value, so rate of change of current decreases in time. Hence induced EMF decreases in time. $\endgroup$ – Ján Lalinský May 20 '18 at 20:48

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