# Current through inductor

When there is an abrupt change in current through an inductor, there will be an equal and opposite voltage developed across it.

1. So how does current actually flow if there is no potential difference(inductor voltage equals and opposed battery voltage) across inductor.

$$\int_0^t \frac{V}L dt = i(t)$$

2. According to the equation however, when there is a constant DC voltage applied to inductor the resulting current is ramp.

i.e., $$i(t) = \frac{V}L t$$

What is the physical meaning of this?

( When I've a capacitor connected to constant current source, the voltage across it would be ramp since the charges are continuously being provided) Is there an explanation like this for inductor when a constant voltage is applied to it?

• In very simple terms, you simply can't have an abrupt change in current through an inductor. If you try to change the current quickly (eg by removing a voltage source) it just generates higher and higher voltages until something gives - sparks, insulation failure, or damage to what is driving it. This is analogous to a capacitor, which effectively cannot have abrupt voltage changes forced upon it. – Icy Oct 8 '15 at 15:27
• 2. A constant voltage applied to a pure inductance (no resistance, if that's possible!) will result in a ramp of current, with slope proportional to the voltage. A capacitor charging at a constant current is accumulating charge at a constant rate therefore the voltage is increasing a constant rate, since V=Q/C and C is constant. – Chu Oct 8 '15 at 15:35
• What is the explanation for 2.? – Aditya Patil Oct 8 '15 at 15:39
• It's in the equation: $e=L\dfrac{di}{dt}$; for a constant e, $\dfrac{di}{dt}$ must be constant. – Chu Oct 8 '15 at 15:50
• I think the stumbling block here may be the fact that the voltage source is exactly balanced by the induced emf, and yet a current flows. This is counter-intuitive and arises when dealing with 'ideal' sources and components. A way around this difficulty is to include a resistance in the circuit and let this resistance tend to zero. – Chu Oct 8 '15 at 17:26