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I read in my textbook that when dc voltage source like a battery is connected to inductor,it develops an emf which opposes the increase of current in the circuit.On applying kirchoff's law in the loop,I can say that the back emf developed in inductor is equal to voltage of the battery ,since there is no other element in circuit where voltage drop can take place.

But then why does the current flow because there is no potential difference as the cell and inductor both have their positive plates connected to each other and have equal emf?Isn't this scenario equivalent to two cells having equal emf connected to each other in such a way that positive terminals of both cells are connected together?

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  • \$\begingroup\$ there is no potential difference But there is, the voltage of the voltage source (the battery) will set the voltage across the inductor. Plates is a term we use in conjunction with a capacitor, what you mean are terminals. The situation is simple: you apply a DC voltage to the inductor, a current starts flowing through the inductor, charging (magnetically) the inductor. In an ideal case the current will increase linearly to an infinite value (or for as long as you have it connected to the voltage source). \$\endgroup\$ – Bimpelrekkie Mar 17 '16 at 9:47
  • \$\begingroup\$ After a while the inductor ceases hostilities vs power source and lets the current to flow. \$\endgroup\$ – Marko Buršič Mar 17 '16 at 13:09
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In your question, you say 'it develops an EMF which opposes the increase of current'. That's not quite correct.

A better way of putting it would be 'when a voltage is applied to an inductor, the inductor develops an EMF which is due to the increase of current.'

We notice that if we connect a voltage to an inductor, then after some time t, a current is flowing, given by the equation \$I_t = \dfrac{Vt}{L}\$. This is the experimental observation. It is this observation that we have to work from, to derive the rules.

Inductors can seem a bit 'spooky', but the situation is exactly the same for a simple resistor. Does the voltage across it cause the current to flow? Or does the current flowing through it cause the voltage? They both happen.

We could say the change in current causes the voltage, or the voltage causes the change in current. Neither statement is 'more true' than the other. One statement may be more useful than the other, depending on what we are trying to analyse. If we are looking at a boost converter for instance, the 'voltage causes current change' is better for the inductor charge up phase, the 'change of current causes voltage' is better for the inductor run down phase.

If we start with the rules and try to work out what is happening, we will come up short, because physics only describes what happens, not why.

We know enough about what is going on with electrical phenomena to know that we don't understand it truly at a level where we can say this or that causes this or that to happen. Voltage, for instance, is a measure of the energy it takes to move charge around. To say then that voltage 'causes' something is to put the cart before the horse. All we can say is that when this happens, say current increases through an inductor, we also notice that we can measure that, which is a voltage between the terminals. Don't try to overthink electrical causes and effects.

You might like to look up Zeno's Arrow Paradox, to see what happens when you try to overthink mechanics.

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  • \$\begingroup\$ So I cannot equate the two scenarios mentioned in details, that is 2 cells joint together with positive terminals connected together because in this case both the cells already have a potential difference .But in the case of solenoid and battery,the emf developed in solenoid is due to the increase in current in circuit,so in a way current is responsible for the equal and opposing emf of solenoid? \$\endgroup\$ – Raksh23 Mar 17 '16 at 13:14
  • \$\begingroup\$ Also my teacher gave me an alternate explanation that even if there is no potential difference,there can be current in circuit as V=IR and in this case both V and R are equal to zero,but we cannot comment about the current in circuit.Is this correct? \$\endgroup\$ – Raksh23 Mar 17 '16 at 13:19
  • \$\begingroup\$ 'Opposes' is more accurate than 'due to', as the former indicates polarity. \$\endgroup\$ – Chu Mar 17 '16 at 16:47
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Good morning! Look, your analizis is correct, but only for the inicial instant, called transient (in that moment, the inductor will opose to the current change and generate a voltage Vl=L*(di/dt)). After that, the circuit will get to the permanent state (in dc, the inductor is just a wire in permanent state). You can see this, by solvig the circuit:

enter image description here

In your case, R is the resistences of the wires and inductor.

Lets analize the result:

For t=0:

i=0, so, if there is no current flowing, it means that the voltage induced in the inductor plus the voltage drop in the wires is equal to the voltage supply.

For t=tends to infinite:

i=V/R, that means that permanent current is just limited by the resistences of the wire and inductor. If R is small, the the permanent current is going to be high (short circuit)

Have a good day!

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