Inductors are simply coils which store energy in form of magnetic field. They are denoted by L and are measured in Henry (H).
A current carrying coil/conductor produces magnetic field and if the current is a time varying then it produces time varying magnetic field. The reason for this will be studied in EMF tutorial. A time varying magnetic flux inside a coil induces an EMF across it, according to faraday’s law of electromagnetic induction. And this change created by induced voltage opposes the current that created it, according to Lenz’s law.
The induced Emf or the voltage drop is given by -L (di/dt). Thus if the current is constant, the voltage across an ideal inductor is expected to be zero. But practically, every inductor always exhibits some resistance. Hence there’s always some voltage drop across inductor even if the current is not varying. Inductor coils can be observed in chokes, transformers etc. While solving problems, the equivalent inductance can be found similarly as done for resistors. We'll show you how to deal with inductors in serial and parallel.
While solving problems, the equivalent inductance can be found similarly as done for resistors. The series and parallel combination of the inductors are resolved as given by the formulae. L(series) = L1 + L2 + L3 and L(parallel) = 1/((1/L1)+(1/L2)+...) - ie, same as the formulas for resistors.
The change in magnetic flux inside a coil is responsible for inducing the voltage across an induction coil. This magnetic flux could’ve been generated by current flowing in that same coil itself (cause for self-inductance) or by another coil placed near to it (mutual inductance).
Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction. If the current is steady, then the magnetic flux through the loop will remain constant. However, suppose the current I changes with time, then according to Faraday’s law, an induced emf will arise to oppose the change. The direction of induced emf is opposite to the direction of source emf. The induced current will flow clockwise if dI/dt >0, and
counterclockwise if dI/dt <0. The property of the coil in which its own magnetic field opposes any change in current is called “self-inductance,” and the emf generated is called the self-induced emf or back emf . All current carrying loops exhibit this property.
Suppose two coils X, Y are placed near each other. The X coil has N1turns and carries a current I1 which gives rise to a magnetic field. Since the two coils are close to each other, some of the magnetic field lines through coil 1 will also flow through coil2. There will be a magnetic flux per turn of coil 2 due to I1. Now, by varying I1 with time, there will be an induced emf associated with the varying magnetic flux in the second coil. Also, the time rate of change of magnetic flux 21 in coil 2 is proportional to the time rate of change of the current in coil 1. M depends only on the geometrical factors Ra and Rb and is independent of current flowing.
Dot Convention: The tutorial will also cover the dot convention which is used to denote the voltage polarity of two mutually inductive components. Fewer times delta can also be seen in place of dot. Both means the same that the two inductive elements are coupled and indicates polarity. The polarity of all dotted terminals will be the same at any particular time. Assuming an ideal transformer with no leakage inductance, when the current in a loop enters the dot, then positive voltage is induced at the dot of the other loop.
Alternately, when current in a loop leaves the dot, negative voltage is induced at the dot of the other loop. Similarly if two inductors are in series the dot convention can be
used in the same manner as in case of transformer.
As the inductor opposes any change in the current through it, work must be done by an external source so as set up current in the inductor. Energy storage in inductor is analogous to that of the capacitor but the form of energy stored is magnetic in inductor and electrical in later. The power, or rate at which an external emf E(ext) works to overcome the self-induced emf E(L) and establish current I in the current is the product I x E(ext). If only the external emf and the inductor are present, then this equals I x L x dI/dt
If the current is increasing with dI/dt > 0 then P>0, which means that the external source is doing positive work to transfer energy to the inductor. Thus, the internal energy U of the inductor is increased. On the other hand, if the current is decreasing with dI/dt < 0, we have P<0. In this case, the external source takes energy away from the inductor, causing its internal energy to go down. The total work done by the external source to increase the current from zero to I is then W(ext) = LI2/2. This is equal to the magnetic energy stored in the inductor.
Here are some of the problems which have been solved in this tutorial :
Q: Solving given circuits for equivalent inductance.
Q: Find the inductance of a uniformly wound solenoid having N turns and length L. Assume that L is much longer than the radius of the windings and the core of the solenoid is air.
Q: What would happen if a ferromagnetic material is placed inside the solenoid?
Ans. The inductance would increase. For a given current, the magnetic flux is now much greater because of the increase in the field originating from the magnetization of the ferromagnetic material.
Q: Calculate the inductance of solenoid with a material of 100 o in its core. It has 100 turns and the length of the solenoid is 50cm and its cross sectional area is 10sqcm.
Q: A long solenoid with length l and a radius R consists of N turns of wire. A current I passes through the coil. Find the energy stored in the system.
Q: If the flux through a 200-turn coil changes steadily from 1 Wb to 4 Wb in one second, what is the voltage induced?
Q: If the current through a 5-mH inductance changes at the rate of 1000 A/s, what is the voltage induced?
Q: A long solenoid with length l and a cross-sectional area A consists of N1 turns of wire. Another solenoid of N2 turns is around it.
(1) Calculate the mutual inductance M, assuming that all the flux from the solenoid passes through the outer coil.
(2) Relate the mutual inductance M to the self-inductances and of the solenoid and the coil.
Q: Consider a coil made of thin copper wire (radius ~ 0.25 mm) and has about 600 turns of average diameter 25 mm over a length of 25 mm. What approximately should the resistance and inductance of the coil be? The resistivity of copper at room
temperature is around 20 nΩ-m. Note that your calculations can only be approximate because this is not at all an ideal solenoid (where length >> diameter).
Q: Calculate the mutual inductance of two solenoids in the above case if N1= 1500 turns, N2= 800 turns, A= 1sqcm, l = 0.02m
Q: An inductor has a negligible resistance and an inductance of 200mH and is connected is series with a 1KΩ resistor to a 24V, d.c supply. Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. Find a) the current flowing in the circuit at a time equal to one time constant b) the voltage drop across the inductor at a time equal to two time constants and c) the voltage drop across the resistor after a time equal to a three time constants.
Q: In a series R-L circuit, the application of a direct voltage results in a steady state current of 0.632 I in 1 second. I being the final steady state value of the current. However, after the current has reached it final value, a sudden short circuit is applied against the source. What would be the value of the current after one second?
Q: Find the current in a series RL circuit having R=2 and L=10 while a dc voltage of 100V is applied. What is the value of this current after 5sec of switching on ?
Complete Tutorial with Solved Problems :
Related Tutorials ( Introduction to Electrical Circuits - DC ) :