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### Inductance.

posted Oct 23, 2017, 11:18 AM by Upali Salpadoru   [ updated Oct 23, 2017, 9:29 PM ]

In electricity there are two forms of inductance. Mutual inductance and self inductance.

Mutual inductance is what happens in a transformer. When changing current in one  oil causes a current in another coil. The unit of mutual inductance is Henry (H)

Change in Magnetic field is proportional to current.

That is  φ / I is a constant (M)     Therefore  φ = m X i.

We know that  Volts = change in flux / time.  That is          V =  N x  Δ φ  / Δ t.

( N is the number of turns)

Substituting the value of φ  in the above we get :-            V = M . Δ I  /   Δ t.

Uses.

Inductors are used to tune radio circuits, filter out unwanted noise and in various other electronic circuits.

 Faraday’s LawThe rate of change in magnetic flux is proportional to voltage of the induced current Lenz’s LawThe direction of the induced current causes a force to oppose The change that caused it.

Example 1.

The current in primary increases from zero to 0.8A in 2.5 s. The secondary coil Voltage was 1.2 v.

 Find the following:-Rate of increasing current.I/tThe Mutual inductance.The time taken for the current to drop to zero  at the rate of 0.2A/s.The induced voltage during the drop. Answers:-Rate  =  Δ I  /   Δ t.                 = 0.8 / 2.5…... = 0.32A/s.     b)        M= V/ 0.32…….= 3.75H.

Inductor and self induction..

An inductor can simply be a coil of wire with a several attached to an electrical circuit.  When a current is passed it develops a strong magnetic field. According to Faraday’s law , when the current is building up , as there is a change in magnetic flux, a current is induced. According to Lenz’s law the induced current will be in the opposite direction to the current that caused the change. So this is sometimes referred to as the back emf.

When the inductor is charging  the equation  V= ir  has to be modified as

Supplied voltage  - Inductor voltage = ir.

This can be written as  Vs - VL  = ir

Charging and Discharging an Inductor.  When the circuit is switched on,  the bulb glows brightly and becomes dim. Then again when switched off the bulb becomes bright and goes off.

Let us use the following data for a calculation:-

Supply voltage   =  1.5 V.

Normal resistance of circuit =  0.3 Ω.

Time taken for current to be steady  = 0.2 seconds.

Inductance of coil   = 0.3 H

 Calculating the steady current in the circuit   Using V - ir   we get   1.5 = i x 0.3 ∴   i =  1.5 / 0.3   =  5.0 A. (amperes) Induced voltage in resistor VL.   =  L xI  / t.V      = 0.3 x 5.0 / 0.2  =  7.5 VIn the absence of supply current this can give a current of  I=  v/r                  =7.5 x 0.3 = 25 AThis is the reason for the bulb to be bright on switching on and off.

Energy stored in an inductor

J = ½ L x I 2

Unit for Inductance.

The basic unit of measurement for inductance is the Henry, ( H )

It can also be given as  Webers per Ampere ( 1 H = 1 Wb/A ).

Example 2

 a.Find the induced voltage when a current of 5A in a inductor of 3 H. is reduced to zero in  0.2 seconds.b. Find the energy stored i the capacitor. a) V =  3 x 5 /0.2  =  350/2   =  175 v.b) Energy   =  ½  L x 5 x 5                 = 0 .5  x  3 x 5x5  =  37.5
Inductor s too have a resistance.
Her the inductor and the resistance symbol are enclosed inside a rectangle . That shows it is the resistance of the coil.

Inductance 0.5 H  Current 6 .0A

Becomes zero in 0.05 s

1. a) Find the induced voltage.

2. b) Find the resistance.

1. a) v= L x I/t

V = 0.5 x 6 /0.05s

V = 3 /.05 =  60 v

b) V=ir

12 = 6 x r

r = 2 Ω.

1.0 A diagram of an induction coil is given here. This can step up a direct current to a required voltage.
a)    Label the numbered parts.      You may use the words from this list.
spark gap,    primary coil,    electro magnet,   vibrator ,  linear spring, adjustable screw,               secondary coil
b)   Primary coil has 120 turns and the secondary 3600 turns.
a) If the voltage supplied to primary is 12 V find the induced voltage in the secondary.