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Electromagnetic Induction

Transformer action.

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Power output = Power input

E2I2 = E1I1

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Emf induced in a coil rotating in a uniform

magnetic field.

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The magnetic flux passing through the coil is

given by,

φ = nAB cos θ = nAB cos ωt

The induced emf is given by,

If f is the frequency of rotation of the coil, ω = 2πf

∴ e = 2πfnABsin ωt As 2πfnAB are constant

∴ e = E0 sin ωt where E0 = 2πfnAB

θ = ωt 0 π/2 π 3π/2 2π 5π/2 3π

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E 0 E0 0 -E0 0 E0 0

Applied ac voltage and current to resistor

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The figure shows a resistor of value R connected

across an ac source of emf e.

∴ e = E0 sin ωt and current i is given as

where I0 = E0 / R is called peak value of current.

Hence, current is in phase of emf. The variation of

current and emf with respect to time is shown

below

i. Peak value :

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The magnitude of maximum value of

alternating emf or current is called its peak

value. The peak values of emf and currents

are E0 and I0 respectively. The time interval

between two successive positive or negative

peak values is called periodic time T of the

emf or current.

ii. Average (mean) value:

The constant value of current or voltage read

by a dc ammeter or voltmeter over one

complete cycle of the current or emf is called

its average value.

Mathematically, the average value of

alternating emf is given by,

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Similarly, Iav = 0. Hence, the dc meters read

zero emf or current.

iii. Root Mean Square (rms) value :

The rms value of alternating current is that

steady current, when flows through a given

resistor, for a given time, produces that same

heat that is produced by the alternating

current, in the same resistor in same time.

Mathematically it is given as,

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Q.43  In an ideal transformer, the primary and the secondary voltages always have          (a.43) Equal magnitude  (b.43) The same phase   (c.43) A phase difference of 900          (d.43) A phase difference of 1800

Q.44  An a.c. ammeter measures the          (a.44) Peak value of current          (b.44) Average value of current          (c.44) r.m.s. value of current          (d.44) Mean square currentQ.37  A step down transformer works on 220volts a.c. mains. It is used to light a

100W, 20V bulb. The main current is 0.5A. What is the efficiency of the transformer?

          (a.37) 91%   (b.37) 80%   (c.37) 71%   (d.37) 51%

Inductive reactance of a pure inductor

∴ e’ = - ωL I0 cos ωt

∴ e’ = - E0 cos ωt where E0 = ωL I0 is the peak

value of emf.

For ideal inductor e’ = e

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Thus, the value of φ is

found to be .

emf leads current by .

As E0 = ωL I0,

XL is called inductive reactance. As ω = 2πf,

XL = 2πfL

If L is expressed in henry and frequency is

expressed in hertz, XL is expressed in ohm.

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Capacitive reactance of a pure capacitor

Application of A. C. to a capacitor:

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This is equal to the instantaneous value of the

applied e. m. f. (e)

This is equation of the instantaneous current.

If cos ωt = 1 then i = imax = i0 = peak value of

current = e0 ωC

Where i0 = e0 ωC …….(2)

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Comparing e = e0 sinωt with equation (2) we

conclude that

i. e. m. f. or voltage across the capacitor

lags behind the current by phase angle

rad or current leads the e. m. f. or voltage

by rad.

ii. Current and e. m. f. both are sinusoidal in

nature of same frequency.

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We know that

“capacitive reactance” (XC)

…….(3)

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Capacitive reactance is defined as the ratio of

r. m. s. voltage across the capacitor to the r.

m. s. current. S. I. unit of XC is ohm.

Q.23  The coefficient of mutual induction of two coils is 10mH. If the current flowing in one coil is 4A then the induced e.m.f. in the second coil will be

          (a.23) 40mV                   (b.23) 20mV          (c.23) Zero  (d.23) 10mVQ.24  The current in a coil decreases from 5A to 0 in 0.1sec. If the average e.m.f.

induced in the coil is 50V, then the self inductance of the coil is          (a.24) 0.25H          (b.24) 0.5H  (c.24) 1H     (d.24) 2H

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