important notes jee - physics -electromagnetic induction part 2
DESCRIPTION
Important notes JEE - Physics - Electromagnetic Induction Part 2TRANSCRIPT
9011041155 / 9011031155
• Live Webinars (online lectures) with
recordings.
• Online Query Solving
• Online MCQ tests with detailed
solutions
• Online Notes and Solved Exercises
• Career Counseling
Electromagnetic Induction
Transformer action.
1
9011041155 / 9011031155
Power output = Power input
E2I2 = E1I1
2
9011041155 / 9011031155
Emf induced in a coil rotating in a uniform
magnetic field.
3
9011041155 / 9011031155
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
4
9011041155 / 9011031155
E 0 E0 0 -E0 0 E0 0
Applied ac voltage and current to resistor
5
9011041155 / 9011031155
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 :
6
9011041155 / 9011031155
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,
7
9011041155 / 9011031155
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,
8
9011041155 / 9011031155
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
9
9011041155 / 9011031155
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.
10
9011041155 / 9011031155
Capacitive reactance of a pure capacitor
Application of A. C. to a capacitor:
11
9011041155 / 9011031155
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)
12
9011041155 / 9011031155
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.
13
9011041155 / 9011031155
We know that
“capacitive reactance” (XC)
…….(3)
14
9011041155 / 9011031155
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
15
9011041155 / 9011031155
16
9011041155 / 9011031155
• Ask Your Doubts
• For inquiry and registration, call 9011041155 /
9011031155.
17