(i) R is not affected by frequency so current does not change on an increasing frequency.
(ii) Inductive reactance X_L = 2π γL γ when frequency increases ~ increases and hence current in the circuit decreases.
(iii) Capacitive reactance X_c = \({ X }_{ c }=\frac { 1 }{ 2\pi \gamma c } \)
As the frequency ⋎ is increased, X_c decreases, and hence current increases.

The current in the coil will flow clockwise the direction of current will be from P to Q in coil PQ and from C to D in coil CD.

The current in the series RLC circuit becomes maximum at resonance. Due to the increase in current, the voltage across L and C are also increased. This magnification of voltages at series resonance is termed as Q-factor. It gives a measure of sharpness of current peak in the resanance condition of LCT circuit

RLC circuits have many applications like filter circuits, oscillators, voltage multipliers, etc. An important use of series RLC resonant circuits is in the tuning circuits of radio and TV systems.

At the transmitting point, the voltage is increased and the corresponding current is decreased by using a step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using a step-down transformer, and then it is given to consumers.

Power factor will be maximum, when Φ = 0 i.e., \(\tan ^{-1}\left(\frac{X_L-X_C}{R}\right)=0\)
\(\therefore X_L=X_C \Rightarrow L \omega=\frac{1}{C \omega}  \)
\(\therefore \omega=\frac{1}{2 \pi \sqrt{L C}}\)
∴ Current I be max \(I_m=\frac{V_m}{R}\)
Hence power factor of a RLC series circuit becomes maximum, when
(i) \(X_L=X_C\)
(ii) Current \(I_m=\frac{V_m}{R}\) will be maximum
(iii) Frequency \(\omega_r=\frac{1}{2 \pi \sqrt{L C}}\)

Change in magnetic field, dB = (8000 - 2000) = 6000 Wb
Change in time, dt = 6s
Radius of the coil, r = 2 x 10^-2 m
Induced emf, e = 44 V,
Area of the coil A =πr² = \(\frac { 22 }{ 7 } \) x (2 x 10^-2)²
= 12.56 x 10^-4 m²
∴ Number of turns in the coil,
\(N=\frac { e }{ A\frac {dB}{dt }} \frac{44}{12.56\times10^{-4} \times(6000/6)}\)
\(N=\frac { 44 \times10 }{12.56 }=3.503\times10=35 \)
∴ Number of turns in the coil = 35

Q factor is defined as the ratio of voltage across L or C to resonance to the applied voltage
Q - factor = \(\frac{Voltage \ across \ L \ or \ C \ resonance }{Applied \ voltage}\)
\(Q-factor=\frac{X_{L}}{R}=\frac{1}{R}\sqrt\frac{{L}}{C}\)

(i) If Ns>Np(K>1),thenVs>Vp and Is (ii) If NsIp. This is step-down transformer, where voltage is decreased and the current is increased.

AC generator work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf

Emf can be produced by changing magnetic flux in any of the following ways:
(i) By changing the magnetic field B
(ii) By changing the area A of the coil and
(iii) By changing the relative orientation θ of the coil with magnetic field.

Yes! An emf will be induced in the wire because it moves perpendicular to the horizontal component of Earth’s magnetic field and hence it cuts the magnetic lines of Earth's magnetic field.

The magnetic field of a cylindrical magnet is symmetrical about its axis. As the magnet is rotated along the axis of the solenoid, there is no induced current in the solenoid because the flux linked with the solenoid does not change due to the rotation of the magnet.

Current leads the applied voltage by \(\frac{\pi}{2}\) in a capacitive circuit.
This is the resistance offered by the capacitor, called capacitive reactance (X_c). It measured in ohm.
\({ X }_{ c }=\frac { 1 }{ \omega C } \)
The capacitive reactance (X_c) varies inversely as the frequency. For a steady current, f = 0
∴\({ X }_{ c }=\frac { 1 }{ \omega C } -\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 0 } =\infty \)
Thus a capacitive circuit offers infinite resistance to the steady current.

\(N_1  =4 \times 10^4, \mathrm{~N}_2=100, \mathrm{I}=2 \mathrm{~A}, \mathrm{~A}=4 \times 10^{-4} \mathrm{~m}^2  \)
\(\mathrm{t}  =0.04 \mathrm{~s}, \mathrm{e}=?  \)
\(\mathrm{~B}  =\mu_0 \mathrm{n}_1 \mathrm{I}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{I}}{l}=\frac{4 \pi \times 10^{-7} \times 4 \times 10^4 \times 2}{1}  \)
\(\mathrm{~B}  =100.48 \times 10^{-3} \mathrm{~T}  \)
\(\phi  =\mathrm{BA}=100.48 \times 10^{-3} \times 4 \times 10^{-4}  \)
\(\therefore \phi_1  =\phi_2=\phi=401.92 \times 10^{-7} \mathrm{wb}  \)
\(\mathrm{d} \phi  =\phi_1-\left(-\phi_2\right)=2 \phi=2 \times 401.92 \times 10^{-7}  \)
\(\therefore \mathrm{d} \phi  =803.84 \times 10^{-7} \mathrm{wb}\)
EMF induced in the coil,
\(e=-N_2 \frac{d \phi}{d t}  \)
\(e=\frac{-100 \times 803.84 \times 10^{-7}}{0.04}=-200.96 \times 10^{-3} \mathrm{~V}\)
\(e=-0.2 \mathrm{~V}\)

Self inductance of a coil is the inductance that enables to poduce an opposing induced emf in it, when the current in the coil changes 
Physical significances :
The inductance plays the same role in a circuit as mass and moment of inertia play at a mechanical motion. When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit (Figure (a)). Likewise, when circuit is broken, the decreasing current induces an emf in the reverse direction. This emf now opposes the decay of current (Figure (b)).
Thus, the inductance of the coil opposes any change in current and tries to maintain the original state

X_L = 184 Ω; X_C = 144 Ω
R = 30 Ω
(i ) The impedance is
\(Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \)
\(=\sqrt { { 30 }^{ 2 }+{ (184-144) }^{ 2 } } \)
\(=\sqrt { 900+1600 } \)
Z = 50Ω
\(\text { (ii) } \phi =\tan ^{-1}\left(\frac{X_1-X_c}{R}\right) \)
\( \phi =\tan ^{-1}\left(\frac{184-144}{30}\right) \)
\(\phi =53.1^{\circ} \)

N_P = 460 turns; N_S = 40,000 turns
V_P = 230 V; R_S = 10⁴ Ω
(i) Secondary voltage,
\({ V }_{ s }=\frac { { V }_{ p }{ N }_{ s } }{ { N }_{ p } } =\frac { 230\times 40,000 }{ 460 } \)
= 20,000V
Secondary voltage per turn,\(\frac { { V }_{ S } }{ { N }_{ S } } =\frac { 20,000 }{ 40,000 } =0.5V\)
(ii) Power delivered
= \({ V }_{ s }{ I }_{ s }=\frac { { V }_{ s }^{ 2 } }{ { R }_{ s } } =\frac { 20,000\times 20,000 }{ { 10 }^{ 4 } } =40kW\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

The direction of conventional current is always opposite to the direction of flow of electrons.

Given: N = 1000
A = 100 cm² = 10-2 m²
v = 100 rpm = \(\frac{100}{60}\) rps
B = 3.6 x 10^-27
em = ?
em = NBAω = NBA (2πγ)
= 1000 x 3.6 x 10^-2 x 10^-2 x 2 x \(\frac{22}{7}\times\frac{100}{60}\)
e = 5.77 V

Given :Primary voltage E₁ = 300 V
Secondary voltage E₂ = 3000 V
To find:
Primary current I₁ = ?
Secondary current I₂ = ?
Solution:
\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } =\frac { { E }_{ 2 } }{ { E }_{ 1 } } =\frac { 3000 }{ 300 } \) = 10 : 1

(i) The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get \(\frac { dE }{ dt } =\frac { 1 }{ 2 } \left( 2v\frac { dv }{ dt } \right) +\frac { 1 }{ 2 } k\left( \frac { dx }{ dt } \right) =0\)
or m \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } +kx=0\)
since \(\frac { dx }{ dt } =\nu \) and \(\frac { dv }{ dt } =\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } \)
(ii) This is the differential equation of the oscillations of the spring-mass system. The general solution of an equation is of the form
x(t) = X_m cos (ωt + Φ)
where X is the maximum value of x(t), ω the angular frequency, and Φ the phase constant.
(iii) Similarly, the electromagnetic energy of the LC system is given by
\(U=\frac { 1 }{ 2 } { Li }^{ 2 }+\frac { 1 }{ 2 } \left( \frac { 1 }{ C } \right) { q }^{ 2 }\) = constant
Differentiating U with respect to time, we get
\(\frac { dU }{ dt } =\frac { 1 }{ 2 } L\left( 2i\frac { di }{ dt } \right) +\frac { 1 }{ 2C } \left( 2q\frac { dq }{ dt } \right) =0\)
(or) \(\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } +\frac { 1 }{ C } q=0\) ....(1)
since \(i=\frac { dq }{ dt } \frac { di }{ dt } =\frac { { d }^{ 2 }q }{ d{ t }^{ 2 } } \)
(iv) The general solution of equation (1) is of the form
q(t) = Q_m cos (ωt + Φ)
(v) where Q_m is the maximum value of q(t), ω the angular frequency, and Φ the phase constant.

(i) Consider a circuit containing a resistor of resistance R, a inductor of inductance L and a capacitor of capacitance C connected across an alternating voltage source (Figure ). The instantaneous value of the alternating voltage is given by the equation
ሀ = V_m sin ωt   ......(1)

(ii) Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, Land C.
(iii) We know that voltage across R (V_R) is in phase with i, voltage across L (V_L) leads i by \(\frac { \pi }{ 2 } \) and voltage across C (V_c) lags behind i  by \(\frac { \pi }{ 2 } \)
(iv) The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor \(\vec { OI } \), V_R by \(\vec { OA } \); V_L by \(\vec { OB } \); V_c by \(\vec { OC } \) as shown in Figure.
(v) The length of these phasors are OI = I_m, OA = I_mR, OB = I_mX_L; OC = I_mX_C 
The circuit is either effectively inductive or capacitive or resistive that depends on the  value of V_L or V_C. Let us assume that V_L>V_C so that nef voltage drop across L-C combination is V_L - V_C which is represented by a phasor \(\vec { OD } \)
(vi) By parallelogram law, the diagonal \(\vec { OE } \) gives the resultant voltage ሀ of V_R and (V_L - V_C) and its length OE is equal to V_m Therefore,
V²_m = V²_R + (V_L - V_C)² = \(\sqrt { { ({ { I }_{ m }R) } }^{ 2 }+{ ({ I }_{ m }{ X }_{ L }-{ I }_{ m }{ X }_{ C }) }^{ 2 } }=I_m \sqrt {R^2+({X_L-X_C)}^2}\) 
or  \({ I }_{ m }=\frac { { V }_{ m } }{ \sqrt { R^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } } \)   ......(2)
\((or) { I }_{ m }=\frac { { V }_{ m } }{ Z } \) where z = \(\sqrt { { R }^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } \)   ......(3)
(vii) Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the Figure.


(viii) From phasor diagram, the phase angle between v and i is found out from the following relation
\(tan\phi =\frac { V_{ L }-{ V }_{ C } }{ { V }_{ R } } =\frac { X_{ L }-{ V }_{ C } }{ R } \)
Special cases:
(i) If X_L > X_C (X_L - X_C) is positive and phase angle \(\phi \) is also positive. It means that the applied voltage leads the current by \(\phi \) (or current lags behind voltage by \(\phi\)). The circuit is inductive.
∴v = V_m sin ωt; i = I_m sin(ωt - \(\phi \))
(ii) If X_L < X_C (X_L - X_C) is negative and\(\phi \) is also negative. Therefore current leads voltage by \(\phi \) (or voltage lags behind current by\(\phi \)) and the circuit is capacitive.
∴ = V_m sin ωt; i = I_m sin(ωt + \(\phi \))
(ii) If X_L = X_C \(\phi \) is zero. Therefore current and voltage are in the same phase and the circuit is resistive
∴v = V_m sin ωt, i = I_m sinωt 

Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction:
(i) In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core.
(ii) The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.
(iii) The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.
Working:
(i) If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core.
(ii) If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil.
(iii) This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.
(iv) As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil v_p or back of ε_p is given by,
v_p = ε_p = -N_p \(\frac{dФ_B}{dt}\)             ........(1)
(vi) The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by ,
ε_s = -N_s\(\frac{dФ_B}{dt}\)  
Where N_p and N_s are the number of turns in the primary and secondary coil respectively. If the secondary circuit is open, then ε_s = ሀ_s  where u is the voltage  ሀ_s across secondary coil
ሀ_s = ε_s = -N_s \(\frac{dФ_B}{dt}\)               ........(2)
From equations (1) and (2),
\(\frac{v_s}{v_p}=\frac{N_s}{N_p}\) = K                    ........(3)
(vii) This constant K is known as voltage transformation ratio. For an ideal transformer, input power v_pi_p= Output power v_si_s 
where i_p and i_s are the currents in the primary and secondary coil respectively.
(ix) Therefore,
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\)             ........(4)
Equation (4) is written in terms of amplitude of corresponding quantities
\(\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}\) = K
i) If N_s> N_p (or K > 1), ∴ V_s > V_p and I_s <  I_p This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.
ii) If N_s< N_p (or K < 1), ∴ V_s < V_p and I_s >  I_p This is step-down transformer where voltage is decreased and the current is increased.
Efficiency of a transformer:
The efficiency η of a transformer is defined as the ratio of the useful output power to the input power. Thus,
\(η=\frac{Outpur \ power}{Input \ power}\times100% \) %          ....(5)

(i) Consider a rectangular coil of turns kept in a uniform magnetic field \(\vec{B}\) as shown in Figure (a). The coil rotates in anti-clockwise direction with an angular velocity  ω about an axis, perpendicular to the field, and to the plane of the paper.
(ii) At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value  \({ \Phi }_{ m }=BA\) (where A is the area of the coil).
(iii) In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is NBA cos ωt, is due to the component of B normal to the plane of the coil (Figure(b)). The component parallel to the plane  (B sin ωt) has no role in electromagnetic induction. Therefore, the flux linkage with the coil  at this deflected position is
NΦ_B = NBA cosθ = NBA cos ωt
According to Faraday's law, the emf induced at that instant is

\(ε=- \frac { d }{ dt } (N{ \Phi }_{ B })=-\frac { d }{ dt } (NBAcos\omega t)\)
= -NBA (-sin ωt)ω
= NBAω sin ωt
(iv) When the coil is rotated through 90⁰ from initial position, sin ωt = 1. Then the maximum value of induced emf is
ε_m = NBAω
Therefore, the value of induced emf at that instant is then given by,
ε = ε_m sinωt   ....(1)

(vii) It is seen that the induced emf varies as sine function of the time angle ωt. The graph between induced emf and time angle for one rotation of coil will be a sine curve (Figure) and the emf varying in this manner is called sinusoidal emf or alternating emf.
If this alternating voltage is given to a closed circuit, a sinusoidally varying current flows in it. This current is called alternating current and is given by,
i = I_m sinωt ....(2)
Where, I_m is the maximum value of induced current.

(i) Self inductance of a coil is desined as the flux linkage with the coil, when a current of 1 A flows through it.
\(L=\frac{{ N\Phi }_{ B }}{i}\)
\(L={ N\Phi }_{ B }\)
(ii) Self inductance of a coil is defined as the opposing emf induced in the coil when the rate of change of current through the coil is 1 A s^-1
\(\varepsilon =\frac { d{ (N\Phi }_{ B }) }{ dt } \) 
\(=-\frac { d(Li) }{ dt }\)
\(\varepsilon =-L\frac { di }{ dt } \)
L = -e

Conservation of energy:
(i) The truth of Lenz's law can be established on the basis of the law of conservation of energy. The explanation is as follows:
(ii) According to Lenz's law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion
(iii) As a result, there will always be a resisting force on the moving magnet.
(iv) Work has to be done by some external agency to move the magnet against this resisting force
(v) Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.
(vi) On the contrary to Lenz's law, let us assume that the induced current helps the cause responsible for its production.
(vii) Now When we push the magnet litle bit towards the coil, the induced current helps the movement of the magnet towards the coil.
(viii) Then the magnet starts moving towards the coil without any expense of energy. This, becomes a perpetual motion machine.
(ix) In practice, no such machine is possible. Therefore, the assumption that the induced current helps the cause is wrong.