Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase difference between V and I
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can ve obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.

The back emf in the coil opposes the change in the current as per Lenz's law.

Self-Inductance When the current in a coil is changed, a back emf is induced in the same coil. This phenomenon is called self-inductance. If Lis self-inductance of coil, then
\(N\phi \propto I\Rightarrow N\phi =LI\Rightarrow L=\frac { N\phi }{ I } \)
The SI unit of self-inductance is Henry (H).
(ii) Mutual inductance of solenoid coil system
\(M=\frac { { \mu }_{ 0 }{ N }_{ 1 }{ N }_{ 2 }{ A }_{ 2 } }{ l } \)
Here, N₁ = 15, N₂ = 1, l = 1cm = 10^-2m,
A = 2.0cm² = 20 x 10^-4m²
∴ \(M=\frac { 4\pi \times { 10 }^{ -7 }\times 15\times 1\times 2.0\times { 10 }^{ -4 } }{ { 10 }^{ -2 } } \)
\(=120\pi \times { 10 }^{ -9 }H\)
Induced emf in the loop
\({ \varepsilon }_{ 2 }=M\frac { { \Delta I }_{ 1 } }{ { \Delta t } } (numerically)=20\pi \times { 10 }^{ -9 }\frac { \left( 4-2 \right) }{ 0.1 } \)
\(=120\times 3.14\times { 10 }^{ -9 }\times \frac { 2 }{ 0.1 } =7.5\times { 10 }^{ -6 }V=7.5\mu V\)

\(X_L=\omega L=2 \pi \nu L=2 \pi \times 50 \times 0.25\)
\(=25 \pi \mathrm{ohm}\)

High frequency alternating current passed through the coil surrounding the metal piece produces eddy currents in the metal piece. The eddy currents produce joule heating in the metal piece on account of its resistance.

1 weber = 10⁸ Maxwell

\(\Phi=B A \cos \Phi=\left(\frac{F}{I l}\right) A=\frac{M L T^{-2}}{A L} \cdot L^2\)
\(=\left[M L^2 T^{-2} A^{-1}\right]\)

The magnetic flux is maximum when area is held perpendicular to the direction of magnetic field.

Given
R = 200Ω, C = 15.0μF = 15.0 x 10^-6F
V = 220 V, ν = 50 Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
\(Z=\sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+(2 \pi v C)^{-2}}\)
\(=\sqrt{(200 \Omega)^{2}+\left(2 \times 3.14 \times 50 \times 15.0 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
\(=\sqrt{(200 \Omega)^{2}+(212.3 \Omega)^{2}}\)
= 291.67Ω
Therefore, the current in the circuit is
\(I=\frac{V}{Z}=\frac{220 \mathrm{~V}}{291.5 \Omega}=0.755 \mathrm{~A}\)
(b) Since the current is the same throughout the circuit, we have
\(V_{R}=I R=(0.755 \mathrm{~A})(200 \Omega)=151 \mathrm{~V}\)
\(V_{C}=I X_{C}=(0.755 \mathrm{~A})(212.3 \Omega)=160.3 \mathrm{~V}\)
The algebraic sum of the two voltages, V_R and V_C is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
\(V_{R+C}=\sqrt{V_{R}^{2}+V_{C}^{2}}\)
= 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Step down transformer:

Principle: When the current flowing through the primary coil changes, an emf is induced in the secondary coil due to the change in magnetic flux linked with it i.e., it works on the principle of mutual induction. For step down transformer,
N_s < N_p, hence \({\epsilon}_{s}<{\epsilon}_{p}\).
\({ { {\epsilon}_{s} }\over{ {\epsilon}_{p} } }={{{N}_{s}}\over{{N}_{p}}}\)
For an ideal transformer, P_in = P_out
\({\epsilon }_{p }{ I}_{p }={ \epsilon}_{s }{I }_{s }\Rightarrow{{{I}_{p}}\over{{I}_{s}}}={{{\epsilon}_{s}}\over{{\epsilon}_{p}}}={{{N}_{s}}\over{{N}_{p}}}\)
P_in = P_out = 550 W = \({\epsilon}_{p}{I}_{p}\) = 550
\(220\times{I}_{p}=550\Rightarrow{I}_{p}={{550}\over{220}}={{5}\over{2}}=2.5 \ A\)

Let VL, VR, Vc and V represent the voltage across the inductor, resistor, capacitor and the source respectively. VR is parallel to I. Vc is pi/2 behind I and VL is pi/2 ahead of I.
Clearly,
\(={ i }_{ 0 }^{ 2 }[{ R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 }]\)
\({ i }_{ o }=\frac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } } \)
\(Impedence=\frac { { V }_{ 0 } }{ { i }_{ 0 } } =\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^{ 2 } } \)
\(\\ =\sqrt { { R }^{ 2 }+\left( \omega L-\frac { 1 }{ \omega C } \right) ^{ 2 } } \)

The capacitance of a capacitor in the tuning circuit is varied such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal to be received. When this happens, the amplitude of the current becomes maximum in the receiving circuit.

\(Here, \ { I }_{ v }=5.0A, \ v=50Hz\)
\((i) \ { I }_{ 0 }=\sqrt { 2 } { I }_{ v }=1.414\times 5.0=7.07A\)
\((ii) \ { I }_{ m }=\frac { 2 }{ \pi } { I }_{ 0 }=\frac { 2 }{ 3.14 } \times 7.07=4.5A\)
\((iii) \ From \ I={ I }_{ 0 } \ sin \ \omega t={ I }_{ 0 }sin2\pi \ vt\)
\(=7.07sin2\pi \times 50\times \frac { 1 }{ 300 } =\frac { 7.07\sqrt { 3 } }{ 2 } =6.12A\)

(i) (b): The relative motion between the coil and the magnet produces change in the magnetic flux in the coil. The induced emf is always in such a direction that it opposes the change in the flux.
(ii) (c)
(iii) (d)
(iv) (a): When an electron is moving from right to left, the flux linked with loop (which is going into the page) will first increase and then decrease as the electron passes by. So the induced current I, in the loop will be first clockwise and will change direction (i.e. will become anticlockwise) as the electron passes by.

(v) (c): When current in coil A increases with time, there will be a change of flux in coil B which will induce a current in B. Now, according to Lenz's law, the direction of induced current in B will be opposite to the direction of current in A. Thus, if two loops carry current in opposite direction they will repel each other.

(a): The capacitive reactance of capacitor is given by \(X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\) 
So this is infinite for D.C (f = 0) and has a finite value for A.C Therefore a capacitor blocks D.C and offers an easy path for A.C.

(c): The manner in which the two coils are oriented, determines the coefficient of coupling between them \(\text { i.e., } K=\sqrt{\frac{M}{L_{1} L_{2}}}\), where L, and L2 are self-inductance of two coils. When the two coils are wound on each other, the coefficient of coupling is maximum and hence mutual inductance between the coil is maximum.