(i)  γ =  γ_r occurs when X_L = X_C. Then the circuit becomes purely resistive. So current and voltage will be in the same phase.
(ii) X_L = 2π γL and \({ X }_{ c }=\frac { 1 }{ 2\pi \gamma c } \)
When γ =  γ_r, XL is small and Xc is large the circuit is capacitive, so current leads the voltage in phase.
The circuit is inductive. So current lags behind the voltage in phase.

A quantity which varies sinusoidally with time and represented as the projection of a rotating vector is called phasor.

The phenomenon of electrical resonance is possible when the circuit contains both L and C. Only then does the voltage across L and C cancels one another when VL and VC are 1800 out of phase and the circuit becomes purely resistive. This implies that resonance will not occur in RL and RC circuits.

The average value of alternating current is defined as the average of all values of current over a positive half-cycle or negative half-cycle.

(i) The current is drawn directly from fixed terminals on the stator without the use of brush contacts.
(ii) The insulation of stationary armature winding is easier.
(iii) The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.
(iv) Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.

N = 40 turns; B = 2 Wb m^-2
A = 200 cm² = 200 x 10^-4 m²;
Initial flux, \(\Phi_i\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos60^o
since θ = 90°− 30°= 60°
\(\Phi_i\)= 2 x 10^-2 Wb
Final flux, \(\Phi_f\) = BA cos\(\theta\)
= 2 x 200 x 10^-4 x cos0^o since \(\theta\) = 0^o
\(\Phi_f\) = 4 x 10^-2Wb
Magnitude of the induced emf is
\(ε =N\frac { d{ \Phi }_{ B } }{ dt } \)
\(=\frac { 40\times (4\times { 10 }^{ -2 }-2\times { 10 }^{ -2 }) }{ 0.2 } =4V\)

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.

(i) The total resistance offered to the flow of current due to resistance R, inductive resistance X_L, and capacitive reactance X_c in a circuit is called impedance.
It is given by \(z=\sqrt { { R }^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } \)
(ii) At resonance, when X_L= X_C

(i) The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
(ii) There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor; the other magnetic energy of the inductor carrying current.
(iii) Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
(iv) By examining, the analogies between the various quantities can be understood and these correspondences.
(v) The angular frequency of oscillations of a spring-mass is given by equation
\(\omega =\sqrt { \frac { k }{ m } } \)
k ⟶ \(\frac { 1 }{ C } \) and m ⟶ L. Therefore, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \)

L = \(\frac { 2 }{ \pi } \)H; C = \(\frac { { 10 }^{ -4 } }{ \pi } F\); R = 100Ω
V_RMS = 220 V; f = 50Hz
X_L= 2πfl = 2π x 50 x \(\frac { 2 }{ \pi } \) = 200Ω
X_c = \(\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 2\pi \times 50\times \frac { 10^{ -4 } }{ \pi } } 100 \Omega\)
(i) Impedance, Z = \(\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^2 } \)
=\(\sqrt { 100^{ 2 }+(200-100)^{ 2 } } \) = 141.4Ω
(ii) Peak value of current,
I_m = \(\frac { { v }_{ m } }{ Z } =\frac { \sqrt { 2 } V_{ RMS } }{ Z } \)
= \(\frac { \sqrt { 2 } \times 220 }{ 141.4 } \) = 2.2 A
(iii) Power factor of the circuit
\(cos\phi =\frac { R }{ Z } =\frac { 100 }{ 141.4 } \)= 0.707
(iv) Power factor at resonance
\(cos\phi =\frac { R }{ Z } =\frac { R }{ R } \) = 1

f = 50Hz ; V_m = 20V
Instantaneous voltage, υ = V_m sinωt
= V_m sin2πvt
= 20sin(2π x 50)t = 20sin(100 x 3.14)t
υ = 20sin 314t
Time for one cycle, \(T=\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
= 20 x 10⁻³ s = 20ms
The wave form is given below

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

\(\phi_B =10 t^2-50 t+250  \)
\(e  =\frac{-d \phi_B}{d t}  \)
\(=\frac{-d}{d t}\left(10 t^2-50 t+250\right) \)
=-(20 t - 50) 
=-20 t + 50
When, t = 3 s, e =-20(3) + 50 = -60 + 50
e = -10V

Given: E_v = 220 V
⋎ = 50 Hz
E = ?
To find:
We can use E_v = \(\frac { { E }_{ m } }{ \sqrt { 2 } } \)
E_m = \(\sqrt2\) x E_v
= \(\sqrt2\) x 220
= 311.08 V
∴ Instantaneous Voltage E = E_m sin ωt
E = E_m sin (2πγ)t
= 311 sin (2ㅠ x 50)t
= 311 sin 100 πt.

Given: The maximum value of voltage Vm = 300V
Angle frequency of ac voltage w = 21t ⋎ = 314
RMS value of the source V_rms = ?
V = 300 sin 314t.
To find:
Peak voltage V_m= 300V, frequency ⋎ =?
Solution:

⋎ = 50 Hz.
RMS voltage, V_rm = \(\frac { { V }_{ m } }{ \sqrt { 2 } } \)
= 0.707 X 300 = 212.1 V
V_rms = 212.1 V.