The current in an a.c circuit is waltless if the average power consumed in the circuit is zero.

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.

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.

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.

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

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.

It is defined as the ratio of true power to the apparent power of an a.c circuit. It is equal to the cosine of the phase angle between current and voltage in the a.c circuit.
It is given by
cos Φ = \(\frac{True \ power}{Apparent \ power}\)
= \(\frac { { P }_{ average } }{ { V }_{ rms }-{ I }_{ rms } } \)

(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

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.

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

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

\(\omega=250 \mathrm{rad} / \mathrm{s}, C=4 \times 10^{-} \mathrm{F}  \)
\(R=100 \Omega, \quad \mathrm{V}_{\mathrm{R}}=40 \mathrm{~V}  \)
\(\therefore I_{\mathrm{R}}=\frac{V_R}{100}=\frac{40}{100}=0.4 \mathrm{~A}  \)
\(\omega=\frac{1}{\sqrt{L C}}  \)
\(\omega^2=\frac{1}{L C}  \)
\((250)^2=\frac{1}{L \times 4 \times 10^{-6}}  \)
\(L=\frac{1}{4 \times(250)^2 \times 10^{-6}}  \)
\(=\frac{1}{4 \times 250 \times 250 \times 10^{-6}}  \)
\(=\frac{1}{1000 \times 10^{-6} \times 250}  \)
\(=\frac{10^3}{250}=\frac{1000}{250}=4 \mathrm{H}\)
Voltage acnoss L, V_t = IX_L
V_L = l x L x ω
= 0.4 x 4 x 250
0.4 x 1000 = 400 V

\(\mathrm{V}_{\mathrm{P}}=220 \mathrm{~V}, \mathrm{~V}_{\mathrm{s}}=11 \mathrm{~V}  \)
\(\mathrm{I}_{\mathrm{P}}=6 \mathrm{~A}, \mathrm{I}_{\mathrm{s}}=100 \mathrm{~A} .  \)
\(\text {Efficiency }=\frac{\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}  \)
\(=\frac{11 \times 100}{220 \times 6}=\frac{1100}{220 \times 6}=\frac{5}{6}=0.83\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-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

Phase difference between V and i = \(\frac{\pi}{2}\)red
∴ P_average = \(\frac { { V }_{ m }{ I }_{ m } }{ 2 } \). cos Φ = \(\frac { 50\times 10 }{ 2 } \) cos 0^o

Given: Voltage across an inductor
V_L=10V
Voltage across a capacitor V_c = 10V
Voltage across resistance V_e = 30
The phase difference between applied voltage and
Solution:
current in the circuit tan Φ = \(\frac { { V }_{ L }-{ V }_{ C } }{ { V }_{ R } } \)
tan Φ = \(\frac { 10-10 }{ 30 } \)
tan Φ = 0
∴ phase difference Φ = 0