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

P _Average = V_rms. I_rms. cost
= V_rms. I_rms. \(\frac{R}{z}\)
= 200 x 1.5 x \(\frac45\)
P_average = 240W.

(i) The electric power generated in a power station situated in a remote place is transmitted to different regions for domestic and industrial use.
(ii) For long-distance transmission, power lines are made of conducting material like aluminum. There is always some power loss associated with these lines.
(iii) If I is the current through the wire and R the resistance, a considerable amount of electric power FR is dissipated as heat. Hence, the power at the receiving end will be much lesser than the actual power generated.
(iv) However, by transmitting the electrical energy at a higher voltage, the power loss can be controlled as is evident from the following two cases

Case (1) A power of 11,000 W is transmitted at 220 V.
Power P = VI
I = \(\frac{P}{V}=\frac{11,000}{220}\) = 0.5A
If R is the resistance of line wires,
Power loss = I²R = 50²R = 2500(R) watts.
Case (2) 11,000 W power is transmitted at 22,000 V.
I = \(\frac{P}{V}=\frac{11,000}{22,000}\)= 0.5A
Power loss = I²R = (0.5)²R = 0.25(R) watts. Hence it is evident that if power is transmitted at a higher voltage the loss of energy in the form of heat can be considerably reduced.

(i) S₁ and S₂ are two long solenoids each length l. The solenoid S₂ is wound closely over the solenoid S_1.
(ii) N₁ and N₂ are the number of turns in the solenoids S₁ and S₂ respectively. Both the solenoids are considered to have the same area of cross-section A as they are closely wound together.

(iii) I₁ is the current flowing through the solenoid S₁ The magnetic field B₁ produced at any point inside the solenoid S₁ due to the current I₁ is
\({ B }_{ 1 }={ \mu }_{ o }\frac { { N }_{ i } }{ l } { I }_{ 1 }\) ...(1)
(iv) The magnetic flux linked with each turn of S₂ is equal to B₁A.
Total magnetic flux linked with solenoid S₂ having N₂ turns is
Φ₂ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \) ...(2)
But Φ₂ = MI₁ ...(3) 
where M is the coefficient of mutual induction between S₁ and S₂
From equations (2) and (3),
MI₁ = \(\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }{ AI }_{ 1 } }{ l } \); M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)
(v) If the core is filled with a magnetic material of permeability μ,
M = \(​​\frac { { \mu }_{ o }{ N }_{ 1 }N_{ 2 }A }{ l } \)

By differentiating equation twice, we get
q(t) Q_m cos(ωt + Φ) ...(1)
\(\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } \) = - Q_m ω² cos (ωt + Φ) ...(2)
Substituting equations (1) and (2) in equation
\(\frac { dU }{ dt } =\frac { 1 }{ 2 } L\left( 2i\frac { di }{ dt } \right) +\frac { 1 }{ 2C } \left( 2q\frac { dq }{ dt } \right) =0\)
(or) \(L\frac { { d }^{ 2 }q }{ { dt }^{ 2 } } +\frac { 1 }{ C } q=0\)
we obtain
L[- Q_m ω² cos (ωt + Φ)] + \(\frac { 1 }{ C } \) Qm cos (ωt + Φ) = 0
Rearranging the terms, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \) 
This equation is the same as that obtained from qualitative analogy.

(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) During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor.
(ii) Although these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy, U= U_E + U_B = \(\frac { { q }^{ 2 } }{ 2C } +\frac { 1 }{ 2 } { Li }^{ 2 }\)
(iii) consider 3 different stages of LC oscillations and calculate the total energy of the system.
Case (i) When the charge in the capacitor, q Q_m = and the current through the inductor, i = 0, the total energy is given by
\(U=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } +0=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
The total energy is wholly electrical
Case (ii) When charge = 0; current = I_m, the total energy is
\(U=0+\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }=\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }\)
 \(=\frac { L }{ 2 } \times \left( \frac { { { Q }_{ m } }^{ 2 } }{ LC } \right) { Q }_{ m }\omega =\frac { { Q }_{ m } }{ \sqrt { LC } } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
Case (ii) When charge = 0 ; current = I_m the total energy is
\(U=0+\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }=\frac { 1 }{ 2 } { { Li }^{ 2 } }_{ m }\)
\(=\frac { L }{ 2 } \times \left( \frac { { { Q }_{ m } }^{ 2 } }{ LC } \right) \) since I_m= \({ Q }_{ m }\omega =\frac { { Q }_{ m } }{ \sqrt { LC } } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
Case (iii) When charge = q; current = t. the total energy is
U = \(\frac { { q }^{ 2 } }{ 2C } +\frac { 1 }{ 2 } { Li }^{ 2 }\)
(iv) Since q = Qm = cos ωt, i = \(\frac { dq }{ dt } ={ Q }_{ m }=\omega \) sin ωt. The negative sign in current indicates that the charge in the capacitor decreases with time.
\(U=\frac { { { Q }_{ m } }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\frac { { { L{ \omega }^{ 2 }Q }_{ m } }^{ 2 }{ sin }^{ 2 }\omega t }{ 2 } \)
\(U=\frac { { { Q }_{ m } }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\frac { { { L{ \omega }^{ 2 }Q }_{ m } }^{ 2 }{ sin }^{ 2 }\omega t }{ 2LC } \)
since \({ \omega }^{ 2 }=\frac { 1 }{ LC } \)
= \(\frac { { { Q }_{ m } }^{ 2 } }{ 2C } { (cos }^{ 2 }\omega t+{ sin }^{ 2 }\omega t)\)
\(U=\frac { { { Q }_{ m } }^{ 2 } }{ 2C } \)
From the above three cases, it is clear that the total energy of the system remains constant.

(i) Power of a circuit is defined as the rate of consumption of electric energy in that circuit. It is given by the product of the voltage and current. In an AC circuit, the voltage and current vary continuously with time.
(ii) The alternating voltage and alternating current in the series RLC circuit at an instant are given by
v = Vm sin ωt and i = I_m sin(ωt + Φ)
(iii) Where Φ is is the phase angle between v and i. The instantaneous power is then written as
P = v_i
= V_mI_m sin ω sin(ωt + Φ)
= V_mI_msin ωt[sin ωt cos Φ - cos ωt sin Φ)
P = V_mI_m[cos Φ sin² ωt - sin ωt cos ωt sin Φ]
(iv) Here the average of sin²ωt over a cycle is \(\frac { 1 }{ 2 } \) and that of sin wt cos wt is zero. Substituting these values, we obtain average power over a cycle.
= \(\frac { { V }_{ m } }{ \sqrt { 2 } } \frac { { I }_{ m } }{ \sqrt { 2 } } \) cos Φ
Pav = V_RMSI_RMS cosΦ
(v) where V_RMS I_RMS is called apparent power and cos Φ is power factor. The average power of an AC circuit is also known as the true power of the circuit.

(i) RMS value is also defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time. The effective value of an alternating voltage is represented by V_eff
(ii) The alternating current i = I_m sin ωt or i = I_m sin θ, is represented graphically in Figure. The corresponding squared current wave is also shown by the dotted lines.
(iii) The sum of the squares of all currents over one cycle is given by the area of one cycle of squared wave. Therefore,
I_RMS = \(\sqrt{\frac{Area \ of \ one \ cycle \ of \ squared \ wave}{Base\ length \ of \ one\ cycle}}\)  ...(1)
(iv) An elementary area of thickness dθ is considered in the first half-cycle of the squared current wave as shown in Figure. Let e be the mid-ordinate of the element. Area of the element = l²dθ
Area of one cycle of squared
wave = \(\int _{ 0 }^{ 2\pi }{ { i }^{ 2 }d\theta } \)

= \(\int _{ 0 }^{ 2\pi }{ { { I }^{ 2 } }_{ m } } { sin }^{ 2 }\theta d\theta ={ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }\theta d\theta } \)
\(={ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ \left[ \frac { 1-cos2\theta }{ 2 } \right] d\theta } \)
since \({ sin }^{ 2 }\theta =\frac { 1-cos2\theta }{ 2 } \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \int _{ 0 }^{ \pi }{ id\theta } =\int _{ 0 }^{ \pi }{ { I }_{ m }cos2\theta d\theta } \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } { \left[ \theta -\frac { sin2\theta }{ 2 } \right] }_{ 0 }^{ 2\pi }\)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \left( 2\pi -\frac { sin2\times 2\pi }{ 2 } \right) -\left( 0-\frac { sin0 }{ 2 } \right) \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \)x2π = I²_m π     [∵ sin 0 = sin 4π = 0]
Substituting this in equation'(1), we get
\(\sqrt { \frac { { { I }^{ 2 } }_{ m }\pi }{ 2\pi } } =\frac { { { I }^{ 2 } }_{ m } }{ \sqrt { 2 } } \)
[Base length of cone cycle is 2π]
I_RMS = 0.707 I_m
Thus we find that for a symmetrical sinusoidal current RMS value of current is 70.7 % of its peak value. Similarly for alternating voltage, it can be shown that
V_rms = 0.707 I_m

(i) The term RMS refers to time-varying sinusoidal currents and voltages and is not used in DC systems.
(ii) The root mean square value of an alternating current is defined as the square root of the mean of the squares of all currents over one cycle. It is denoted by I_RMS. For alternating voltages, the RMS value is given by I_VRM^.
(iii) The alternating current i = I_m sin ωt or i = I_m sin θ, is represented graphically in Figure. The corresponding squared current wave is also shown by the dotted lines.
(iv) The sum of the squares of all currents over one cycle is given by the area of one cycle of the squared wave. Therefore,
I_RMS = \(\sqrt{\frac{Area \ of \ one \ cycle \ of \ squared \ wave}{Base\ length \ of \ one\ cycle}}\) ....(1)
(v) An elementary area of thickness dθ is considered in the first half-cycle of the squared current wave as shown in Figure. Let i² be the mid-ordinate of the element. Area of the element = i2dθ
Area of one cycle of squared.
wave = \(\int _{ 0 }^{ 2\pi }{ { i }^{ 2 }d\theta } \)

= \(\int _{ 0 }^{ 2\pi }{ { { I }^{ 2 } }_{ m } } { sin }^{ 2 }\theta d\theta ={ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }\theta d\theta } \)
= \({ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ \left[ \frac { 1-cos2\theta }{ 2 } \right] d\theta } \)
since \({ sin }^{ 2 }\theta =\frac { 1-cos2\theta }{ 2 } \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \int _{ 0 }^{ \pi }{ id\theta } =\int _{ 0 }^{ \pi }{ { I }_{ m }cos2\theta d\theta } \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } { \left[ \theta -\frac { sin2\theta }{ 2 } \right] }_{ 0 }^{ 2\pi }\)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \left( 2\pi -\frac { sin2\times 2\pi }{ 2 } \right) -\left( 0-\frac { sin0 }{ 2 } \right) \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \times \pi ={ { I }^{ 2 } }_{ m }\pi \)    [∵ sin 0 = sin 4π = 0]
Substituting this in equation (1), we get
\(\sqrt { \frac { { { I }^{ 2 } }_{ m }\pi }{ 2\pi } } =\frac { { { I }^{ 2 } }_{ m } }{ \sqrt { 2 } } \)    [Base length of cone cycle is 2π]
I_RMS = 0.707 I_m

(i) The magnitude of an alternating current in a circuit changes from one instant to other instant and its direction also reverses for every half cycle.
(ii) During positive half cycle, current is taken as positive and during negative cycle it is negative. Therefore mean or average value of symmetrical alternating current over one complete cycle is zero.
(iii) Therefore the average or mean value is measured over one half of a cycle. These electrical terms, average current and average voltage can be used in both AC and DC circuit analysis and calculations.
(iv) 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.
(v) The instantaneous value of sinusoidal alternating current is given by the equation
i = I_m sin ωt or i = I_m sinθ (where θ = ωt) whose graphical representation is given in Figure.
(vi) The sum of all currents over a half-cycle is given by area of positive half-cycle (or negative half-cycle). Therefore,
I_av = \(\frac{Area \ of \ positive \ half-cycle (or \ negative \ half -cycle)}{Base \ lenght \ of \ half - cycle)}\) ...(1)

(vii) Consider an elementary strip of thickness dθ in the positive half-cycle of the current wave. Let i be the mid-ordinate of that strip.
Area of the elementary strip = i dθ
Area of positive half-cycle
= \(\int _{ 0 }^{ \pi }{ id\theta } =\int _{ 0 }^{ \pi }{ { I }_{ m }sin\theta d\theta } \)
= \({ I }_{ m }{ [-cos\theta ] }_{ 0 }^{ \pi }={ -I }_{ m }[cos\pi -cos0]=2{ I }_{ m }\)
Substituting this in equation (1), we get (The base length of half-cycle is π)
Average value of AC, I_av = \(\frac { { 2I }_{ m } }{ \pi } \)
I_av = 0.637 I_m
(viii) Hence the average value of AC is 0.637 times the maximum value I_m of the alternating current. For negative half cycle, I_av = -0.637 I_m

(i) During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy, \(U={ U }_{ E }+{ U }_{ B }=\cfrac { { q }^{ 2 } }{ 2C } +\cfrac { 1 }{ 2 } { Li }^{ 2 }\)
(ii) Let us consider 3 different stages of LC oscillations and calculate the total energy of the system. Case (i) When the charge in the capacitor, q = Qm and the current through the inductor, i = 0, the total energy is given by
\(U=\cfrac { { Q }_{ m }^{ 2 } }{ 2C } +0=\cfrac { { Q }_{ m }^{ 2 } }{ 2C } \)
The total energy is wholly electrical.
Case (ii) When charge = 0 ; current = I_m ' the total energy is 
\(U=0+\cfrac { { Q }_{ m }^{ 2 } }{ 2C } { LI }_{ 3 }^{ 2 }=\cfrac { 1 }{ 2 } { LI }_{ m }^{ 2 }\)
= \(\cfrac { 1 }{ 2 } \times \left( \cfrac { { Q }_{ m }^{ 2 } }{ LC } \right) Since\quad { I }_{ m }={ Q }_{ m }\omega =\cfrac { { Q }_{ m } }{ \sqrt { LC } } \)
= \(\cfrac { { Q }_{ m }^{ 2 } }{ LC } \)
The total energy is wholly magnetic.
Case (iii) When charge = q; current = i, the total energy is
\(U=\cfrac { { q }^{ 2 } }{ 2C } +\cfrac { 1 }{ 2 } { Li }^{ 2 }\)
Since \(q={ Q }_{ m }\)
\(i=-\cfrac { dq }{ dt } ={ Q }_{ m }\)
The negative sign in current indicates that the charge in the capacitor decreases with time
\(U=\cfrac { { Q }_{ m }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\cfrac { { L\omega }^{ 2 }{ Q }_{ m }^{ 2 }\omega t }{ 2 } \)
= \(U=\cfrac { { Q }_{ m }^{ 2 }{ cos }^{ 2 }\omega t }{ 2C } +\cfrac { { L\omega }^{ 2 }{ Q }_{ m }^{ 2 }\omega t }{ 2 } \)
= \(\cfrac { { Q }_{ m }^{ 2 } }{ 2C } \left( { cos }^{ 2 }+\omega t+{ sin }^{ 2 }\omega t \right) \)
\(U=\cfrac { { Q }_{ m }^{ 2 } }{ 2C } \)
From above three cases, it is clear that the total energy of the system remains constant.

(i) Consider a rectangular conducting loop of width 1 in a uniform magnetic field \(\vec { B } \)  which is perpendicular to the plane of the loop and is directed inwards.
(ii) A part of the loop is in the magnetic field while the remaining part is outside the loop as shown in Figure.
(iii) When the loop is pulled with a constant velocity \(\vec { v } \) to the right, the area of the portion of the loop within the magnetic field will decrease.
(iv) Thus, the flux linked with the loop will also decrease. According to Faraday's law, an electric current is induced in the loop which flows in a direction so as to oppose the pull of the loop.
(v) Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is
\({ \phi }_{ B }=\int _{ A }^{ }{ B.d } \vec { A } =BAcos\theta \)
Here θ = 0^o and cos  0^o = 1
= BA
\({ \phi }_{ B }=Blx\)
(vi) As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
\(\varepsilon =\frac { d{ \Phi }_{ B } }{ dt } =\frac { d }{ dt } (Blx)\)
(vii) Here, both B and 1 are constants. Therefore,
\(\varepsilon =Bl\frac { dx }{ dt } \)
ε = Blv
where \(v=\frac { dx }{ dt } \) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.