\(N_z=300, N_s=1200, R_p=0.82 \Omega, R_s=6.2 \Omega,  \)
\(\text {Output power }=32 \mathrm{~kW}, \mathrm{~V}_{\mathrm{s}}=1600 \mathrm{~V}, \eta=80 \%  \)
\(V_s I_s=32000 \mathrm{~W}  \)
\(1600 \times I_s=32000  \)
\(\therefore I_s=\frac{32000}{1600}  \)
\(I_s=20 \mathrm{~A}\)    ......(1)
∴ Power loss in secondary coil \(P  =I_{\mathrm{s}}^2 \times R_{\mathrm{s}}  \)    ......(2)
\(=(20)^2 \times 6.2  \)
\(=400 \times 6.2=2480.0  \)
\(=2480 \mathrm{~W}=2.48 \mathrm{~kW}\)
∴ Power loss in secondary coil P = 2.48 kW   ......(3)
Efficiency \( \eta  =\frac{V_s I_s}{V_P I_P} \)
\(\therefore \frac{80}{100}  =\frac{32 \times 10^3}{V_P I_P}  \)
\(\therefore V_P I_p =\frac{32 \times 10^3 \times 100}{80}  \)
\(=\frac{3200 \times 10^3}{80}  \)
\(V_P I_P =40 \times 10^3 \mathrm{~W}\)     ......(4)
We know that, \(\frac{N_p}{N_s}=\frac{V_P}{V_S}\)     ......(5)
\(\frac{300}{1200}=\frac{V_p}{1600} \Rightarrow V_p  =\frac{300 \times 1600}{1200}  \)
\(=400 \mathrm{~V}\)     ......(6)
From (4), V_pI_p = 40 x 10³
400 I_p = 40 x 10³
\(\therefore I_p =\frac{40 \times 10^3} { 400}={100A}   \)
∴ Power loss in the primary coil = I²_pR_p
= (100)² x 0.82
= 10⁴ x 0.82 = 8200 W
∴ Power loss in the primary coil = 8.2 kW

V_p = 200 V; V_s = 120V;
Efficiency, η = 90% \(= \frac{90}{100}\)
Resistance, R = 40Ω;
Efficiency. \(\eta =\frac { { V }_{ s }{ I }_{ s } }{ { V }_{ p }{ I }_{ p } }\)
\(=\frac { 90 }{ 100 } = \frac{ 120 }{ 200 }\times \frac{I_s}{I_p}\)
\(=\frac { I_s }{ I_p } =\frac{90 \times 200}{120 \times 100}=\frac{18000}{12000}=\frac{3}{2}=1.5\)
\(\therefore I_p=\frac{I_s}{1.5}\) 
\(But,{ I }_{ s }=\frac { 120 }{ 40} =3\)
\(\therefore I_p =\frac { 3}{1.5}=2A\)
Current drawn by the primary I_p = 2A

Transformer ratio, \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ N_p } \)
Here V_s = 11 V,  V_r = 220 V, P = 88 W
(i) Voltage Transformation ratio = \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { 11 } }{ 220 } \)
\(K =\frac { { 1 } }{ 20 } \)
(ii) Current in the primary, \(I_p =\frac { { P } }{ V_p } \)
\(=\frac { { 88 } }{220 } =\frac{4}{10}=0.4 A\)
∴ I_p = 0.4 A

Relative permeability of an iron core = μ_r = 900
Area of cross-section, A₁ = A₂ = 4 x 10^-4 m^-2
Current in solenoid, \(\mathrm{I}=1_2-I_1=8-2=6 \mathrm{~A}, Length, l_1=l_2=4 \times 10^{-2} \mathrm{~m}=0.04 \mathrm{~m} \)
Number of turns in S₁ = N₁ = 200, Number of turns in S₂ = N₂ = 800,
\(\text {Mutual inductance } M  =\frac{\mu_0 \mu, \mathrm{N}_1 \mathrm{~N}_2 A_2}{l} \)
\(\therefore M  =\frac{4 \times 3.14 \times 10^{-7} \times 900 \times 200 \times 800 \times 4 \times 10^{-4}}{4 \times 10^{-2}} \)
M  = 1.81 H
\(\text {Induced emf } e  =-M \frac{d I}{d t}  \)
\(=-1.81 \times \frac{6}{0.04}=-271.5 \mathrm{~V}  \)
\(M=1.81 \mathrm{H}, \text { Induced } \mathrm{emf} e  =-271.5 \mathrm{~V}\)

No of turns of small solenoid, N₂ = 200
Radius of small solenoid r = 2 cm = 2 x 10^-2 m
Area of small solenoid, A₂= πr² = 3.14 x 4 x 10^-4 m²
Turn density of long solenoid, \(\frac { { N }_{ 1 } }{ l } =90\times10^2m^{-1}, M=?\)
M = μ₀\(\frac { { N }_{ 1 } }{ l } \)N₂A₂
=4 x 3.14 x 10^-7 x 90 x 10² x 200 x 3.14 x 4 x 10^-4 = 2.84 x 10^-3 H
M = 2.84 mH

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

μ_r = 1 (air core)
Length of two solenoids, \(l_1=l_2=80 \times 10^{-2} \mathrm{~m}\)
Cross-sectional area of two solenoids, \(A_1=A_2=5 \times 10^{-4} \mathrm{~m}^2\)
Number of turns in the first coll, N₁=1200 turns
Number of turns in the second coil, N₂ = 400 turns
\(n_1=\frac{N_1}{l}=\frac{1200}{80 \times 10^{-2}}=15 \times 10^{-2}=1500  \)
\(n_2=\frac{N_2}{l}=\frac{400}{80 \times 10^{-2}}=500\)
Mutual inductance, \(M= μ_0μ_r n_1 n_2 A_2 l\)
\(M  =4 \times 3.14 \times 10^{-7} \times 1 \times 1500 \times 500 \times 5 \times 10^{-4} \times 80 \times 10^{-2}  \)
\(=12.56 \times 10^{-7} \times 75 \times 10^4 \times 5 \times 10^{-4} \times 80 \times 10^{-2}  \)
\(=3.76,800 \times 10^{-9}=0.376 \times 10^{-3} \mathrm{H}=0.376 \mathrm{mH}  \)
\(=0.38 \mathrm{mH}  \)
\(\therefore \text {Mutual inductance } =0.38 \mathrm{mH}\)

Current, I = 0.4 A, Magnetic flux, Φ = 4 x 10^-3 Wb, Number of turns, N = 200
Inductance, L = \(\frac { N\Phi }{ I }\)
\(L=\frac { 200\times 4\times 10^{ -3 } }{ 0.4 } =200 \times10^{-3+1}\)
= 200 x 10^-2 = 2H.
∴ Inductance of the coil = 2 H.

μ₀ = 4π x 10^-7 H/m, I = 1 A, n = 400 turns/cm = 40000 turns/m
Area A = πr² = 3.14 x (0.02)² = 12.56 x 10^-4 m²
Magnetite flux of n turns of the coil, Φ =μ₀nlA
4π x 10^-7 x 40000 x 1x 12.56 x 10^-4 = 4 x 3.14 x 4 x 12.56 x 10^-7
Φ = 0.631 x 10^-4 Wb

(i) Consider a straight conducting rod AB of length I in a uniform magnetic field \(\vec { B } \) which is directed perpendicularly into the plane of the paper.
(ii) The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec { v } \) towards right side
(iii) When the rod moves, the free electrons present in it also move with same velocity \(\vec { v } \) in \(\vec { B } \). As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation

\({ \vec { F } }_{ B }=-e(\vec v\times \vec { B } )\)
(iv) The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field \(\vec { E } \) directed along BA 
(v) Due to the electric field \(\vec { E } \), the coulomb force starts acting on the free electrons along AB and is given by
\({ \vec { F } }_{ E }=-e\vec { F } \)
(vi) The magnitude of the electric field \(\vec { E } \) keeps on increasing as long as accumulation of electrons at the end  A continues. The force \({ \vec { F } }_{ E }\) also increases until equilibrium is reached.
(vii) At equilibrium, the magnetic  Lorentz force \({ \vec { F } }_{ B }\) and the coulomb force \({ \vec { F } }_{ E }\) balance each other and no further accumulation of free electrons at the end A takes place.

The potential difference between two ends of the rod is
V = El
V = vBl
Thus, the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf
ε = Blv
As this emf is produced due to the movement of the rod, it is often called as motional emf. If the ends A and B are connected by an external circuit or total resistance R, then current \(i=\frac {ε}{R}=\frac{Blv}{R}\)flows in it. The direction of the current is found from right-hand thumb rule.

The given equation is v = 10 sin ( 3π x 10⁴ t)
(i) At t = 0 s,
v = 10 sin0^o = 0
(ii) At t = 50 µs,
v = 10sin( 3π x 10⁴ x 50 x 10^-6)
= \(10 \sin \left(150 \pi \times 10^{-2} \times \frac{180^{\circ}}{\pi}\right)\) 
= 10sin (270^o) = 10 x -1
= -10V
(iii) At t = 75 µs,
\(v=10 \sin \left(3 \pi \times 10^{4} \times 75 \times 10^{-6}\right)\) 
= \(10 \sin \left(225 \pi \times 10^{-2} \times \frac{180^{\circ}}{\pi}\right)\) 
= 10sin (405^o) = 10sin45^o
= \(10 \times 1 / \sqrt{2}=7.07 \mathrm{~V}\) 

L = 500 x 10^-6H; C\(=\frac { 80 }{ { \pi }^{ 2 } } \times { 10 }^{ -12 }F;R=628\Omega \)
(i) Resonant frequency is
\({ f }_{ r }=\frac { 1 }{ 2\pi \sqrt { LC } } =\frac { 1 }{ 2\pi \sqrt { 500\times { 10 }^{ -6 }\times \frac { 80 }{ { \pi }^{ 2 } } \times { 10 }^{ -12 } } } \)
\(=\frac { 1 }{ 2\sqrt { 40,000\times { 10 }^{ -18 } } } \)
\(=\frac { 10,000\times { 10 }^{ 3 } }{ 4 } = 2500\)
f_r = 2500 KHz
(ii) Q–factor
\(=\frac { { \omega }_{ r }L }{ R } =\frac { 2\times 3.14\times 2500\times { 10 }^{ 3 }\times 500\times { 10 }^{ -6 } }{ 628 } \)
Q = 12.5

L = 40 x 10^-3 H; i = 0.3 sin (200t – 40^o)
X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90^o Therefore,
v = V_msin (ωt + 90^o)
v = 2.4sin (200t - 40^o + 90^o)
v = 2.4sin (200t + 50^o)volt

L = 80 x 10^-6H; C = 2000 x 10^-12 F
R = 50 Ω; f_r = 400 x 10³Hz
(i) Q-factor, \(Q_1=\frac { 1 }{ R } \sqrt { \frac { L }{ C } } \)
=\(\frac { 1 }{ 50 } \sqrt { \frac { 80\times { 10 }^{ -6 } }{ 200\times 10^{ -12 } } } \)=4
(ii) When L₂ = 2 L
= 2 x 80 x 10^-6 H
= 160 x 10^-6 H,
C₂ = \(\frac { 1 }{ 4{ \pi }^{ 2 }{ f }_{ r }^{ 2 }{ L }_{ 2 } } \)
=\(\frac { 1 }{ 4\times 3.14^{ 2 }\times (400\times 10^{ 3 })^2\times 160\times 10^{ -6 } } \)
\(\simeq \) 1000 x 10^-12 F
C₂ = 1000 pF
(iii) Q₂ = \(\frac { 1 }{ R } \sqrt { \frac { { L }_{ 2 } }{ { C }_{ 2 } } } =\frac { 1 }{ 50 } \sqrt { \frac { 160\times 10^{ -6 } }{ 1000\times 10^{ -12 } } } \)
= \(\frac { 1 }{ 50 } \sqrt { \frac { 16\times { 10 }^{ -5 } }{ { 10 }^{ -9 } } } =\frac { 4\times { 10 }^{ 2 } }{ 50 } \) = 8

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
\(\text { (ii) } \phi =\tan ^{-1}\left(\frac{X_1-X_c}{R}\right) \)
\( \phi =\tan ^{-1}\left(\frac{184-144}{30}\right) \)
\(\phi =53.1^{\circ} \)

i = 77 sin 314t ; t = 2 ms = 2 x 10^-3 s
The general equation of an alternating current is i = I_m sinωt. On comparison,
(i) Peak current, I_m = 77A
(ii) Frequency, \(f=\frac { \omega }{ 2\pi } =\frac { 314 }{ 2\times 3.14 } =50Hz\)
(iii) Time period, \(T-\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s\)
(iv) At t = 2 m s, Instantaneous current, i = 77sin(314 x 2 x 10⁻³)
\( =77 \sin \left(314 \times 2 \times 10^{-3} \times \frac{180^{\circ}}{3.14}\right) \)
\(=77 \sin 36^{\circ}=77 \times 0.5878 \)
= 45.26 A

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

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

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
= \({ V }_{ s }{ I }_{ s }=\frac { { V }_{ s }^{ 2 } }{ { R }_{ s } } =\frac { 20,000\times 20,000 }{ { 10 }^{ 4 } } =40kW\)

A = 70 x 10^-4m²; N = 600 turns
B = 0.4 Wbm^-2; f = 500 rpm
The instantaneous emf is
ε = ε_msinωt
since \({ \epsilon }_{ m }=N{ \Phi }_{ m }\omega =N(BA)(2\pi f)\)
ε = NBA x 2\(\pi\)f x sinωt
(i) When ωt = 0^o,
ε = ε_m sin0 = 0
(ii) When ωt = 90^o,
ε = ε_msin90^o = NBA x 2\(\pi\)f x 1
= 600 x 0.4 x 70 x 10^-4 x 2 x \(\frac{22}{7}\times(\frac{500}{60})\)
= 88V
(iii) When ωt = 90° – 60° = 30°,
ε = ε_m sin30^o = 88 x \(\frac{1}{2}\) = 44V

N_A = 200 turns; N_B = 1000 turns;
r_A = 20 x 10^-2 m; r_B = 2 x 10^-2 m;
dt = 0.04 s; di_A = 6−2 = 4A
Let i_A be the current flowing in coil A, then the magnetic field B_A at the centre of the circular coil A is
\({ B }_{ A }=\frac { { \mu }_{ o }{ N }_{ A }{ i }_{ A } }{ { 2r }_{ A } } =\frac { 4\pi \times { 10 }^{ -7 }{ N }_{ A }{ i }_{ A } }{ { 2r }_{ A } } \)
\(=\frac { { 10 }^{ -7 }\times 2\times 3.14\times 200 }{ 20\times { 10 }^{ -2 } } \times { i }_{ A }\)
= 6.28 x 10^-4i_A Wbm^-2
The magnetic flux linkage of coil B is
\({ N }_{ B }{ \Phi }_{ B }={ N }_{ B }{ B }_{ A }{ A }_{ B }\)
= 1000 x 6.28 x 10^-4 x i_A x 3.14 x (2 x 10^-2)²
= 7.89 x 10^-4i_AWb turns
The mutual inductance between the coils
\(\\ { M }=\frac { { N }_{ B }{ \Phi }_{ B } }{ { i }_{ A } } =7.89\times { 10 }^{ -4 }H\)
Induced emf in coil B is
ε_B = - M\(\frac { { di }_{ A } }{ dt } \)
ε_B = \(\frac { 7.89\times { 10 }^{ -4 }\times (6-2) }{ 0.04 } \)(magnitude only)
ε_B = 78.9mV
The rate of change of magnetic flux of coil B is
\(\frac { d\left( { N }_{ B }{ \Phi }_{ B } \right) }{ dt } ={ \epsilon }_{ B }=78.9m{ Wbs }^{ -1 }\)

N = 500 turns; μ_r = 800;
l = 40 cm = 0.4 m; r = 3 cm = 0.03 m;
di = 3 – 0 = 3 A; dt = 0.4 s
Self inductance,
\(L=\mu { n }^{ 2 }Al\left( \because \mu ={ \mu }_{ o }{ \mu }_{ r };A={ \pi r }^{ 2 };n=\frac { N }{ l } \right) \)
\(=\frac { { \mu }_{ 0 }{ \mu }_{ r }{ N }^{ 2 }\pi { r }^{ 2 } }{ l } \)
\(=\frac { 4\times 3.14\times { 10 }^{ -7 }\times 800\times { 500 }^{ 2 }\times 3.14\times { (3\times 10 }^{ -2 }{ ) }^{ 2 } }{ 0.4 } \)
L=1.77H
Induced emf, ε = -L\(\frac{di}{dt}\)
\(=\frac{1.77\times3}{0.4}\)
ε = -13.275V

l = 0.5 m; h = 7.2 m; u = 0 m s^–1;
g = 10 m s^–2; B_H = 4.04 x 10^–5 T
The final velocity of the rod is
V² = u² = + 2g h = 0 + (2 x 10 x 7.2) =144
v = 12 ms^-1
The magnitude of the induced emf when the rod is about to touch the ground is
ε = B_H lv = 4.04 × 10^–5 × 0.5 × 12
= 242.4 µV

A = 5 x 10^-2 m²; B = 0.2 T
(i) θ = 0°;
\({ \Phi }_{ B }=BAcos\theta =0.2\times 5\times { 10 }^{ -2 }\times { cos }0^{ o }\)
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }Wb\)
(ii) θ = 90° – 60° = 30°;
\(\Phi_B\) = BAcosθ  = 0.2 x 5 x 10^-2 x cos 30^o
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }\times \frac { \sqrt { 3 } }{ 2 } =8.66\times { 10 }^{ -3 }Wb\)
(iii) θ = 90°;
\({ \Phi }_{ B }\) = BA cos90^o = 0

(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.