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

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

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 = 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

\(C=\frac { { 10 }^{ 2 } }{ \pi } \times { 10 }^{ -6 }F,{ V }_{ RMS }=220V;f=50Hz\)
(i) Capacitive reactance,
\({ X }_{ c }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi fC } \)
\(=\frac { 1 }{ 2\times \pi \times 50\times \frac { { 10 }^{ -4 } }{ \pi } } =100\Omega \)
(ii) RMS value of current,
\(\\ { I }_{ RMS }=\frac { { V }_{ RMS } }{ { X }_{ c } } =\frac { 220 }{ 100 } =2.2\ A\)
(iii) V_m = 220 x \(\sqrt{2}\) = 311V
I_m = 2.2 x \(\sqrt{2}\) = 3.1A
Therefore,
v = 311sin314t
i = 3.1sin\((314t+\frac { \pi }{ 2 } )\)