\(L=\frac{20}{\pi^2} \mathrm{H}, \mathrm{f}=50 \mathrm{~Hz}  \)
\(f=\frac{1}{2 \pi \sqrt{L C}}  \)
\(50=\frac{1}{2 \pi \sqrt{\frac{20}{\pi^2} \times C}}  \)
\(50=\frac{1}{2 \times \sqrt{20 C}}  \)
\(\therefore(50)^2=\frac{1}{4 \times 20 C}  \)
\(\therefore C=\frac{1}{2500 \times 4 \times 20}=5 \times 10^{-6}=5 \mu \mathrm{F}\)

\(Q_{midpoint}=\frac{Q}{\sqrt{1^2+1^2}}=\frac{Q}{\sqrt2}\)

P_av = \(\frac{1}{2}\)V₀I₀cosΦ
\(= \frac{1}{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}cos\times\frac{\pi}{3}\times \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

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

In RL circuit, tanΦ = \(\frac{X_l}{R}\)
If R = X₁, then tanΦ = \(\frac{X_l}{X_l}=1\)
∴ Φ = tan^-1(1)=\(\frac{\pi}{4}\)
∴ Phase difference \(=\frac{\pi}{4}\)

\(\Phi = \frac{\pi}{3}-\frac{\pi}{3}=0 \)
Power factor = cosФ = cos 0 = 1

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

\(A_1  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_1  =10 \text { turns }  \)
\(A_2  =4 \times 10^{-4} \mathrm{~m}^2  \)
\(N_2  =15 \times 10^{-}=1500 \mathrm{turn} / \mathrm{m}  \)
\(\phi  =B_2 A_2=\left(\mu_0 \mathrm{n}_2 I_2\right) \mathrm{A}_1 \)
\(Where, \mathrm{n}_2=\frac{\mathrm{N}_2}{l}=1500 \mathrm{turn} / \mathrm{m}\)
The mutual Inductance is,
\(M  =\frac{N_1 o_{12}}{I_2}=\mu_0 n_2 N_1 A_1  \)
\(=4 \pi \times 10^{-7} \times 1500 \times 10 \times 4 \times 10^{-4}  \)
\(=7.54 \times 10^{-6} \mathrm{H}=7.54 \mu \mathrm{H}\)

Solution
e = -L\(\frac{dl}{dt}\)

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

The direction of conventional current is always opposite to the direction of flow of electrons.