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

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

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 \(=\frac{\pi}{4}\)

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

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