Across resistor, $$I = \frac{V}{R} = \frac{{100}}{{1000}} = 0.1A$$
At resonance, $${X_L} = {X_C} = \frac{1}{{\omega C}} = \frac{1}{{200 \times 2 \times {{10}^{ - 6}}}} = 2500$$
Voltage across $$L$$ is $$I\,{X_L} = 0.1 \times 2500 = 250V$$

Here
$$\eqalign{ & i = \frac{e}{{\sqrt {{R^2} + X_L^2} }} = \frac{e}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} = \frac{e}{{\sqrt {{R^2} + 4{\pi ^2}{v^2}{L^2}} }} \cr & 10 = \frac{{220}}{{\sqrt {64 + 4{\pi ^2}{{\left( {50} \right)}^2}L} }}\,\,\left[ {\because R = \frac{V}{I} = \frac{{80}}{{10}} = 8} \right] \cr} $$
On solving we get $$L= 0.065\,H$$

$$\eqalign{ & \frac{{{I_s}}}{{{I_p}}} = \frac{{{n_p}}}{{{n_s}}};\frac{{80}}{{{I_p}}} = \frac{{20}}{1} \cr & {\text{or}}\,{I_p} = 4\,amp. \cr} $$

In an $$ac$$  circuit, a pure indcutor does not consume any power. Therefore, power is consumed by the resistor only.
$$\eqalign{ & \therefore P = I_v^2R \cr & {\text{or}}\,\,108 = {\left( 3 \right)^2}R\,\,{\text{or}}\,\,R = 12\,\Omega \cr} $$

We define the quality factor of the circuit as follows
Quality factor, $$Q = 2\pi \times \frac{{{\text{total energy stored in the circuit}}}}{{{\text{loss in energy in each cycle}}}}$$
But the total energy stored in circuit $$ = Li_{rms}^2$$
and the energy loss per second $$ = i_{rms}^2R$$
So, loss in energy per cycle $$ = \frac{{i_{rms}^2R}}{f}$$
Hence, quality factor, $$Q = 2\pi \times \frac{{Li_{rms}^2}}{{\frac{{i_{rms}^2R}}{f}}}$$
$$ = \frac{{2\pi fL}}{R} = \frac{{\omega L}}{R}$$

Since the phase difference between $$L\,\& \,C$$  is $$\pi ,$$
∴ net voltage difference across $$LC = 50 - 50 = 0$$

$$\eqalign{ & {\text{Given}}:L = 20\,mH;C = 50\,\mu F;R = 40\Omega \cr & V = 10\sin 340\,t \cr & \therefore {V_{{\text{runs}}}} = \frac{{10}}{{\sqrt 2 }} \cr & {X_C} = \frac{1}{{\omega C}} = \frac{1}{{340 \times 50 \times {{10}^{ - 6}}}} = 58.8\,\Omega \cr & {X_L} = \omega L = 340 \times 20 \times {10^{ - 3}} = 6.8\,\Omega \cr & {\text{Impedance, }}Z = \sqrt {{R^2} + {{\left( {{X_C} - {X_L}} \right)}^2}} \cr & = \sqrt {{{40}^2} + {{(58.8 - 6.8)}^2}} = \sqrt {4304} \,\Omega \cr} $$
Power loss in $$A.C.$$  circuit,
$$\eqalign{ & P = i_{rms}^2R = {\left( {\frac{{{V_{rms}}}}{Z}} \right)^2}R = {\left( {\frac{{\frac{{10}}{{\sqrt 2 }}}}{{\sqrt {4304} }}} \right)^2} \times 40 \cr & = \frac{{50 \times 40}}{{4304}} \simeq 0.51\,W \cr} $$

Given,
Inductance $$L = 20\,mH$$
Capacitance, $$C = 50\,\mu F$$
Resistance, $$R = 40\,\Omega $$
emf, $$V = 10\sin 340\,t$$
$$\because $$ Power loss in $$AC$$ circuit will be given as
$$\eqalign{ & {P_{av}} = I_V^2R = {\left[ {\frac{{{E_V}}}{Z}} \right]^2} \cdot R \cr & = {\left( {\frac{{10}}{{\sqrt 2 }}} \right)^2} \cdot \left[ {40\frac{1}{{{{40}^2} + {{\left( {\frac{{340 \times 20 \times {{10}^3} - 1}}{{340 \times 50 \times {{10}^{ - 6}}}}} \right)}^2}}}} \right] \cr & = \frac{{100}}{2} \times 40 \times \frac{1}{{1600 + {{\left( {6.8 - 58.8} \right)}^2}}} \cr & = \frac{{2000}}{{1600 + 2704}} \cr & \approx 0.46\,W \cr & \approx 0.51\,W \cr} $$

Root mean square value of an alternating current is defined as the square root of the average of $${i^2},$$ during a complete cycle it may be taken by
$$\eqalign{ & \overline {{i^2}} = \frac{{\int_0^{\frac{{2\pi }}{\omega }} {{i^2}} dt}}{{\frac{{2\pi }}{\omega }}} \cr & = \frac{{\int_0^{\frac{{2\pi }}{\omega }} {i_0^2} {{\sin }^2}\omega t\,dt}}{{\frac{{2\pi }}{\omega }}} \cr & = \frac{{i_0^2\omega }}{{2\pi }}\int_0^{\frac{{2\pi }}{\omega }} {\frac{1}{2}} \left( {1 - \cos 2\omega t} \right)dt \cr & = \frac{{i_0^2\omega }}{{4\pi }}\left[ {t - \frac{{\sin 2\omega t}}{{2\omega }}} \right]_0^{\frac{{2\pi }}{\omega }} \cr & = \frac{{i_0^2\omega }}{{4\pi }}\left( {\frac{{2\pi }}{\omega }} \right) = \frac{{i_0^2}}{2} \cr & \therefore {i_{rms}} = \sqrt {\overline {{i^2}} } = \frac{{{i_0}}}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & \phi = \vec B \cdot \vec A;\phi = BA\cos \omega t \cr & \varepsilon = - \frac{{d\phi }}{{dt}} = \omega BA\sin \omega t;\,i = \frac{{\omega BA}}{R}\sin \omega t \cr & {P_{{\text{inst}}}} = {i^2}R = {\left( {\frac{{\omega BA}}{R}} \right)^2} \times R{\sin ^2}\omega t \cr & {P_{{\text{avg}}}} = \frac{{\int\limits_0^T {{P_{{\text{inst}}}} \times dt} }}{{\int\limits_0^T {dt} }} = \frac{{{{\left( {\omega BA} \right)}^2}}}{R}\frac{{\int\limits_0^T {{{\sin }^2}\omega tdt} }}{{\int\limits_0^T {dt} }} = \frac{1}{2}\frac{{{{\left( {\omega BA} \right)}^2}}}{R} \cr & \therefore {P_{{\text{avg}}}} = \frac{{{{\left( {\omega B\pi {r^2}} \right)}^2}}}{{8R}}\,\,\left[ {A = \frac{{\pi {r^2}}}{2}} \right] \cr} $$