$${V_{rms}} = \sqrt {\frac{{\left( {\frac{T}{2}} \right)V_0^2 + 0}}{T}} = \frac{{{V_0}}}{{\sqrt 2 }}.$$

In an $$AC$$ circuit with voltage $$V$$ and current $$i,$$ the power dissipated is given by
$$P = Vi\cos \phi $$
where, $$\phi $$ is the phase and $$\cos \phi $$  is the power factor. Thus, the power dissipated, depends upon the phase between voltage $$V$$ and current $$i.$$

For resonant frequency to remain same
$$LC$$ should be const. $$LC$$ = const
$$ \Rightarrow LC = L' \times 2C \Rightarrow L' = \frac{L}{2}$$

Energy stored in magnetic field $$ = \frac{1}{2}L{i^2}$$
Energy stored in electric field $$ = \frac{1}{2}\frac{{{q^2}}}{C}$$
$$\therefore \frac{1}{2}L{i^2} = \frac{1}{2}\frac{{{q^2}}}{C}$$
$${\text{Also}}\,q = {q_0}\cos \omega t\,{\text{and}}\,\omega = \frac{1}{{\sqrt {LC} }}$$
On solving $$t = \frac{\pi }{4}\sqrt {LC} $$

Capacitive reactance, $${X_C} = \frac{1}{{\omega C}} = \frac{1}{{2\pi \upsilon C}}$$
$$ \Rightarrow {X_L} \propto \frac{1}{\upsilon }$$
With increases in frequency, $${X_C}$$ decreases. Hence, option (C) represents the correct graph.

For maximum power, $${X_L} = {X_C},$$   which yields
$$\eqalign{ & C = \frac{1}{{{{\left( {2\pi n} \right)}^2}L}} = \frac{1}{{4{\pi ^2} \times 50 \times 50 \times 10}} \cr & \therefore C = 0.1 \times {10^{ - 5}}F = 1\mu F \cr} $$

In series $$RLC$$  circuit,
Voltage, $$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $$
And, at resonance, $${V_L} = {V_C}$$
Hence, $$V = {V_R}$$

Total effective resistance of $$LCR$$  circuit is called impedance of the $$LCR$$  series circuit. It is represented by $$Z.$$
where $$Z = \frac{{{V_0}}}{{{i_0}}} = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} $$
Given, $${V_{AC}} = 200\,V$$
Resistance offered by inductor $${X_L} = 50\,\Omega $$
Resistance offered by capacitance $${X_C} = 50\,\Omega $$
$$\eqalign{ & R = 10\,\Omega \cr & Z = ? \cr & \therefore Z = \sqrt {{{\left( {10} \right)}^2} + {{\left( {50 - 50} \right)}^2}} \cr & {\text{or}}\,\,Z = 10\,\Omega \cr} $$

When magnetic flux linked with a coil changes, induced emf is produced in it and the induced current flows through the wire forming the coil. In 1895, Focault experimentally found that these induced currents are set up in the conductor in the form of closed loops. These currents look like eddies or whirlpools and likewise are known as eddy curents. They are also known as Focault’s current. These currents oppose the cause of their origin, therefore, due to eddy currents, a great amount of energy is wasted in form of heat energy. If core of transformer is laminated, then their effect can be minimised.

Problem Solving Strategy
Compare the given equation with the equation of alternating voltage i.e. $$e = {e_m}\sin \omega t$$
where, $${e_m} = {e_{rms}}$$
Given,
emf, $$e = 200\sqrt 2 \sin 100\,t$$
and $$C = 1\mu F = 1 \times {10^{ - 6}}F$$
As $${e_{rms}} = 200\,V\,{\text{and}}\,\omega = 100$$
$$\eqalign{ & \therefore {X_C} = \frac{1}{{\omega C}} = \frac{1}{{1 \times {{10}^{ - 6}} \times 100}} = {10^4}\,\Omega \cr & {i_{rms}} = \frac{{{e_{rms}}}}{{{X_C}}} = \frac{{200}}{{{{10}^4}}} = 2 \times {10^{ - 2}}A = 20\,mA \cr} $$