$$\eqalign{ & {V_{rms}} = \frac{{200\sqrt 2 }}{{\sqrt 2 }} = 200V \cr & {I_{rms}} = \frac{{{V_{rms}}}}{{{X_C}}} = \frac{1}{{100 \times {{10}^{ - 6}}}} = 2 \times {10^{ - 2}} = 20\,mA \cr} $$

$$\eqalign{ & \frac{{{V_2}}}{{{V_1}}} = 0.8\frac{{{I_1}}}{{{I_2}}} \Rightarrow \frac{{{V_2}{I_2}}}{{{V_1}{I_1}}} = 0.8 \cr & {V_1} = 220\,V,{I_2} = 2.0\,A,{V_2} = 440\,V \cr & {I_1} = \frac{{{V_2}{I_2}}}{{{V_1}}} \times \frac{{10}}{8} = \frac{{440 \times 2 \times 10}}{{220 \times 8}} = 5\,A \cr} $$

When capacitance is taken out, the circuit is $$LR.$$
$$\eqalign{ & \therefore \tan \phi = \frac{{\omega L}}{R} \cr & \Rightarrow \omega L = R\tan \phi = 200 \times \frac{1}{{\sqrt 3 }} = \frac{{200}}{{\sqrt 3 }} \cr} $$
Again , when inductor is taken out, the circuit is $$CR.$$
$$\eqalign{ & \therefore \tan \phi = \frac{1}{{\omega CR}} \cr & \frac{1}{{\omega c}} = R\tan \phi = 200 \times \frac{1}{{\sqrt 3 }} = \frac{{200}}{{\sqrt 3 }} \cr & {\text{Now, }}Z = \sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}} - \omega L} \right)}^2}} \cr & = \sqrt {{{(200)}^2} + {{\left( {\frac{{200}}{{\sqrt 3 }} - \frac{{200}}{{\sqrt 3 }}} \right)}^2}} = 200\Omega \cr & {\text{Power dissipated }} = {V_{rms}}{I_{rms}}\cos \phi \cr & = {V_{rms}}\frac{{{V_{rms}}}}{Z}.\frac{R}{Z}\left( {\because \cos \phi = \frac{R}{Z}} \right) \cr & = \frac{{{V^2}_{rms}R}}{{{Z^2}}} = \frac{{{{\left( {220} \right)}^2} \times 200}}{{{{\left( {200} \right)}^2}}} = \frac{{220 \times 220}}{{200}} = 242W \cr} $$

$$D.C.$$  ammeter measure average current in $$AC$$ current, average current is zero for complete cycle. Hence reading will be zero.

A transformer is essentially an $$AC$$  device. $$DC$$  source so no mutual induction between coils
$$ \Rightarrow {E_2} = 0\,\,{\text{and}}\,\,{I_2} = 0$$

Since the capacitor is connected in series to the resistor, current $${I_C}$$ from the supply and $${I_R}$$ through the resistor is in phase as represented by choice (A).

$$\eqalign{ & f = \frac{1}{{2\pi \sqrt {LC} }} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,f \propto \frac{1}{{\sqrt C }} \to \frac{1}{{\sqrt 4 }} = \frac{1}{2}\,{\text{time}} \cr} $$

$${I_{rms}} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}}} \right)}^2}} }}$$
As $$\omega $$ increases, $${I_{rms}}$$  increases. Therefore the bulb glows brighter.

A circuit equivalent to the given circuit can be drawn as follows :

Where
$$\eqalign{ & {R_{{\text{eq}}}} = R + \frac{{2R \times 2R}}{{4R}} = R + R \cr & {R_{{\text{eq}}}} = 2R \cr} $$
∴ Time constant $$ = \frac{L}{{2R}}$$

as current any instant in the circuit will be
$$I = {I_{dc}} + {I_{ac}} = a + b\sin \omega t$$
$$\eqalign{ & {\text{so,}}\,\,{I_{{\text{eff}}}} = {\left[ {\frac{{\int_0^T {{I^2}dt} }}{{\int_0^T {dt} }}} \right]^{\frac{1}{2}}} = {\left[ {\frac{1}{T}\int_0^T {{{\left( {a + b\sin \omega t} \right)}^2}dt} } \right]^{\frac{1}{2}}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,{I_{{\text{eff}}}} = {\left[ {\frac{1}{T}\int_0^T {{{\left( {{a^2} + 2ab\sin \omega t + {b^2}{{\sin }^2}\omega t} \right)}^2}dt} } \right]^{\frac{1}{2}}} \cr} $$
but as $$\frac{1}{T}\int_0^T {\sin \omega tdt} = 0\,{\text{and}}\,\frac{1}{T}\int_0^T {{{\sin }^2}\omega t\,dt = \frac{1}{2}} $$
So, $$\,{I_{{\text{eff}}}} = {\left[ {{a^2} + \frac{1}{2}{b^2}} \right]^{\frac{1}{2}}}$$