$$\eqalign{ & {\text{Power}}\,{\text{facto}}{{\text{r}}_{\left( {{\text{old}}} \right)}} = \frac{R}{{\sqrt {{R^2} + X_L^2} }} = \frac{R}{{\sqrt {{R^2} + {{\left( {2R} \right)}^2}} }} = \frac{R}{{\sqrt 5 R}} \cr & {\text{Power}}\,{\text{facto}}{{\text{r}}_{\left( {{\text{new}}} \right)}} = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }} = \frac{R}{{\sqrt {{R^2} + {{\left( {2R - R} \right)}^2}} }} = \frac{R}{{\sqrt 2 R}} \cr & \therefore \frac{{{\text{New power factor}}}}{{{\text{Old power factor}}}} = \frac{{\frac{R}{{\sqrt 2 R}}}}{{\frac{R}{{\sqrt 5 R}}}} = \sqrt {\frac{5}{2}} \cr} $$

For series $$LCR$$  circuit
Voltage $$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $$
Since, $${V_L} = {V_C}$$
Hence $$V = {V_R} = 220\,V$$
Also, current $$i = \frac{V}{R} = \frac{{220}}{{100}} = 2.2\,A$$

Power dissipated in series $$LCR.$$
$$\eqalign{ & P = I_{rms}^2R \cr & = \frac{{\varepsilon _{rms}^2R}}{{{{\left| Z \right|}^2}}} = \frac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \right]}} \cr} $$
As $${I_{rms}} = \frac{{{V_{rms}}}}{Z}$$

As it can be easily seen by the direction of $$I$$ that $$Q$$ is decreasing thus, energy of capacitor is decreasing and hence, energy of inductance is increasing or $$\left( {\frac{1}{2}L{I^2}} \right)$$  gives that $$I$$ is increasing.

Net reactive capacitance,

$${X_C} = \frac{1}{{2\pi fC}}$$
So, current in circuit, $$I = \frac{V}{Z} = \frac{V}{{\sqrt {{R^2} + {{\left( {\frac{1}{{2\pi fC}}} \right)}^2}} }}$$
$$ \Rightarrow I = \frac{{2\pi fC}}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }} \times V$$
Voltage drop across capacitor, $${V_C} = I \times {X_C}$$
$$ = \frac{{2\pi fC}}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }} \times \frac{1}{{2\pi fC}}$$
i.e. $${V_C} = \frac{V}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }}$$
When mica is introduced, capacitance will increase hence, voltage across capacitor get decrease.

$$\eqalign{ & {\text{Given,}}\,{V_L}:{V_C}:{V_R} = 1:2:3 \cr & V = 100\,V \cr & {V_R} = ? \cr} $$
As we know, $$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $$
Solving we get, $${V_R} \simeq 90\,V$$

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

There is no $$DC$$ current in inductive circuit because it does not allow $$DC$$ current to flow and maximum $$DC$$ current in capacitive circuit. Hence, the current is zero in $${A_2}$$ and maximum in $${A_1}.$$

Resonance frequency,
$$\eqalign{ & {\omega _0} = \frac{1}{{\sqrt {LC} }} = \frac{1}{{\sqrt {L'\left( {2C} \right)} }} \cr & \therefore L' = \frac{L}{2}. \cr} $$

Charge on he capacitor at any time $$t$$ is given by $$q = CV\left( {1 - {e^{\frac{t}{\tau }}}} \right)\,\,{\text{at}}\,\,t = 2\tau $$
$$q = CV\left( {1 - {e^{ - 2}}} \right)$$