We know that frequency of electrical oscillation in $$L.C.$$  circuit is
$$\eqalign{ & f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}}} \cr & {\text{Now,}}\,L = 2L\,\,\& \,\,C = 4C \cr & f' = \frac{1}{{2\pi }}\sqrt {\frac{1}{{2L \cdot 4C}}} = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}}} \times \frac{1}{{2\sqrt 2 }} \Rightarrow f' \cr & = \frac{1}{{2\sqrt 2 }} \times f \cr} $$

$$\eqalign{ & {i_{1rms}} = \frac{{{E_{rms}}}}{{\sqrt {X_C^2 + R_1^2} }} = \frac{{130}}{{13}} = 10\,A \cr & {i_{2rms}} = \frac{{{E_{rms}}}}{{\sqrt {X_L^2 + R_2^2} }} = 13\,A \cr} $$
Power dissipated $$ = i_{1rms}^2{R_1} + i_{2rms}^2{R_2} = {10^2} \times 5 + {13^2} + 6$$
= power delivered by battery
$$ = 500 + 169 \times 6 = 1514\,watt$$

The equation of a semi - circular wave is
$$\eqalign{ & {x^2} + {y^2} = {a^2}\,\,{\text{or}}\,\,{y^2} = {a^2} - {x^2} \cr & {I_{rms}} = \sqrt {\frac{1}{{2a}}\int_{ - a}^{ + a} {{y^2}} dx} \cr & I_{rms}^2 = \frac{1}{{2a}}\int_{ - a}^{ + a} {\left( {{a^2} - {x^2}} \right)} dx \cr & = \frac{1}{{2a}}\int_{ - a}^{ + a} {\left( {{a^2} - {x^2}} \right)} dx = \frac{1}{{2a}}\left| {{a^2}x - \frac{{{x^3}}}{3}} \right|_{ - a}^{ + a} \cr & = \frac{1}{{2a}}\left( {{a^3} - \frac{{{a^3}}}{3} + {a^3} - \frac{{{a^3}}}{3}} \right) = \frac{{2{a^2}}}{3} \cr & {I_{rms}} = \sqrt {\frac{{2{a^2}}}{3}} = \sqrt {\frac{2}{3}} a \cr} $$

Time period,
$$\eqalign{ & {T_1} = 2\pi \sqrt {LC} , \cr & {T_2} = 2\pi \sqrt {\frac{{LC}}{2}} \cr & {T_3} = 2\pi \sqrt {2LC} \cr} $$
Clearly $${t_2} < {t_1} < {t_3}.$$

$${\text{Since}}\,\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}$$
where
$${{N_s}}$$ = No. of turns across primary coil $$= 50$$
$${{N_p}}$$ = No. of turns across secondary coil $$= 1500$$
$$\eqalign{ & {\text{and}}\,{V_p} = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {{\phi _0} + 4t} \right) = 4 \cr & \Rightarrow {V_s} = \frac{{1500}}{{50}} \times 4 \cr & = 120\,V \cr} $$

When a resistor is connected to an $$AC$$ source. The power drawn will be
$$P = {V_{rms}} \cdot {I_{rms}} = {V_{rms}} \cdot \frac{{{V_{rms}}}}{R} \Rightarrow V_{rms}^2 = PR$$
When an inductor is connected in series with the resistor, then the power drawn will be
$$P' = {V_{rms}} \cdot {I_{rms}}\cos \phi $$
where, $$\phi = $$ phase difference
$$\eqalign{ & \therefore P' = \frac{{V_{rms}^2}}{R} \cdot \frac{{{R^2}}}{{{Z^2}}} = P \cdot R \cdot \frac{R}{{{Z^2}}} \cr & \Rightarrow P' = \frac{{P \cdot {R^2}}}{{{Z^2}}} = P{\left( {\frac{R}{Z}} \right)^2} \cr} $$

At $$t = 0,$$  no current will flow through $$L$$ and $${{R_1}}$$
∴ Current through battery $$ = \frac{V}{{{R_2}}}$$
$${\text{At}}\,\,t = \infty ,$$
effective resistance, $${R_{{\text{eff}}}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$$
∴ Current through battery $$ = \frac{V}{{{R_{{\text{eff}}}}}} = \frac{{V\left( {{R_1} + {R_2}} \right)}}{{{R_1}{R_2}}}$$

$${P_{av}} = {V_{rms}} \cdot {i_{rms}}\cos \phi = \frac{1}{2}{V_0}{i_0}\cos \phi \,\,\left[ {\because {V_{rms}} = \frac{{{V_0}}}{{\sqrt 2 }},{i_{rms}} = \frac{{{i_0}}}{{\sqrt 2 }}} \right]$$
where, $$\cos \phi = $$  power factor

A circuit in which inductance $$L,$$ capacitance $$C$$ and resistance $$R$$ are connected in series, and the circuit admits maximum current corresponding to a given frequency of $$AC$$ is called series resonance circuit. The impedance $$\left( Z \right)$$ of an $$RLC$$  circuit is given by
$$Z = \sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} $$
At resonance $${X_L} = {X_C}$$
i.e. $$\omega L = \frac{1}{{\omega C}},Z = R$$
So, circuit behaves as if it contains $$R$$ only. So, phase difference $$= 0.$$
Frequency of resonating $$LCR$$  circuit is given by
$$\eqalign{ & {\omega ^2} = \frac{1}{{LC}} \cr & \Rightarrow f = \frac{1}{{2\pi \sqrt {LC} }} \cr} $$

Impedence of the $$R-C$$  circuit, $$Z = \sqrt {{R^2} + X_C^2} $$
where, $$R = 100\,\Omega \,\,{\text{and}}\,\,{X_C} = 100\,\Omega $$
$$ \Rightarrow \tau = \sqrt {{{\left( {100} \right)}^2} + {{\left( {100} \right)}^2}} = 100\sqrt 2 \,\Omega $$
Peak value of the current,
$${\operatorname{I} _{max}} = \frac{{{V_{\max }}}}{Z} = \frac{{220\sqrt 2 }}{{100\sqrt 2 }} = 2.2\,A$$