KEY CONCEPT: We know that power consumed in $$a.c.$$  circuit is given by, $$P = {E_{rms}}.{I_{rms}}\cos \phi $$
Here, $$E = {E_0}\sin \omega t$$
$$I = {I_0}\sin \left( {\omega t - \frac{\pi }{2}} \right)$$
which implies that the phase difference, $$\phi = \frac{\pi }{2}$$
$$\therefore P = {E_{rms}}.{I_{rms}}\cos \frac{\pi }{2} = 0\,\,\,\,\,\left( {\because \cos \frac{\pi }{2} = 0} \right)$$

Equivalent circuit can be drawn as below

$$\eqalign{ & RC = 3 \times {10^{ - 3}}\sec \cr & R \times 3 \times {10^{ - 6}} = 3 \times {10^{ - 3}} \cr & R = 1000\Omega \cr} $$

The transformer converts $$AC$$ high voltage into $$AC$$ low voltage, but it does not cause any change in frequency.
The ratio of voltage across input with output voltage is given by
$$\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}$$
$${{N_s}} =$$  No. of turns in secondary coil
$${{N_p}} =$$  No. of turns in primary coil
Making substitution, we obtain
$$\eqalign{ & {V_s} = \frac{{{N_s}}}{{{N_p}}}{V_p} \cr & = \frac{{5000}}{{500}} \times 20 \cr & = 200\,V \cr} $$
Thus, output has voltage $$200\,V$$  and frequency $$50\,Hz.$$

$$\eqalign{ & I = {I_0}\left( {I - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {\text{where}}\,\,\tau \to {\text{time constant}} \cr & \therefore \frac{3}{4}{I_0} = {I_0}\left( {I - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & \Rightarrow \frac{3}{4} = l - {e^{\frac{{ - t}}{\tau }}} \cr & \Rightarrow {e^{\frac{{ - t}}{\tau }}} = \frac{1}{4} \cr & \Rightarrow \frac{{ - t}}{\tau }\ln e = \ln \frac{1}{4} \cr & \Rightarrow \frac{{ - 4}}{\tau } = - 2\ln 2 \cr & \Rightarrow \tau = \frac{2}{{\ln 2}} \cr} $$

When a bulb of resistance $$R$$ is connected in series with a coil of self-inductance $$L,$$ then current in the circuit is given by
$$I = \frac{E}{{\sqrt {{\omega ^2}{L^2} + {R^2}} }},$$     where $$E$$ is the voltage of an $$AC$$  source
As $$L = \frac{{{\mu _0}{\mu _r}{N^2}A}}{l}$$
$$ \Rightarrow L \propto {\mu _r}$$
When iron rod is inserted, $$L$$ increases, therefore current $$I$$ decreases.