Given, 

$$\begin{gathered} \mathrm{Resistance}, \mathrm{R} = 100 \mathrm{\Omega } \\ \mathrm{Voltage} \mathrm{applied}, \mathrm{V} = 220 V \\ \mathrm{Frequency} \mathrm{of} \mathrm{the} \mathrm{AC} \mathrm{supply}, \mathrm{f} = 50 \mathrm{Hz} \end{gathered}$$ 

$\therefore$ $\mathrm{\omega } = 2\mathrm{\pi } \mathrm{f} = 2 \times 3.14 \times 50 = 314$ 

 Using the relation
                    ${\mathrm{I}}_{\mathrm{eff}} = \frac{{\mathrm{V}}_{\mathrm{eff}}}{\mathrm{R}}$
Putting values,
                       ${\mathrm{I}}_{\mathrm{eff}} =\frac{220}{100} = 2.2 \mathrm{A}$

Given,
Inductance, L = 44 mH = $44 \times {10}^{-3}\mathrm{H}$

$$\begin{gathered} \mathrm{Frequency} \mathrm{of} \mathrm{the} \mathrm{Ac} \mathrm{supply}, \mathrm{f} = 50 \mathrm{Hz} \\ \mathrm{Rms} \mathrm{value} \mathrm{of} \mathrm{voltage}, {\mathrm{E}}_{\mathrm{rms}} = 220 \mathrm{V} \end{gathered}$$ 

$$\begin{gathered} \mathrm{Inductive} \mathrm{reactance} \mathrm{is} \mathrm{given} \mathrm{by}, \boxed{{\mathrm{X}}_{\mathrm{L}} = \mathrm{L}.\mathrm{\omega }} \\ = \mathrm{L}. 2\mathrm{\pi f} \\ = 44 \times {10}^{-3} \times 2 \times 3.14 \times 50 \end{gathered}$$
Now,
                    $$\begin{gathered} \boxed{{\mathrm{I}}_{\mathrm{rms}} = \frac{{\mathrm{E}}_{\mathrm{rms}}}{{\mathrm{X}}_{\mathrm{L}}}} \\ = \frac{220}{44 \times {10}^{-3}\times 2\times 3.14\times 50} \end{gathered}$$
                         $= 15.9 \mathrm{A}.$ 

Given,

$$\begin{gathered} \mathrm{Peak} \mathrm{volatge}, {\mathrm{E}}_0 = 300 \mathrm{V} \\ \mathrm{Rms} \mathrm{value} \mathrm{of} \mathrm{current}, {\mathrm{I}}_{\mathrm{rms}} = 10 \mathrm{A} \end{gathered}$$ 

Now,
Using the relation,

                     $\boxed{{\mathrm{E}}_{\mathrm{rms}} = \frac{{\mathrm{E}}_0}{\sqrt{2}}}$ 

                           $$\begin{gathered} = \frac{300}{\sqrt{2}} \\ = \frac{300}{1.414} \\ = 212.13 \mathrm{V} \end{gathered}$$

Given, 

$$\begin{gathered} \mathrm{Peak} \mathrm{voltage}, {\mathrm{E}}_0 = 300 \mathrm{V} \\ \mathrm{Rms} \mathrm{value} \mathrm{of} \mathrm{current}, {\mathrm{I}}_{\mathrm{rms}} = 10 \mathrm{A} \end{gathered}$$ 

Hence, using the formula,

              $\boxed{{\mathrm{I}}_{\mathrm{rms}} = \frac{{\mathrm{I}}_0}{\sqrt{2}}}$
         
                 ${\mathrm{I}}_0 = \sqrt{2} {\mathrm{I}}_{\mathrm{rms}}$

                    $$\begin{gathered} = \sqrt{2} \times 10 \\ = 14.1 \mathrm{A}. \end{gathered}$$ 
which is the required value of peak current. 

Given, 

$$\begin{gathered} \mathrm{Resistance}, \mathrm{R} = 100 \mathrm{\Omega } \\ \mathrm{Potential} \mathrm{applied}, \mathrm{V} = 220 ( effective voltage {\mathrm{V}}_{\mathrm{eff}}) \\ \mathrm{frequency}, \mathrm{f} = 50 \mathrm{Hz} \end{gathered}$$ 

Since, $\mathrm{\omega } = 2\mathrm{\pi } \mathrm{f} = 2 \times 3.14 \times 50 = 314$ 
Therefore,
Power consumed = Current x Voltage
                        = ${\mathrm{I}}_{\mathrm{eff}} \times {\mathrm{V}}_{\mathrm{eff}}$ 

                        = 2.2 x 200
                        = 484 watt.