Correct Answer - Option 1 : P

_{1} = 16 W, P

_{2} = 4 W

**Concept:**

Resistance of a bulb of power P and with a voltage source V is given by the formula:

\(R = \frac{{{V^2}}}{P}\)

**Calculation:**

Resistance of the given two bulbs are given as:

\({R_1} = \frac{{{V^2}}}{{{P_1}}} = \frac{{{{(220)}^2}}}{{25}}\)

\({R_2} = \frac{{{V^2}}}{{{P_2}}} = \frac{{{{(220)}^2}}}{{100}}\)

Since, bulbs are connected in series. This means same amount of current flows through them.

The current in circuit is given as:

\(i = \frac{V}{{{R_{{\rm{total\;}}}}}} = \frac{{220}}{{\frac{{{{(220)}^2}}}{{25}} + \frac{{{{(220)}^2}}}{{100}}}} = \frac{1}{{11}}{\rm{\;A}}\)

The power drawn by bulbs are given as:

\({{\rm{P}}_1} = {{\rm{i}}^2}{{\rm{R}}_1} = {\left( {\frac{1}{{11}}} \right)^2} \times \frac{{220 \times 220}}{{25}} = 16{\rm{\;W}}\)

\({{\rm{P}}_2} = {{\rm{i}}^2}{{\rm{R}}_2} = {\left( {\frac{1}{{11}}} \right)^2} \times \frac{{220 \times 220}}{{100}} = 4{\rm{\;W}}\)