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:
\({{\rm{P}}_1} = {{\rm{i}}^2}{{\rm{R}}_1} = {\left( {\frac{1}{{11}}} \right)^2} \times \frac{{220 \times 220}}{{25}} = 16{\rm{\;W}}\)
\({{\rm{P}}_2} = {{\rm{i}}^2}{{\rm{R}}_2} = {\left( {\frac{1}{{11}}} \right)^2} \times \frac{{220 \times 220}}{{100}} = 4{\rm{\;W}}\)