Correct Answer - Option 3 : 200/3 W
Concept:
Two bulbs are in series, having wattage rating and voltage rating then,
1st bulb rating = P1/ V1
Now, the resistance of 1st bulb
\({R_1} = \frac{{{{V_1}^2}}}{{P_1}}\)
2nd bulb rating = P2/ V2
The resistance of 2nd bulb
\({R_2} = \frac{{{{V_2}^2}}}{{P_2}}\)
Both the bulbs are connected in series across VTotal V.
Now the total power consumed is,
\(P = \frac{{{V^2}}}{{{R_1} + {R_2}}}\) Watt
Calculation:
1st bulb rating = 200 W/200 V
Now, the resistance of 1st bulb
\({R_1} = \frac{{{{200}^2}}}{{200}} = 200\; Ω\)
2nd bulb rating = 100 W/200 V
The resistance of 2nd bulb
\({R_2} = \frac{{{{200}^2}}}{{100}} = 400\; Ω\)
Both the bulbs are connected in series across 200 V.
Now the total power consumed is,
\(P = \frac{{{V^2}}}{{{R_1} + {R_2}}}\) Watt
\( = \frac{{{{200}^2}}}{{600}} = 200/3\;W\)