Correct Answer - Option 3 : 200/3 W

**Concept:**

Two bulbs are in series, having wattage rating and voltage rating then,

1st bulb rating = P_{1}/ V_{1}

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 V_{Total} 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

\( = \frac{{{{200}^2}}}{{600}} = 200/3\;W\)