Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is:

800 W

Let, the bulb 400 W is having resistance value of R. For 200 W, necessary value of resistance will be 2R.

Total value of resistance in the circuit will be R + R = 2R

If I is the maximum current in the circuit, then I²R = 400 through 200 W each, current will be I/2 and power will be

$\frac{I_2}{4} \times 2 R = \frac{I^2 R}{2}=\frac{400}{2}$ = 200 W

Power circuit as a whole = I₂ × 2R = 2 × I²R = 2 × 400 = 800W