Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is:
Answer & explanation
Correct answer: option 1
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