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  • Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is: <img alt="" src="https://cdn.entrance360.com/media/uploads/2023

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

Option: 1

800 W


Option: 2

400 W


Option: 3

200 W


Option: 4

600 W


Answers (1)

best_answer

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 \mathrm{R+R=2 R}

If I is the maximum current in the circuit, then \mathrm{\mathrm{I}^2 \mathrm{R}=400} through 200 W each, current will be I/2 and power will be 

\frac{\mathrm{I}_2}{4} \times 2 \mathrm{R}=\frac{\mathrm{I}^2 \mathrm{R}}{2}=\frac{400}{2}=200 \mathrm{~W}

Power circuit as a whole =\mathrm{I}_2 \times 2 \mathrm{R}=2 \times \mathrm{I}^2 \mathrm{R}=2 \times 400=800 \mathrm{~W}

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