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Question = Mangal's room heater is marked as 1,000 W - 200 V find the percentage change in power of heater, if the voltage drops to 160 V
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Answers
Answered by
71
Answer. Here, Power of the heater at 200V
P = 1,000 W
Resistance of the Heater
[texR \: = \frac{ {V}^{2} }{P} \: = \frac{ {(200)}^{2} }{1000} = \frac{40000}{1000} = 40 ohm \\ \\ Power of the heater at V' = 160 V, P' = \frac{ {V'}^{2} }{R}
= \frac{ {(160)}^{2} }{40}
= \frac{25600}{40}
= 160W
Percentage fall in the power of heater \frac{P' - P}{P} \ \times 100 \\ = \frac{1000 - 640}{1000}\times100 = 36%[/tex]
Answered by
0
Answer:
160 watt
Step-by-step explanation:
36%[/Tex]
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