Math, asked by tanu4781, 1 year ago

Its Very urgent please

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 Anonymous
71

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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 roopesh9240
0

Answer:

160 watt

Step-by-step explanation:

36%[/Tex]

mark this as brainliest

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