Question

An electric bulb of resistance 500 S2 draws current
0.4 A from the source. Calculate : (a) the power of
bulb and (b) the potential difference at its end.​

Answer 1

Answer:

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Explanation:

Given : x sin3 θ + y cos3 θ = sin θ cos θ

⇒ (x sin θ) sin2 θ + (y cos θ) cos2 θ = sin θ cos θ

⇒ (x sin θ) sin2 θ + (x sin θ) cos2 θ = sin θ cos θ (∵ y cos θ = x sin θ

⇒ x sin θ(sin2 θ + cos2 θ) = sin θ cos θ

⇒ x sin θ = sin θ cos θ

⇒ x = cos θ ….(1)

Again x sin θ = y cos θ

⇒ cos θ sin θ = y cos θ

⇒ y = sin θ ….(2)

Squaring and adding (1) and (2) we get the required result

Answer 2

Answer:

please send the picture of the question