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2 years ago

5

A lightbulb has a resistance of 285 ω when operating with a potential difference of 110 v across it. what is the current in the

lightbulb?

1 answer:

Doss [256]2 years ago

5 0

Current = voltage / resistance

= 110 v / 285 Ω

= 0.386 Ampere (rounded)

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Answer:

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function = x^2 + y^2 ≤ 6^2

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solution

first we consider the polar coordinate (a,θ)

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I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

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I(o) = 2 × 1444×k × ${\pi}$

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