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The Flow of Electrons as Electric Current

If electric c...

Question

If electric current is $i = t^{2}$ and 0 < t < T then ,r.m.s. value of current is

A

\frac{T^{2}}{\sqrt{2}}

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B

\frac{T^{2}}{2}

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C

\frac{T^{2}}{\sqrt{5}}

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D

N o n e o f t h e s e

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Solution

The correct option is C $\frac{T^{2}}{\sqrt{5}}$
$i_{r m s} = \sqrt{\frac{1}{T} \int_{0}^{T} i^{2} d t} = \frac{T^{2}}{\sqrt{5}}$

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Similar questions

x = log t

y = t^{2} - 1,

y^{''} (1)

t = 1

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

then the value of

\frac{d^{2} y}{d x^{2}}

t = 0

t^{2} + t + 1 = 0

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

t^{2} + t + 1 = 0,

then the value of the expression

(t + \frac{1}{t}) + (t^{2} + \frac{1}{t^{2}}) + \dots + (t^{27} + \frac{1}{t^{^{27}}})

t^{2} + t + 1

= 0 then prove that the value of

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . {(t^{27} + \frac{1}{t^{27}})}^{2} = 54

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