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If the voltage of an AC source varies with time according to the relation $E = 6 \sin (200 πt)$ , then:

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the peak voltage of the source is 120V.

the peak voltage is 6V.

the peak voltage $\frac{120}{\sqrt{2}} V$ .

the frequency of source voltage is 100 Hz.

answer is B, D.

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Detailed Solution

i) peak emf is $E_{0} = 6 V;$

ii) $ω = 2 πϑ = 200 π \Rightarrow frequency, ϑ = 100 Hz$

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