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The equation for an alternating currentis given by i=77sin 314t. Find the peak value . |
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Answer» Solution :i=77 SIN 314, t = t =2 ms = `2xx 10^(-3) s ` The generaleqution of an alternatingcurrentis ` I = I_(s)`sin `omegat` . On comparison . (i) Peak VALUE, ` I_(m) = 77A` (ii)Frequency , `f = (omega)/(2pi) = (314)/(2 xx 3.14) = 50Hz` (iii)Time period,` T= (1)/(f)= (1)/(50) = 0.02s` (iv) At `""t = 2 ms` Instataneous value` i= 77 sin (314 xx 2 xx 10^(-3)) = 42.24 A` |
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