Question

The r.m.s. value of an ac of 50 Hz is 10 amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be(a) 2 × 10⁻² sec and 14.14 amp(b) 1 × 10⁻² sec and 7.07 amp(c) 5 × 10⁻³ sec and 7.07 amp(d) 5 × 10⁻³ sec and 14.14 amp

Answer 1

answer : (d) 5 × 10⁻³ sec and 14.14 Amp

explanation : given , frequency, f = 50Hz

so, time period , T = 1/50 = 2 × 10^-2 sec

rms value of current of an ac circuit , $i_{rms}$ = 10 Amp

we know, alternating current reaches the maximum value when time taken by it is one fourth of time period.

e.g., $t=\frac{T}{4}$

or, t = 2 × 10^-2/4 = 5 × 10^-3 sec

now, peak value of current, $i_p=\sqrt{2}i_{rms}$

here given, $i_{peak}=\sqrt{2}\times10=14.14$

hence, peak value of current will be 14.14Amp

Answer 2

Explanation:

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