Question

In an a.c. circuit, the r.m.s. value of current, Iᵣₘₛ is related to
the peak current, I₀ by the relation
(a) Iᵣₘₛ = √2 I₀ (b) Iᵣₘₛ = πI₀
(c) Iᵣₘₛ = 1/π I₀ I 1
(d) Iᵣₘₛ = 1/√2 I₀

Answer 1

The alternating current is represented as

I=I0sin(ωt) ..........................................(1)

where, I0 is the peak value of current

Now, formula for a continuous function (or waveform) f(t) defined over the interval t1≤t≤t2 is

frms=t2−t11∫t1t2f(t)2dt

Therefore,

Irms=t2−t11∫t1t2(I0sin(ωt))2dt

since, I0 is positive, we can write

Irms=I0t2−t11

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Answer 2

Answer:

(d). Lrms = 1/Π1•l1

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