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Asked in GSEB Board March 2023 · Instantaneous values and timing
Given: Iᵣₘₛ = 5 A, f = 50 Hz, and I = 0 at t = 0, so (taking it to rise from zero) I = I₀ sin ω t.
Peak value: I₀ = √2 Iᵣₘₛ = 5√2 A.
Phase at t = 1/(300) s: ω t = 2π×50×1/(300) = π/3.
I = 5√2 sin π/3 = 5√2×(√3)/2 = 5√3/2 A ≈ 6.1 A.
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