An alternating voltage V(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:
Options

Question

An alternating voltage V(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:

G1 Physics · Accepted Answer

For a purely resistive load, current and voltage are in phase. The current rises from I₀/2 to I₀ as phase changes from: sinθ = 1/2 ⇒ θ = π/6 to θ = π/2 Phase change: Δθ = π/2 - π/6 = π/3 Given, ω = 100π rad/s Therefore, t = Δθ/ω = (π/3)/(100π) = 1/300 s = 3.33 ms