An alternating voltage V(t)=220 sin 100 π t 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:

Solution

Physics sir · 3 steps

Ignore the 50 ohms; just ask how much phase the sine sweeps from 1/2 to 1, then divide by ω.

1. Current follows voltage. In a purely resistive load the current is in phase with the voltage, so i = I₀ sin 100π t. The value of R only sets I₀ and does not change the timing.
2. Find the two phases. Half the peak means sin 100π t = 1/2, so 100π t₁ = π/6. The peak means sin 100π t = 1, so 100π t₂ = π/2.

   Δ t = Δθ/ω = (π/2-π/6)/100π = (π/3)/100π = 1/300 s

3. So. Δ t = 1/300 s ≈ 3.3 ms, which is option (b).

The idea to keep. For timing questions on a sinusoid, turn each value into a phase angle, subtract, and divide by ω. In a pure resistor, current and voltage share the same phase.

Answer from the NTA final answer key. Solution by PrepWiser.