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PHY 102 Test Compilation PHY 102 17 Objective Question

Question

An a.c. waveform is given as v = 290sin ωt. If ω = 1256 rad/s, find the r.m.s and the period.
Options
A 205V, 5ms
Correct Answer
B 289V, 500ms
C 290V, 0.5ms
D 290V, 50ms
Correct Answer

Option A is the correct answer.

Detailed Explanation

The peak voltage (V_peak) from the waveform equation v = V_peak sin(ωt) is 290 V.The root mean square (r.m.s) voltage is calculated as V_rms = V_peak / √2. V_rms = 290 V / √2 ≈ 205.06 V ≈ 205 V.The period (T) is related to the angular frequency (ω) by the formula T = 2π / ω. Given ω = 1256 rad/s:T = 2π / 1256 ≈ 6.283 / 1256 ≈ 0.005 s = 5 ms.Therefore, the r.m.s voltage is approximately 205 V and the period is 5 ms.

Hint

Remember the relationships between peak voltage and RMS voltage, and between angular frequency and period for an AC waveform.

Question Info