FCC Exam Question: 3-12B2

At 150 degrees, what is the amplitude of a sine-wave having a peak value of 5 volts?

A. -4.3 volts. C. +2.5 volts.
B. -2.5 volts. D. +4.3 volts.
C.
D.
Correct Answer: C

Explanation: The instantaneous amplitude of a sine wave is determined by its peak value and the sine of its phase angle. The formula is: Amplitude = Peak Value × sin(angle) In this case, the peak value is 5 volts and the angle is 150 degrees. * Calculate sin(150°): sin(150°) = 0.5 * Multiply by the peak value: 5 volts × 0.5 = 2.5 volts Therefore, at 150 degrees, the amplitude of the sine wave is +2.5 volts. Option B (-2.5 volts) would be correct if the angle were 210 degrees (or -30 degrees), where sin(210°) = -0.5. Options A and D represent different magnitudes, which would occur at angles where the sine function has a different absolute value (e.g., approximately 60 or 120 degrees for +4.3 volts).

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Includes Elements 1, 3, 6, 7R, 8, and 9.