FCC Exam Question: 3-12B3

At 240 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: A

Explanation: The instantaneous amplitude of a sine wave at any given phase angle can be calculated using the formula: Amplitude = Peak Value × sin(angle). In this case, the peak value is 5 volts and the angle is 240 degrees. Amplitude = 5V × sin(240°) The sine of 240 degrees is -0.866 (since 240 degrees is in the third quadrant, where sine values are negative, and sin(240°) = sin(180° + 60°) = -sin(60°)). Amplitude = 5V × (-0.866) Amplitude = -4.33 volts. Therefore, the amplitude at 240 degrees is approximately -4.3 volts. Option B (-2.5 volts) would correspond to an angle where sin(angle) = -0.5, or if you mistakenly used cosine. Options C (+2.5 volts) and D (+4.3 volts) are incorrect because the sine wave is in a negative phase at 240 degrees.

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