Sine-wave evaluation: For a 6 V peak sinusoid v(t) = 6 * sin(θ), what is the instantaneous peak voltage at an electrical angle of 250° (report polarity and magnitude)?
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A+5.64 V
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B–5.64 V
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C+26.13 V
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D–26.13 V
Answer
Correct Answer: –5.64 V
Explanation
Introduction:Evaluating sinusoidal signals at specific phase angles is a routine task in AC circuit analysis and signal processing. This item reinforces the relationship between angle, sign, and amplitude, particularly in quadrants where the sine function is negative.
Given Data / Assumptions:
- Peak amplitude Vp = 6 V.
- Instantaneous form: v(θ) = Vp * sin(θ).
- Angle θ = 250 degrees.
Concept / Approach:The sine of an angle in degrees can be related to a known reference angle using quadrant identities. Since 250° = 180° + 70°, the value sin(250°) = −sin(70°). Knowing sin(70°) ≈ 0.9397 allows a quick numerical evaluation. The polarity is negative in the third quadrant for sine.
Step-by-Step Solution:
Write v(θ) = 6 * sin(250°).Use identity: sin(250°) = −sin(70°).Compute sin(70°) ≈ 0.9397.v(250°) ≈ 6 * (−0.9397) ≈ −5.6382 V.Round to two decimals: −5.64 V.Verification / Alternative check:Calculator check using radians: 250° * pi / 180 ≈ 4.3633 rad; sin(4.3633) ≈ −0.9397; result matches.
Why Other Options Are Wrong:
- +5.64 V: Correct magnitude but wrong sign; sine is negative at 250°.
- +26.13 V or −26.13 V: Exceed the peak amplitude of 6 V; impossible for a 6 V peak sine.
Common Pitfalls:Confusing peak with RMS (6/√2 ≈ 4.24 V), using degrees/radians mismatched, or selecting the correct magnitude but not the correct polarity based on the quadrant.
Final Answer:–5.64 V