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)?

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:

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