Question:
An ac source has a peak voltage \( \frac{200}{\sqrt{2}} \) V and frequency 50 Hz. The value of voltage after \( \frac{1}{600} \) s from the start is:
An ac source has a peak voltage \( \frac{200}{\sqrt{2}} \) V and frequency 50 Hz. The value of voltage after \( \frac{1}{600} \) s from the start is:
Show Hint
Always perform angular reductions in radians first before converting to standard degrees. A time of \( \frac{1}{600} \) seconds at 50 Hz corresponds to exactly one-twelfth of a full cycle duration, which maps directly to a \( 30^\circ \) phase angle step.
Updated On: Jun 8, 2026
- 220 V
- \( \frac{200}{\sqrt{2}}\text{V} \)
- \( \frac{100}{\sqrt{2}}\text{V} \)
- 50 V
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The Correct Option is C
Solution and Explanation
Concept:
The instantaneous voltage \( v(t) \) of an alternating current source starting from zero value can be modeled as a sinusoidal wave function:
\[
v(t) = V_m \sin(\omega t) = V_m \sin(2\pi f t)
\]
where \( V_m \) is the peak voltage and \( f \) is the linear frequency.
Step 1: Substituting given values into the wave function.
From the question data:
• Peak voltage value, \( V_m = \frac{200}{\sqrt{2}}~\text{V} \)
• Frequency parameter, \( f = 50~\text{Hz} \)
• Time increment, \( t = \frac{1}{600}~\text{s} \)
\[ v(t) = \left(\frac{200}{\sqrt{2}}\right) \sin\left(2\pi \times 50 \times \frac{1}{600}\right) \]
Step 2: Evaluating the angular argument inside the sine function.
\[ \text{Angle} = 100\pi \times \frac{1}{600} = \frac{\pi}{6}\text{ radians} = 30^\circ \]
Step 3: Computing the final instantaneous voltage value.
Since \( \sin(30^\circ) = \frac{1}{2} \): \[ v(t) = \left(\frac{200}{\sqrt{2}}\right) \times \frac{1}{2} = \frac{100}{\sqrt{2}}~\text{V} \] This matches option (C) perfectly.
Step 1: Substituting given values into the wave function.
From the question data:
• Peak voltage value, \( V_m = \frac{200}{\sqrt{2}}~\text{V} \)
• Frequency parameter, \( f = 50~\text{Hz} \)
• Time increment, \( t = \frac{1}{600}~\text{s} \)
\[ v(t) = \left(\frac{200}{\sqrt{2}}\right) \sin\left(2\pi \times 50 \times \frac{1}{600}\right) \]
Step 2: Evaluating the angular argument inside the sine function.
\[ \text{Angle} = 100\pi \times \frac{1}{600} = \frac{\pi}{6}\text{ radians} = 30^\circ \]
Step 3: Computing the final instantaneous voltage value.
Since \( \sin(30^\circ) = \frac{1}{2} \): \[ v(t) = \left(\frac{200}{\sqrt{2}}\right) \times \frac{1}{2} = \frac{100}{\sqrt{2}}~\text{V} \] This matches option (C) perfectly.
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