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:

• A. 220 V
• B. \( \frac{200}{\sqrt{2}}\text{V} \)
• C. \( \frac{100}{\sqrt{2}}\text{V} \)
• D. 50 V

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.

Answer 1

The Correct Option is C

Step 1: Write the AC voltage equation.
An AC source that starts from zero gives a voltage that follows a sine wave: \[ v(t) = V_m\sin(\omega t) = V_m\sin(2\pi f t) \]
Step 2: List the given values.
Peak voltage $V_m = \dfrac{200}{\sqrt{2}}$ V, frequency $f = 50$ Hz, time $t = \dfrac{1}{600}$ s.
Step 3: Substitute into the equation.
\[ v = \left(\frac{200}{\sqrt{2}}\right)\sin\left(2\pi\times50\times\frac{1}{600}\right) \]
Step 4: Simplify the angle.
\[ 2\pi\times50\times\frac{1}{600} = \frac{100\pi}{600} = \frac{\pi}{6} \] which is $30^{\circ}$.
Step 5: Use the value of sine.
Since $\sin 30^{\circ} = \dfrac{1}{2}$: \[ v = \left(\frac{200}{\sqrt{2}}\right)\times\frac{1}{2} \]
Step 6: Get the final voltage.
\[ v = \frac{100}{\sqrt{2}}\ \text{V} \] \[ \boxed{v = \frac{100}{\sqrt{2}}\ \text{V}} \]