Question

The instantaneous value of an alternating current is given by $i = 50 \sin (100 \pi t)$. It will achieve a value of $25\text{ A}$ after a time interval of ($\sin 30^\circ = 0.5$)

• A. $\frac{1}{300}\text{ s}$
• B. $\frac{1}{100}\text{ s}$
• C. $\frac{1}{200}\text{ s}$
• D. $\frac{1}{600}\text{ s}$

Hint:

Whenever the target value is exactly half of the peak amplitude ($I = \frac{1}{2}I_0$) for a standard sine wave starting from zero, the phase angle is always $\frac{\pi}{6}$. Equating $\omega t = \frac{\pi}{6}$ lets you solve for time instantly without writing out any trigonometric lines!

Answer 1

The Correct Option is D

Step 1: What is given.
An alternating current changes with time as $i = 50\sin(100\pi t)$. We must find the first time at which the current equals $25$ A.
Step 2: Set the target value.
Put $i = 25$ into the equation: \[ 25 = 50\sin(100\pi t) \]
Step 3: Get the sine alone.
Divide both sides by $50$: \[ \sin(100\pi t) = \frac{25}{50} = 0.5 \]
Step 4: Use the known angle.
We are told $\sin 30^\circ = 0.5$. In radians, $30^\circ = \dfrac{\pi}{6}$. So the angle inside the sine must equal $\dfrac{\pi}{6}$: \[ 100\pi t = \frac{\pi}{6} \]
Step 5: Cancel pi.
Divide both sides by $\pi$: \[ 100 t = \frac{1}{6} \]
Step 6: Solve for time.
\[ t = \frac{1}{600}\ \text{s} \] So the current first reaches $25$ A after $\dfrac{1}{600}$ s, which is option (4). \[ \boxed{t = \frac{1}{600}\ \text{s}} \]