Question

The instantaneous value of current in an a.c. circuit is \(I = 3 \sin (50\pi t + \frac{\pi}{4})\text{A}\). The current will be maximum for the first time at

• A. \(\frac{1}{50} \text{ s}\)
• B. \(\frac{1}{100} \text{ s}\)
• C. \(\frac{1}{200} \text{ s}\)
• D. \(\frac{1}{600} \text{ s}\)

Hint:

Always consider phase constant while finding first maximum.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A sinusoidal function \(I(t) = I_0 \sin(\text{Phase})\) reaches its maximum value when its phase angle is \(\pi/2\) radians (or \(90^\circ\)).
Step 2: Key Formula or Approach:
For \(I\) to be maximum: \(50\pi t + \frac{\pi}{4} = \frac{\pi}{2}\).
Step 3: Detailed Explanation:
The expression for current is given as \(I = 3 \sin (50\pi t + \frac{\pi}{4})\).
The first positive maximum occurs when the sine argument is \(\pi/2\): \[ 50\pi t + \frac{\pi}{4} = \frac{\pi}{2} \] Subtract \(\pi/4\) from both sides: \[ 50\pi t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] Divide both sides by \(\pi\): \[ 50t = \frac{1}{4} \] Solve for \(t\): \[ t = \frac{1}{4 \times 50} = \frac{1}{200}\text{ s} \] Thus, the current reaches its peak for the first time at \(0.005\text{ s}\).
Step 4: Final Answer:
The current is maximum for the first time at \(t = \frac{1}{200}\text{ s}\).