Question

An alternating current is given by \( I = I_0 \cos (100\pi t) \). The least time the current takes to decrease from its maximum value to zero will be:

• A. \(\left( \frac{1}{200} \right) \text{s}\)
• B. \(\left( \frac{1}{150} \right) \text{s}\)
• C. \(\left( \frac{1}{100} \right) \text{s}\)
• D. \(\left( \frac{1}{50} \right) \text{s}\)

Hint:

The period is T = 2(pi)/(100(pi)) = 1/50 s. The current falls from its maximum to zero in T/4 = 1/200 s.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The current is \(I = I_0\cos(100\pi t)\). It is largest (equal to \(I_0\)) at \(t = 0\). We need the first time at which it becomes zero.

Step 2: Key Formula or Approach:
Compare with \(I = I_0\cos\omega t\): the angular frequency is \(\omega = 100\pi\) rad/s. The period is \(T = \dfrac{2\pi}{\omega}\).

Step 3: Detailed Explanation:
\[ T = \frac{2\pi}{100\pi} = \frac{1}{50}\ \text{s} \]
The cosine is zero first when its angle is \(\pi/2\):
\[ 100\pi t = \frac{\pi}{2} \quad \Rightarrow \quad t = \frac{1}{200}\ \text{s} \]
This is \(T/4\), a quarter of a period, as we expect, since a cosine falls from its maximum to zero in a quarter cycle.

Step 4: Why the other options are wrong:
Option (B) \(\frac{1}{150}\) s does not match any fixed fraction of the period \(T = \frac{1}{50}\) s.
Option (C) \(\frac{1}{100}\) s is \(T/2\), the time from the maximum value to the minimum value.
Option (D) \(\frac{1}{50}\) s is the full period \(T\), after which the current returns to its maximum.

Final Answer:
The least time is \(\dfrac{1}{200}\) s, which is option (A).
\[ \boxed{t = \frac{1}{200}\ \text{s}} \]