Question:medium

If the r.m.s. current in an a.c. circuit is \(5\) A, having frequency \(50\) Hz. The value of current \(\frac{1}{300}\) s after its value becomes zero (in A) is
\(\{\sin (\pi /3)=(3)^{0.5}/2\}\)

Show Hint

Write the current as \(I_0\sin\omega t\) with \(I_0=\sqrt{2}\,I_{rms}\) and evaluate it at the given time.
Updated On: Nov 12, 2026
  • \(5\sqrt{2}\)
  • \(5\sqrt{\frac{3}{2}}\)
  • \(\frac{5}{6}\)
  • \(\frac{5}{\sqrt{2}}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Fraction of a cycle
The period is $\dfrac{1}{50}$ s, so $\dfrac{1}{300}$ s is $\dfrac{1}{6}$ of a cycle, i.e. $60^{\circ}$.

Step 2: Evaluate
$I=I_0\sin60^{\circ}=5\sqrt{2}\times\dfrac{\sqrt{3}}{2}=\dfrac{5\sqrt{6}}{2}$.

Step 3: Compare the options
$\dfrac{5\sqrt{6}}{2}=5\sqrt{\dfrac{3}{2}}$. The option $5\sqrt{2}$ is the peak, and $\dfrac{5}{\sqrt{2}}$ would be at $30^{\circ}$.

Final Answer:
The current is 5 times the square root of 3/2 ampere. \[ \boxed{\text{(B) }5\sqrt{\dfrac{3}{2}}} \]
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