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The current at the half-power point is:

1. \(I_m/\sqrt{2}\)
2. \(\sqrt{2}I_m\)
3. \(2I_m\)
4. \(I_m/2\)

Correct answer: \(I_m/\sqrt{2}\)

Solution

At the half-power point, the power is half the maximum power. Since \(P \propto I^2\), the current must reduce by a factor of \(\sqrt{2}\). Therefore, \(I = I_m/\sqrt{2}\).

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