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(A) The power dissipated in an $LCR$ circuit is given by $P = I_{rms}^2 R = \frac{1}{2} I_0^2 R$,where $I_0$ is the peak current.At resonance,the powe

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$\frac{1}{\sqrt{2}}$ times its maximum value.

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(A) The power dissipated in an $LCR$ circuit is given by $P = I_{rms}^2 R = \frac{1}{2} I_0^2 R$,where $I_0$ is the peak current.At resonance,the power dissipated is maximum,given by $P_{max} = \frac{1}{2} I_{max}^2 R$.We want the power to be half of the maximum power: $P = \frac{1}{2} P_{max}$.Substituting the expressions,we get $\frac{1}{2} I^2 R = \frac{1}{2} (\frac{1}{2} I_{max}^2 R)$.This simplifies to $I^2 = \frac{1}{2} I_{max}^2$.Taking the square root of both sides,we get $I = \frac{1}{\sqrt{2}} I_{max}$.Thus,the current amplitude must be $\frac{1}{\sqrt{2}}$ times its maximum value.

In a series resonant $LCR$ circuit,for the power dissipated to become half of the maximum power dissipated,the current amplitude is

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