A series LCR circuit consists of an inductor | vedclass

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(B) At resonance,$\omega L = \frac{1}{\omega C}$,the impedance of the circuit is $Z = R$. Therefore,the current is $I_0 = \frac{\varepsilon_0}{R}$.Whe

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$\sqrt{3} R$

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(B) At resonance,$\omega L = \frac{1}{\omega C}$,the impedance of the circuit is $Z = R$. Therefore,the current is $I_0 = \frac{\varepsilon_0}{R}$.When the angular frequency is changed to $\omega^{\prime}$,the new current is $I^{\prime} = \frac{I_0}{2} = \frac{\varepsilon_0}{2R}$.The impedance at frequency $\omega^{\prime}$ is given by $Z^{\prime} = \sqrt{R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2}$.Since $I^{\prime} = \frac{\varepsilon_0}{Z^{\prime}}$,we have $\frac{\varepsilon_0}{2R} = \frac{\varepsilon_0}{\sqrt{R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2}}$.Squaring both sides,we get $4R^2 = R^2 + \left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2$.Rearranging the terms,$\left(\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right)^2 = 3R^2$.Taking the square root,we find $\left|\omega^{\prime} L - \frac{1}{\omega^{\prime} C}\right| = \sqrt{3} R$.

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