In the given figure,the current amplitude is

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(A) The given voltage is $V = V_0 \sin(\omega t) + \frac{V_0}{2} \cos(\omega t)$.This can be written as $V = V_{eq} \sin(\omega t + \phi)$,where the a

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$\sqrt{\frac{5V_0^2}{4(\omega^2L^2 + R^2)}}$

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(A) The given voltage is $V = V_0 \sin(\omega t) + \frac{V_0}{2} \cos(\omega t)$.This can be written as $V = V_{eq} \sin(\omega t + \phi)$,where the amplitude $V_{eq} = \sqrt{V_0^2 + (V_0/2)^2} = \sqrt{V_0^2 + V_0^2/4} = \sqrt{\frac{5V_0^2}{4}} = \frac{\sqrt{5}V_0}{2}$.The impedance of the $LR$ circuit is $Z = \sqrt{R^2 + (\omega L)^2}$.The current amplitude $I_0$ is given by $I_0 = \frac{V_{eq}}{Z}$.Substituting the values,$I_0 = \frac{\sqrt{5}V_0 / 2}{\sqrt{R^2 + \omega^2L^2}} = \sqrt{\frac{5V_0^2}{4(R^2 + \omega^2L^2)}}$.Thus,the correct option is $A$.

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