An inductor of inductance L,a capacitor of ca | vedclass

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(D) Given:Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.Potential difference across inductor,$V_{L} = 40 \, V$.Potential difference across capacitor

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$5$

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(D) Given:Amplitude of current,$I_{0} = 10 \sqrt{2} \, A$.Potential difference across inductor,$V_{L} = 40 \, V$.Potential difference across capacitor,$V_{C} = 10 \, V$.Potential difference across resistor,$V_{R} = 40 \, V$.Step $1$: Calculate the $RMS$ current $(I_{RMS})$.$I_{RMS} = \frac{I_{0}}{\sqrt{2}} = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \, A$.Step $2$: Calculate the $RMS$ voltage of the source $(V_{RMS})$.In an $LCR$ series circuit,the total voltage is given by:$V_{RMS} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$$V_{RMS} = \sqrt{(40)^{2} + (40 - 10)^{2}}$$V_{RMS} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, V$.Step $3$: Calculate the impedance $(Z)$ of the circuit.$Z = \frac{V_{RMS}}{I_{RMS}} = \frac{50 \, V}{10 \, A} = 5 \, \Omega$.Therefore,the impedance of the circuit is $5 \, \Omega$.

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