A circuit has a resistance of 11,Omega,an ind | vedclass

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(D) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Given: $R = 11\,\Omega$,$X_L = 25\,\Omega$,$X_C = 18\,\O

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$20$

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(D) The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Given: $R = 11\,\Omega$,$X_L = 25\,\Omega$,$X_C = 18\,\Omega$.Substituting the values: $Z = \sqrt{11^2 + (25 - 18)^2} = \sqrt{121 + 7^2} = \sqrt{121 + 49} = \sqrt{170}$.Wait,recalculating: $Z = \sqrt{121 + 49} = \sqrt{170} \approx 13.03\,\Omega$. Given the options,let's re-verify the values: $11^2 = 121$,$(25-18)^2 = 7^2 = 49$. $121+49 = 170$. If $R=12$,$Z = \sqrt{144+49} = \sqrt{193}$. If $R=12$ and $X_L-X_C=5$,$Z=13$. Assuming the intended calculation was $Z = \sqrt{12^2 + 5^2} = 13$,but based on the provided values $R=11, X_L=25, X_C=18$,$Z = \sqrt{170} \approx 13.03$. Using $Z \approx 13\,\Omega$,the current $I = V/Z = 260/13 = 20\,A$.

$A$ circuit has a resistance of $11\,\Omega$,an inductive reactance of $25\,\Omega$,and a capacitive reactance of $18\,\Omega$. It is connected to an $AC$ source of $260\,V$ and $50\,Hz$. The current through the circuit (in amperes) is:

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