An A.C. circuit contains a resistance of 12 | vedclass

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(C) Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.In an $L-R$ series circuit,the impedance $Z$ is given by $Z = \sqrt

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{5}{12}ight)$

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(C) Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.In an $L-R$ series circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$.Substituting the values: $Z = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \ \Omega$.The phase angle $\phi$ between the current and the potential difference is given by $	an \phi = \frac{X_L}{R}$.Therefore,$	an \phi = \frac{5}{12}$,which implies $\phi = 	an^{-1}\left(\frac{5}{12}ight)$.Alternatively,using $\cos \phi = \frac{R}{Z} = \frac{12}{13}$,we get $\phi = \cos^{-1}\left(\frac{12}{13}ight)$.Since the provided options in the original question were inconsistent with standard calculations,the correct expression is $	an^{-1}\left(\frac{5}{12}ight)$ or $\cos^{-1}\left(\frac{12}{13}ight)$. Based on the provided choices,option $C$ is the most accurate representation.

An $A.C.$ circuit contains a resistance of $12 \ \Omega$ and an inductive reactance of $5 \ \Omega$. The phase angle between the current and the potential difference will be

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