An a.c. circuit contains a resistance of 12 | vedclass

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(D) In an $LR$ series circuit, the phase angle $\phi$ between the current and the potential difference is given by the formula: $	an \phi = \frac{X_L

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$\cos ^{-1}\left(\frac{12}{13}\right)$

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(D) In an $LR$ series circuit, the phase angle $\phi$ between the current and the potential difference is given by the formula: $	an \phi = \frac{X_L}{R}$.Given: Resistance $R = 12 \ \Omega$ and inductive reactance $X_L = 5 \ \Omega$.Substituting the values: $	an \phi = \frac{5}{12}$.This implies $\phi = 	an^{-1}\left(\frac{5}{12}ight)$.To express this in terms of $\sin$ or $\cos$, we consider a right-angled triangle with opposite side $5$ and adjacent side $12$. The hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.Therefore, $\sin \phi = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{5}{13}$, which means $\phi = \sin^{-1}\left(\frac{5}{13}ight)$.Also, $\cos \phi = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{12}{13}$, which means $\phi = \cos^{-1}\left(\frac{12}{13}ight)$.Comparing this with the given options, the correct option is $\cos^{-1}\left(\frac{12}{13}ight)$.

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

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