An A.C. circuit contains resistance of $12 \Omega$ and inductive reactance $5 \Omega$. The phase angle…

An A.C. circuit contains resistance of $12 \Omega$ and inductive reactance $5 \Omega$. The phase angle between current and potential difference will be
  1. $\cos ^{-1}\left(\frac{12}{13}\right)$
  2. $\sin ^{-1}\left(\frac{12}{13}\right)$
  3. $\cos ^{-1}\left(\frac{5}{12}\right)$
  4. $\sin ^{-1}\left(\frac{5}{12}\right)$

Solution

$\mathrm{R}=12 \Omega, \mathrm{X}=5 \Omega$ $\therefore \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}^{2}}=\sqrt{144+25}=13 \Omega$ $\cos \theta=\frac{\mathrm{R}}{\mathrm{z}}=\frac{12}{13}$ $\therefore \theta=\cos ^{-1} \frac{12}{13}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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