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An A.C. circuit contains resistance of $12 \Omega$ and inductive reactance $5 \Omega$. The phase
angle between current and potential difference will be
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angle between current and potential difference will be
Solution:
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{12}{13}\right)$
(B)
$\mathrm{R}=12 \Omega, \mathrm{X}=5 \Omega$
$\therefore \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}^{2}}=\sqrt{144+25}=13 \Omega$
$\cos \theta=\frac{\mathrm{R}}{\mathrm{z}}=\frac{12}{13}$
$\therefore \theta=\cos ^{-1} \frac{12}{13}$
$\mathrm{R}=12 \Omega, \mathrm{X}=5 \Omega$
$\therefore \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}^{2}}=\sqrt{144+25}=13 \Omega$
$\cos \theta=\frac{\mathrm{R}}{\mathrm{z}}=\frac{12}{13}$
$\therefore \theta=\cos ^{-1} \frac{12}{13}$
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