MHT CET · Physics · Alternating Current
An A.C. circuit contains resistance of \(12 \Omega\) and inductive reactance \(5 \Omega\). The phase
angle between current and potential difference will be
- A \(\cos ^{-1}\left(\frac{12}{13}\right)\)
- B \(\sin ^{-1}\left(\frac{12}{13}\right)\)
- C \(\cos ^{-1}\left(\frac{5}{12}\right)\)
- D \(\sin ^{-1}\left(\frac{5}{12}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{12}{13}\right)\)
Step-by-step Solution
Detailed explanation
(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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