A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{3}\]
Correct Answer: B
Solution :
The phase angle \[(\theta )\]between I and V is given by\[\tan \theta =\frac{{{X}_{L}}-{{X}_{C}}}{R}\]where,\[{{X}_{L}}=2\pi fL\] ?(i) \[=2\pi \times 50\times \left[ \frac{200}{\pi }\times {{10}^{-3}} \right]\] \[=20\Omega \] \[{{X}_{C}}=\frac{1}{2\pi fC}\] \[=\frac{1\times \pi }{2\pi \times 50\times {{10}^{-3}}}\]\[=10\Omega \] and \[R=10\Omega \] Substituting values of \[{{X}_{L}},{{X}_{C}}\]and R in Eq. (i), we get\[\tan \theta =\frac{20-10}{10}=1\] \[\Rightarrow \] \[\tan \theta =\tan \frac{\pi }{4}\] \[\therefore \] \[\theta =\frac{\pi }{4}\] The phase angle of the circuit is \[\frac{\pi }{4}.\]
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