In the given circuit, \[C=\frac{\sqrt{3}}{2}\mu F,{{R}_{2}}=20\Omega ,\]and\[{{R}_{1}}=10\Omega .\]Current in \[L-{{R}_{1}}\]path is \[{{I}_{1}}\]and in \[C-{{R}_{2}}\] path it is \[{{I}_{2}}.\]The voltage of A.C source is given by, \[V=200\sqrt{2}\sin (100t)\]volts. The phase difference between \[{{I}_{1}}\] and \[{{I}_{2}}\]is
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A) 0
B) \[30{}^\circ \]
C) \[90{}^\circ \]
D) \[60{}^\circ \]
E) None of these
Correct Answer: E
Solution :
For current \[{{I}_{1}}\] \[\tan \phi =\frac{{{X}_{L}}}{{{R}_{1}}}=\frac{\omega L}{{{R}_{1}}}=\frac{100\times \frac{\sqrt{3}}{10}}{10}=\sqrt{3}\] \[\phi ={{60}^{o}};\]V leads \[{{I}_{1}}.\] For current \[{{I}_{2}},\] \[\tan \phi '=\frac{{{X}_{C}}}{{{R}_{2}}}=\frac{1}{\omega C{{R}_{2}}}=\frac{1}{100\times \frac{\sqrt{3}}{2}\times {{10}^{-6}}\times 20}=\frac{1000}{\sqrt{3}}\] \[\phi '\simeq {{90}^{o}};V;ags\,{{I}_{2}}.\] The required phases difference between \[{{I}_{1}}\]and \[{{I}_{2}}\] is\[\phi +\phi '={{60}^{o}}+{{90}^{o}}={{150}^{o}}\] *None of the given options is correct
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