A) 25 Hz
B) \[\frac{25}{\pi }\] Hz
C) 50 Hz
D) \[\frac{50}{\pi }\]Hz
Correct Answer: B
Solution :
The Impedance (Z) of an R-L-C series circuit is given by \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] As frequency of alternating emf applied to the circuit is increased, \[{{X}_{L}}\] goes on increasing attd \[{{X}_{C}}\] goes on decreasing. For a particular valtie of \[\omega =({{\omega }_{r}}\,say)\] \[{{X}_{L}}={{X}_{C}}\] i.e., \[{{\omega }_{r}}L=\frac{1}{{{\omega }_{r}}C}\] or \[{{\omega }_{r}}=\frac{1}{\sqrt{LC}}\] or \[2\pi {{v}_{r}}=\frac{1}{\sqrt{LC}}\] or \[{{v}_{r}}=\frac{1}{2\pi \sqrt{LC}}\] \[\therefore \] \[v=\frac{1}{2\times 3.14\times \sqrt{5\times 80\times {{10}^{-6}}}}\] \[=\frac{1}{2\times 3.14\sqrt{(400\times {{10}^{-6}})}}\] \[=\frac{1}{2\times 3.14\times 2\times {{10}^{-2}}}\] \[=\frac{100}{3.14\times 4}\] \[=\frac{25}{3.14}=\frac{25}{\pi }Hz\]You need to login to perform this action.
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