JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा JEE PYQ-Alternating Current

In a series resonant LCR circuit, .the voltage across R is 100 V and\[R=1k\,\Omega \]. With \[C=2\mu F\] The resonant frequency\[\omega \]is 200 rad/s. At resonance, the voltage across L is [AIEEE 2006]

A) \[2.5\times {{10}^{-2}}V\]

B) \[40\,V\]

C) \[250\,V\]

D) \[4\times {{10}^{-3}}\,V\]

Correct Answer: C

Solution :

[c] Trigger to the mind that at resonance, the net impedance of the circuit is equal to net resistance of the circuit i.e., the net reactance of the circuit is must be zero. Remembering that Ohm's law is also valid for AC circuit,

At resonance, \[\omega L=\frac{1}{\omega C}\]

Current flowing through the circuit,

\[l=\frac{{{V}_{R}}}{R}=\frac{100}{1000}=0.1\,A\]

So, voltage across L is given by

\[{{V}_{L}}=l{{X}_{L}}=l\omega L\]

but\[\omega L=\frac{1}{\omega C}\] (at resistance,\[{{X}_{L}}={{X}_{C}}\])

\[\therefore \] \[{{V}_{L}}=\frac{l}{\omega C}=\frac{0.1}{200\times 2\times {{10}^{-6}}}=250\,V\]

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