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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Alternating Current
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An AC Source gives a voltage of $V= 20\; \cos \;(2000 \;t)$. It is connected in a circuit as shown. The voltmeter reading is

$(a)\;4\;V \\ (b)\;4 \sqrt 2\;V \\(c)\;2 \sqrt {3} \\(d)\;3\sqrt {2} V$

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For the given circuit the resonance frequency
$w=\large\frac{1}{\sqrt {LC}} =\frac{1}{\sqrt {5 \times 10^{-3} \times 50 \times 10^{-6}}}$
$\qquad= 2000\;rad/s$
Given AC source has $V=20 \;\cos (2000 t) \;V$
Since the frequency of the ac source is also $2000\;rad/s$ it running under resonance conditions.
$I=\large\frac{V}{R} =\frac{20/\sqrt 2}{6+4}$
$\qquad= \sqrt 2 A$
$V=I \times 4= 4 \sqrt 2 \;V$
Hence b is the correct answer.


answered Mar 25, 2014 by meena.p
edited Sep 24, 2014 by thagee.vedartham

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