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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Alternating Current
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An LCR circuit given below is connected to a variable frequency w and constant peak voltage $V_0$ . at certain $w_0=w_1$ the rms voltage value across L, C and R are respectively $100\;V,60\;V$ and $30\;V$ . At different value of $w_0=w_2$ we find the voltage across resistance is $50\;V$ Then $w_2$ is

$(a)\;\sqrt 3 w_1 \\(b)\;\sqrt {\large\frac{3}{5}}w_1 \\ (c)\;\sqrt 2 w_1 \\ (d)\;w_1 $

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1 Answer

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$v_{rms} =\sqrt {V_R^2+(V_L -V_L)^2}$
$\qquad= \sqrt {(30)^2+(40)^2}$
$\qquad= 50\;V$
Since at $w=w_2$ voltage accross the resistor is $50 V$ this must be resonance condition
$\therefore X_L=X_C$
$w_2L= \large\frac{1}{w_2C}$
$w_2= \large\frac{1}{\sqrt {LC}}$
also at $w=w_1$
$I= \large\frac{100}{X_L}$
$\quad= \large\frac{60}{X_L}$
$C= \large\frac{100}{w_1^2 L \times 60}=\frac{5}{3w_1^2L}$
$w_2=\large\frac{1}{\sqrt { L \times \Large\frac{5}{3 w_1^2 L}}}$
$\quad= \sqrt {\large\frac{3}{5}}$$w_1$
Hence b is the correct answer.
answered Mar 26, 2014 by meena.p
 

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