Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Alternating Current
0 votes

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 $

Can you answer this question?

1 Answer

0 votes
$v_{rms} =\sqrt {V_R^2+(V_L -V_L)^2}$
$\qquad= \sqrt {(30)^2+(40)^2}$
$\qquad= 50\;V$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App