logo

Ask Questions, Get Answers

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

A LC circuit has a $10\;V,50\;Hz$ AC sources, A current of $1A$ flows in the circuit. The voltage leads by u phase angle of $\large\frac{\pi}{3}$ radians. The Value of resistance and inductive resistance are

$(a)\;5 \; \Omega,10\;\Omega \\ (b)\;5\;\Omega,5 \sqrt 3 \Omega \\(c)\;3 \sqrt 3 \Omega , 10 \;\Omega \\(d)\;5 \sqrt {3} \Omega, 2 \sqrt 3 \Omega $

Can you answer this question?
 
 

1 Answer

0 votes
$Z= \sqrt {R^2+X_L^2}$
$E= 10\;V$ and $I= 1A$
$\therefore Z=\large\frac{E}{I}=\frac{10}{1} $$=10 \; \Omega$
$\large\frac{X_L}{R}$$= \tan \phi=\tan \large\frac{\pi}{3} $$=\sqrt 3$
$X_L=\sqrt 3 R$
$Z= \sqrt {R^2+(\sqrt 3R)^2}=10$
=> $2R= 10$
$R= 5 \Omega$
$X_L =\sqrt 3 R =\sqrt 3 \times 5$
$\qquad= 5 \sqrt 3 \Omega$
Hence b is the correct answer.
answered Mar 25, 2014 by meena.p
 

Related questions

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