Electro Magnetic Induction And Alternating Currents Question 148

Question: For a series RLC circuit R = XL = 2XC. The impedance of the circuit and phase difference (between) V and i will be

Options:

A) $ \frac{\sqrt{5}R}{2},,{{\tan }^{-1}}(2) $

B) $ \frac{\sqrt{5}R}{2},{{\tan }^{-1}}( \frac{1}{2} ) $

C) $ \sqrt{5}X_{C},{{\tan }^{-1}}(2) $

D) $ \sqrt{5}R,,{{\tan }^{-1}}( \frac{1}{2} ) $

Show Answer

Answer:

Correct Answer: B

Solution:

$ X_{L}=R,\ \ X_{C}=R/2 $
$ \therefore \tan \varphi =\frac{X_{L}-X_{C}}{R}=\frac{R-\frac{R}{2}}{R}=\frac{1}{2} $
$ \Rightarrow \varphi ={{\tan }^{-1}}(1/2) $

Also $ Z=\sqrt{R^{2}+{{(X_{L}-X_{C})}^{2}}}=\sqrt{R^{2}+\frac{R^{2}}{4}}=\frac{\sqrt{5}}{2}R $



NCERT Chapter Video Solution

Dual Pane