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 $
