Discussion Forum



1)

In the circuit shown L= 1μ H, C= 1μ F, and R= 1 kΩ. They are connected in series with an AC source V= V0  sinωt as shown. Which of the following options is/are correct?

 10122019665_circ.PNGω


A) At $\omega\sim0$ the current flowing through the circuit becomes nearly zero

B) The frequency at which the current will be in phase with the voltage is indepedent of R

C) The current will be in phase with the voltage if$\omega =10^{4} rads^{-1}$

D) At $\omega >>10^{6}rads^{-1}$ the circuit behaves like a capacitor

Answer:

Option A,B

Explanation:

At ω =0, $X_{c}=\frac{1}{\omega C}=\infty$  therefore  current  is  nearly zero.

   Further at the resonance frequency, current and voltage are in phase. This resonance frequency is given by

             $\omega_{r}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10^{-6}\times10^{-6}}}=10^{6}rad/sec$

  We can see that this frequency is independent of R

 Further , XL= ωL , XC = $\frac{1}{\omega C}$

  At ω =ωr   = 106 rad/S, XL=XC . 

 For ω >ωr , XL > XC.  So, the circuit is inductive.