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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?

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 , X_L= ωL , X_C = $\frac{1}{\omega C}$

At ω =ω_r = 10⁶ rad/S, X_L=X_C .

For ω >ω_r , X_L > X_C.So, the circuit is inductive.

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Explanation: