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A.)At $\omega \quad >>{ 10 }^{ 6 }\quad rad.{ r }^{ -1 }$, the circuit behaves like a capacitor

B.)At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero.

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

D.)The frequency at which the current will be in phase with the voltage is independent of R.

Answer
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$Z\quad =\quad R\quad +\quad j\omega L\quad -\dfrac { j }{ \omega C }$

${ \omega }_{ 0 }=\dfrac { 1 }{ \sqrt { LC } }$

$I\quad =\quad \dfrac { V }{ \sqrt { { R }^{ 2 }+\quad { \left( \omega L\quad -\dfrac { 1 }{ \omega C } \right) }^{ 2 } } }$

When L, R and C are connected in series, impedance is given by,

$Z\quad =\quad R\quad +\quad j\omega L\quad -\dfrac { j }{ \omega C }$ …(1)

Resonant frequency in series LCR circuit is given by,

${ \omega }_{ 0 }=\dfrac { 1 }{ \sqrt { LC } }$

Substituting values in above equation we get,

${ \omega }_{ 0 }=\dfrac { 1 }{ \sqrt { { 10 }^{ -6 }\times { 10 }^{ -6 } } }$

$\therefore \quad { \omega }_{ 0 }=\sqrt { { 10 }^{ -12 } }$

$\therefore \quad { \omega }_{ 0 }=\quad { 10 }^{ 6 }rad$

This obtained value is independent of R.

Now, current I in series LCR circuit is given by,

$I\quad =\quad \dfrac { V }{ \sqrt { { R }^{ 2 }+\quad { \left( \omega L\quad -\dfrac { 1 }{ \omega C } \right) }^{ 2 } } }$

At $\omega \sim 0$, above equations become,

$I\quad =\quad \cfrac { V }{ \infty }$

$\therefore \quad I\quad =\quad 0$

Thus, the current flowing through the circuit becomes zero.

At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero and the frequency at which the current will be in phase with the voltage is independent of R respectively.

When current is in phase with voltage, Impedance (I) becomes equal to resistance (R).

Therefore, the equation. (1) becomes,

$j\omega L\quad =\quad \dfrac { j }{ \omega C }$

$\therefore \quad { \omega }^{ 2 }=\quad \dfrac { 1 }{ LC }$

$\therefore \quad { \omega }=\quad \dfrac { 1 }{ \sqrt { LC } }$

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