Alternating Current Test - 55

Consider the parallel resonant circuit shown in adjacent figure. One branch contains an inductor of inductance $$L$$ and a small ohmic resistance $$R$$, whereas the other branch contains a capacitor of capacitance $$C$$. The circuit is fed by a source of alternating emf
$$E={ E }_{ 0 }{ e }^{ i\omega t }={ E }_{ 0 }\sin { \omega t } $$
The impedance of inductor branch, $${ Z }_{ 1 }=R+j\omega L$$
The impedance of capacitor branch, $${ Z }_{ 2 }=\left( 1/j\omega C \right) $$
Net impedance $$Z$$ of the two parallel branches is given by
$$\cfrac { 1 }{ Z } =\cfrac { 1 }{ { Z }_{ 1 } } +\cfrac { 1 }{ { Z }_{ 2 } } =\cfrac { 1 }{ R+j\omega L } +j\omega C=\cfrac { R }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega  }^{ 2 } } +j\omega \left[ C-\cfrac { L }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega  }^{ 2 } }  \right] $$
The current flowing in the circuit
$$I=\cfrac { E }{ Z } =E\left[ \cfrac { R }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega  }^{ 2 } } +j\omega \left[ C-\cfrac { L }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega  }^{ 2 } }  \right]  \right] $$
For resonance to occur, the current must be in phase with the applied emf. For this, the reactive component of current should be zero, ie
$$\omega \left( C-\cfrac { L }{ { R }^{ 2 }+{ L }^{ 2 }{ \omega  }^{ 2 } }  \right) =0$$
This gives resonant angular frequency
$$\quad { \omega  }_{ r }=\sqrt { \cfrac { 1 }{ LC } -\cfrac { { R }^{ 2 } }{ { L }^{ 2 } }  } $$
At parallel circuit resonance, the impedance is maximum and current is minimum. Parallel resonant circuit is sometimes called the anti-resonance in order to distinguish from series resonance.