Q.Determine the following in an alternating L-C-R circuit :

Parallel AC Circuit Impedance and Current Phasors

When branches are in parallel across an ac source $v = v_m \sin \omega t$, every branch shares the same voltage, so the natural quantities to add are the branch currents, not the branch voltages. The total current is the phasor sum $i = i_1 + i_2$.

Branch Currents

A pure resistor draws $i_1$ in phase with $v$, amplitude $v_m/R$. A series $L$–$C$ branch has net reactance $X = X_L - X_C = \omega L - \dfrac{1}{\omega C}$; it draws $i_2$ of amplitude $v_m/|X|$, lagging $v$ by $90^\circ$ when $X_L > X_C$ (net inductive) or leading by $90^\circ$ when $X_C > X_L$ (net capacitive).

Combining the Branch Currents

Because $i_1$ (along $v$) and $i_2$ (perpendicular to $v$) are $90^\circ$ apart, the resultant amplitude is

and the current makes a phase angle $\phi$ with the voltage given by $\tan\phi = i_2/i_1$.

Impedance

Writing $i_m = v_m/Z$, the admittances add in quadrature:

What Happens Near Resonance …