Q.An ac source $v=v_m\sin\omega t$ drives a circuit having two parallel branches connected across it. One branch contains only a resistor $R$; the other branch contains a capacitor $C$ in series with an inductor $L$. Let $i$ be the total current drawn from the source, $i_1$ the current in the resistor branch, and $i_2$ the current in the series capacitor-inductor branch, so that $i=i_1+i_2$. Find the net current $i$ and its phase relative to the applied voltage, show that $i=v/Z$, and find the impedance $Z$ of this circuit. Take $X_L=\omega L$ and $X_C=1/(\omega C)$.

Set-up

The two branches are in parallel, so each has the full source voltage $v=v_m\sin\omega t$ across it, and the source current is $i=i_1+i_2$.

Branch 1 — the resistor

in phase with $v$ (phase angle $0$).

Branch 2 — the series L-C

The net reactance of the series capacitor and inductor is $X_L-X_C$, so its impedance is the purely imaginary $j(X_L-X_C)$. Hence

i.e. $i_2$ is $90^\circ$ out of phase with $v$ (it lags by $90^\circ$ when $X_L>X_C$, and leads when $X_C>X_L$).

Adding the phasors

Since $i_1$ (along $v$) and $i_2$ (perpendicular to $v$) are at right angles, their resultant amplitude is

Writing this as $i_m=v_m/Z$ identifies the impedance:

So indeed $i=\dfrac{v}{Z}$.

Phase of the net current …

Method: Phasor Analysis of Parallel AC Branches

Use this whenever two or more branches sit in parallel across the same AC source and you need the total current, its phase, and the equivalent impedance.

Steps

Step 1: Note that every branch sees the SAME voltage

In a parallel arrangement, the source voltage $v = v_m\sin\omega t$ appears identically across every branch — unlike a series circuit, where it is the current that is common. This is why the natural quantities to combine here are currents, not voltages.

Step 2: Find each branch's current amplitude and phase separately

Treat each branch as an isolated circuit driven by $v$:

A pure resistor gives a current in phase with $v$. A branch containing only reactive elements (inductor and/or capacitor) gives a current $90^\circ$ out of phase — lagging if the branch is net inductive ($X_L>X_C$), leading if net capacitive ($X_C>X_L$).

Step 3: Combine the branch currents as phasors, not scalars

Because one branch current is along $v$ (phase $0$) and the other is perpendicular to it (phase $\pm90^\circ$), they add like the legs of a right triangle:

$i_m = \sqrt{i_{1m}^2 + i_{2m}^2}, \qquad \tan\phi = \frac{i_{2m}}{i_{1m}}$ …