Parallel and series-parallel AC circuits

In a series R-L-C circuit the current is common to every element, so you naturally add impedances. Flip the circuit to parallel branches and voltage becomes the common quantity — adding impedances no longer works, because current, not voltage, is what splits between branches. The fix is to work with the reciprocal of impedance, admittance, which adds directly for parallel branches exactly the way impedance adds for series ones. This is the closing topic of the Single-Phase AC Circuits unit in EE11001, and it leans on everything before it: reactance, the $j$ operator, phasor diagrams and power. The payoff is a single, reliable method — admittance $Y = 1/Z = G + jB$ — for finding branch currents, total current and overall power factor in any parallel or series-parallel AC network, however many branches it has.

The core insight

Parallel AC branches are combined by adding their admittances $Y = 1/Z = G + jB$ (never their impedances), and series-parallel networks are solved by alternating this admittance-addition with ordinary series impedance addition, section by section.

The core idea

For any impedance $Z = R + jX$ , the admittance is its reciprocal, measured in siemens (S):

Y = \frac{1}{Z} = G + jB

where: $G$ is the conductance and $B$ is the susceptance. A common trap is assuming $G = 1/R$ and $B = 1/X$ — that is only true for a purely resistive or purely reactive single-element branch. For a general branch with both $R$ and $X$ , you must rationalise the fraction:

Y = \frac{1}{R+jX} = \frac{R-jX}{R^2+X^2} \;\Rightarrow\; G = \frac{R}{R^2+X^2}, \qquad B = \frac{-X}{R^2+X^2}

Notice the sign flip: an inductive branch has $X > 0$ but contributes negative susceptance $-jB$ to the admittance, while a capacitive branch ( $X<0$ ) contributes positive susceptance $+jB$ . This reversal of sign on taking the reciprocal is the single most common source of errors in this topic.

Why admittance suits parallel circuits. In a parallel circuit the voltage $V$ is common to every branch, and each branch draws its own current $I_k = V Y_k$ . Because all branch currents share the same $V$ , the total admittance of parallel branches simply adds — the direct analogue of impedances adding in series:

Y_{total} = Y_1 + Y_2 + Y_3 + \cdots

The total current then follows as $I_{total} = V\,Y_{total}$ , and the overall power factor of the combination is $\cos\theta = G_{total}/|Y_{total}|$ , the admittance-triangle counterpart of $\cos\theta = R/Z$ for a series circuit.

Series-parallel circuits mix the two techniques in stages. Reduce each parallel section to one equivalent impedance using $Z = 1/Y_{total}$ for that section, then add series impedances in rectangular form as usual, working from the network's innermost sections outward — the same layered strategy used for resistive networks, but carried out with complex numbers throughout. For exactly two parallel branches $Z_1$ and $Z_2$ , a faster shortcut than converting to admittance and back is the product-over-sum rule, which is simply the algebraic rearrangement of $1/Z_{eq} = 1/Z_1 + 1/Z_2$ :

Z_{eq} = \frac{Z_1 Z_2}{Z_1+Z_2}

Figure

Two impedance branches drawn in parallel between a common top and bottom rail, each branch labelled with its impedance and admittance, feeding into a single total current arrow

Derivation

Derivation ATotal admittance of parallel branches

The goal is to show, starting only from Ohm's law in phasor form and Kirchhoff's current law, why admittances of parallel branches add directly to give the total admittance, so that a network of any number of parallel branches can be collapsed to one equivalent admittance and hence one equivalent impedance in a single step.

Step 1. Consider two impedances $Z_1$ and $Z_2$ connected in parallel across a common phasor voltage $V$ (the defining feature of a parallel connection — every branch sees the same $V$ ). By Ohm's law in phasor form, each branch carries its own current $I_1 = V/Z_1$ and $I_2 = V/Z_2$ .

Step 2. Kirchhoff's current law says the total current supplied is the sum of the branch currents: $I_{total} = I_1 + I_2 = V/Z_1 + V/Z_2$ .

Step 3. Factor out the common voltage: $I_{total} = V\left(\dfrac{1}{Z_1} + \dfrac{1}{Z_2}\right) = V\left(Y_1 + Y_2\right)$ , using the definition $Y = 1/Z$ .

Step 4. But $I_{total} = V\,Y_{total}$ by definition of the equivalent admittance of the whole combination, so comparing the two expressions gives $Y_{total} = Y_1 + Y_2$ , and by the same argument extended to any number of branches, $Y_{total} = Y_1+Y_2+Y_3+\cdots$ . The equivalent impedance, if needed, follows as $Z_{total} = 1/Y_{total}$ .

Worked example

Two impedances $Z_1 = 10\angle0^\circ\ \Omega$ (a pure resistor) and $Z_2 = 10\angle90^\circ\ \Omega$ (a pure inductive reactance) are connected in parallel across a 100 V, 50 Hz supply. Find the admittance of each branch, the total admittance, the total current, and the overall power factor.

Step 1. Convert to rectangular form and find each admittance. $Z_1 = 10+j0\ \Omega \Rightarrow Y_1 = 1/10 = 0.1\angle0^\circ = 0.1+j0\ \text{S}$ . $Z_2 = 0+j10\ \Omega \Rightarrow Y_2 = 1/(j10) = -j0.1\ \text{S} = 0.1\angle{-90^\circ}\ \text{S}$ .

Step 2. Add the admittances: $Y_{total} = Y_1+Y_2 = 0.1 - j0.1\ \text{S}$ .

Step 3. Find magnitude and angle: $|Y_{total}| = \sqrt{0.1^2+0.1^2} = \sqrt{0.02} = 0.1414\ \text{S}$ , and $\theta_Y = \tan^{-1}(-0.1/0.1) = -45^\circ$ . Since $I = VY$ , this angle of $Y$ gives the current's angle relative to voltage directly, so the current lags voltage by $45^\circ$ .

Step 4. Compute the total current: $I_{total} = V \times |Y_{total}| = 100 \times 0.1414 = 14.14\ \text{A}$ , lagging the supply voltage by $45^\circ$ .

Step 5. Compute the power factor: $\cos\theta = \cos45^\circ = 0.707$ lagging. As a cross-check, $\cos\theta = G_{total}/|Y_{total}| = 0.1/0.1414 = 0.707$ , which matches.

The total current is 14.14 A at a power factor of 0.707 lagging.

Competitive stretch

Extend the network: put a resistor $R_1=5\ \Omega$ in series with the parallel pair $Z_2 = 8+j6\ \Omega$ and $Z_3 = 8-j6\ \Omega$ . Using product-over-sum, $Z_2 Z_3 = (8+j6)(8-j6) = 64-(j6)^2 = 64+36 = 100$ , and $Z_2+Z_3 = 16+j0$ , so $Z_{23} = 100/16 = 6.25\ \Omega$ — purely real because $Z_2$ and $Z_3$ are complex conjugates and their reactances cancel. Adding the series resistor, $Z_{total} = 5 + 6.25 = 11.25\ \Omega$ , again purely resistive. This conjugate-pair cancellation is a recognisable special case, not a general rule — a generic parallel R-L/R-C mix gives a complex equivalent impedance.

Common traps

Assuming $G = 1/R$ and $B = 1/X$ for a branch that has both resistance and reactance — this shortcut only holds for a purely resistive or purely reactive single-element branch; a combined $R+jX$ branch needs the rationalised form $Y = (R-jX)/(R^2+X^2)$ .
Adding parallel impedances directly as if they were in series, instead of adding admittances or using product-over-sum.
In a series-parallel reduction, forgetting to convert a reduced parallel section's admittance back to an impedance ( $Z=1/Y$ ) before adding it to the series part in rectangular form.
Losing track of the sign of susceptance: an inductive branch's impedance is $+jX$ but its admittance contributes $-jB$ — the sign flips on taking the reciprocal, and forgetting this flips the lag/lead conclusion.

Ask the tutor: if your series-parallel network has three or more nested parallel sections, ask the AI tutor to help you plan the reduction order before diving into the algebra.

Why bother with admittance at all — could I not just use the product-over-sum rule for everything

Product-over-sum works cleanly for exactly two parallel branches, but it gets messy fast with three or more, since you would need repeated pairwise combinations. Admittance sidesteps that: however many branches there are, you just add their admittances in one line, $Y_{total}=Y_1+Y_2+Y_3+\cdots$ , then invert once at the end if you need an impedance. Use product-over-sum as a shortcut for two branches, and admittance addition for anything larger.

Why does the sign of susceptance flip relative to reactance

It comes directly from rationalising $1/(R+jX)$ . Multiplying numerator and denominator by the conjugate $R-jX$ gives $Y = (R-jX)/(R^2+X^2)$ , so the imaginary part is $-X/(R^2+X^2)$ — the negative sign is built into the algebra of division, not a separate physical fact you must memorise blindly. It is worth deriving once so you trust it rather than misremembering it under exam pressure.

In the series-parallel worked example, is it a coincidence that Z23 came out purely real

No, it follows from Z2 and Z3 being complex conjugates of each other, $8+j6$ and $8-j6$ . Whenever two parallel impedances are conjugates, their reactive parts cancel in the sum $Z_2+Z_3$ , leaving a real denominator, and the product $Z_2Z_3 = R^2+X^2$ is also real, so the ratio is real. This is a recognisable special pattern worth spotting quickly, but do not expect it in a generic parallel combination — most series-parallel networks reduce to a complex equivalent impedance.