Circuit Analysis

A. Three-phase circuits

Three-Phase Circuits and Power

Wye and delta relationships, balanced three-phase power, and the √3 conventions that underpin nearly every exam calculation.

Three-phase is the native language of power engineering: three sinusoidal sources of equal magnitude displaced

120^\circ

apart deliver constant (non-pulsating) total power, use conductors more efficiently than single-phase, and create the rotating magnetic field that motors depend on. A large share of PE Power problems — machines, faults, transmission, metering — reduce to fluency with two connections (wye and delta) and one formula family (

S = \sqrt{3}\,V_{LL}I_L

).

Phase sequence and notation

With sequence ABC, phasors follow

V_{an} = V\angle 0^\circ

,

V_{bn} = V\angle{-120^\circ}

,

V_{cn} = V\angle{+120^\circ}

. Line-to-line voltages lead their corresponding line-to-neutral voltages by

30^\circ

and are

\sqrt{3}

larger:

V_{ab} = \sqrt{3}\,V_{an}\angle{+30^\circ}

. Keep 'line' quantities (between conductors, through a conductor) distinct from 'phase' quantities (across or through one element of the source/load) — the whole topic is bookkeeping between those two.

Wye connection

In a wye (star), each phase element sits between a line and the neutral point. The line current necessarily equals the phase current, while the line-line voltage gains the

\sqrt{3}

and

30^\circ

:

I_L = I_\phi, \qquad V_{LL} = \sqrt{3}\,V_{LN}

Delta connection

In a delta, each phase element sits directly between two lines, so phase voltage equals line voltage, and the line current is the vector difference of two phase currents:

V_{LL} = V_\phi, \qquad I_L = \sqrt{3}\,I_\phi

Balanced three-phase power

Whether the load is wye or delta, total power in terms of line quantities is the same — this is the workhorse of the exam. The angle

\theta

is always the angle of the load impedance (between phase voltage and phase current), never the

30^\circ

wye-delta shift.

P = \sqrt{3}\,V_{LL} I_L \cos\theta, \quad Q = \sqrt{3}\,V_{LL} I_L \sin\theta, \quad S = \sqrt{3}\,V_{LL} I_L

Delta–wye conversion and per-phase analysis

A balanced delta load

Z_\Delta

is equivalent to a wye load

Z_Y = Z_\Delta/3

. Converting everything to wye lets you analyze one phase against neutral (the per-phase or single-line equivalent), then multiply powers by three. This is also why a delta bank's per-unit impedance equals the wye equivalent's — and the foundation for the one-line diagrams used everywhere downstream.

Z_Y = \frac{Z_\Delta}{3}

Unbalanced systems — a preview

When magnitudes or angles are unbalanced (faults, unbalanced loads), the per-phase shortcut fails: neutral current flows in a four-wire wye,

I_n = I_a + I_b + I_c

, and a three-wire system forces the load neutral to shift. The systematic tool for unbalance is symmetrical components, covered in its own concept — but simple unbalanced wye problems can be done directly with KCL and phasor arithmetic.

Exam strategy

Anchor every problem with

S = \sqrt{3}V_{LL}I_L

and the power triangle. When a problem says '480 V three-phase', that is line-line by convention. Find line current from power ratings first; only then worry about connection-specific phase quantities. For impedance problems, convert delta to wye and work per-phase with

V_{LN} = V_{LL}/\sqrt{3}

.

Key equations

Wye relationships

V_{LL} = \sqrt{3}\,V_{LN}\angle{+30^\circ}, \quad I_L = I_\phi

Line-line voltage leads line-neutral by 30° (ABC sequence).

Delta relationships

V_{LL} = V_\phi, \quad I_L = \sqrt{3}\,I_\phi\angle{-30^\circ}

Line current lags its phase current by 30° (ABC sequence).

Three-phase apparent power

S = \sqrt{3}\,V_{LL} I_L = 3\,V_\phi I_\phi

Same total from line or phase quantities.

Real and reactive power

P = \sqrt{3}\,V_{LL}I_L\cos\theta, \quad Q = \sqrt{3}\,V_{LL}I_L\sin\theta

θ = load impedance angle = angle between phase voltage and phase current.

Power triangle

S = \sqrt{P^2 + Q^2}, \quad pf = \cos\theta = P/S

Lagging = inductive (Q > 0 absorbed).

Delta→wye impedance

Z_Y = Z_\Delta / 3

Balanced loads only.

Per-phase voltage

V_{LN} = V_{LL}/\sqrt{3}

480 V system → 277 V; 208 V → 120 V; 13.8 kV → 7.97 kV.

Worked examples

Line current of a motor load

Problem. A three-phase motor draws 75 kW at 0.85 PF lagging from a 480 V system. Find the line current and the reactive power.

Solution.

S = P/pf = 75/0.85 = 88.2\,\text{kVA}

.

I_L = S/(\sqrt{3}V_{LL}) = 88{,}235/(\sqrt{3}\cdot 480) = 88{,}235/831.4 = 106\,\text{A}

.

\theta = \cos^{-1}0.85 = 31.79^\circ

, so

Q = S\sin\theta = 88.2 \times 0.527 = 46.5\,\text{kvar}

(lagging/absorbed).

I_L = \frac{75{,}000}{0.85\,\sqrt{3}\,(480)} \approx 106\ \text{A}

Delta load, phase and line currents

Problem. A balanced delta load of $Z = 30\angle 25^\circ\,\Omega$ per phase is connected to a 480 V, three-phase source. Find the phase current, line current, and total real power.

Solution. Phase (element) voltage equals line voltage:

V_\phi = 480

V.

I_\phi = 480/30 = 16.0\,\text{A}

;

I_L = \sqrt{3}(16.0) = 27.7\,\text{A}

.

P = \sqrt{3}\,V_{LL}I_L\cos\theta = \sqrt{3}(480)(27.7)\cos 25^\circ = 23{,}030 \times 0.906 = 20.9\,\text{kW}

(equivalently

3V_\phi I_\phi\cos\theta

).

Wye load per-phase analysis

Problem. A wye-connected load of $Z = 12\angle 40^\circ\,\Omega$ per phase is fed from a 208 V three-phase panel. Find the line current and total apparent power.

Solution.

V_{LN} = 208/\sqrt{3} = 120\,\text{V}

.

I_L = I_\phi = 120/12 = 10.0\,\text{A}

.

S = \sqrt{3}(208)(10.0) = 3.60\,\text{kVA}

;

P = 3.60\cos 40^\circ = 2.76\,\text{kW}

,

Q = 3.60\sin 40^\circ = 2.32\,\text{kvar}

.

Common pitfalls

•Counting the phases twice: $S = \sqrt{3}V_{LL}I_L$ already accounts for three phases — do not also multiply by 3. The factor-of-3 form uses phase quantities: $S = 3V_\phi I_\phi$ .
•Using the 30° wye-delta shift as the power-factor angle. θ in the power equations is the load impedance angle only.
•Forgetting that nameplate three-phase voltages are line-line: 480 V means $V_{LN} = 277$ V, not 480 V, in per-phase work.
•Delta loads: dividing line voltage by impedance gives PHASE current; multiply by $\sqrt{3}$ for line current — not the reverse.
•Treating $Z_\Delta$ and $Z_Y$ as interchangeable: a delta bank rewired to wye draws one-third the power at the same line voltage.
•Assuming a neutral exists: three-wire systems have no return path; unbalanced wye loads then shift the neutral point instead of driving neutral current.
•Mixing sequences: with ACB sequence all the ±30° relationships flip sign. The exam states the sequence when it matters — read for it.

References

NCEES PE Power Reference Handbook §3.1.1 — 3-Phase Circuits
NCEES PE Power Reference Handbook §1.3.1 — Algebra of Complex Numbers
Grainger & Stevenson, Power System Analysis — ch. 1–2 background

B. Symmetrical components

Symmetrical Components

Decomposing any unbalanced three-phase set into positive-, negative-, and zero-sequence sets — the tool behind all unbalanced fault analysis.

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C. Per-unit system

The Per-Unit System

Normalizing voltage, current, power, and impedance to common bases — the change-of-base formula and why transformers vanish from the math.

Power systems span voltage levels from 120 V to 765 kV through dozens of transformers. Working in volts and ohms means endlessly re-referring impedances across turns ratios — tedious and error-prone. The per-unit (pu) system normalizes every quantity to a fraction of a chosen base, and with bases chosen to track transformer ratios, the transformers literally disappear from the impedance diagram. Equipment also behaves comparably in pu (a transformer's impedance on its own rating stays in a narrow band — roughly 4–6% at 500 kVA, above 12% at 500 MVA — while its ohms differ by orders of magnitude), so wrong answers look wrong.

Choosing bases

Pick ONE system-wide apparent-power base

S_{base}

(commonly 100 MVA in utility studies, or the rating of a major transformer) and a voltage base

V_{base}

for each voltage zone, stepping the voltage base across every transformer in proportion to its turns ratio. Everything else follows:

I_{base} = \frac{S_{base}}{\sqrt{3}\,V_{LL,base}}, \qquad Z_{base} = \frac{V_{LL,base}^2}{S_{base}}

Per-unit values

A quantity in per-unit is its actual value divided by the base of the same dimension. Note that with three-phase bases (

S

= total three-phase MVA,

V

= line-line kV) the familiar formulas hold without any

\sqrt{3}

or 3 appearing in pu equations — those constants are absorbed into the bases.

X_{pu} = \frac{X_{actual}}{X_{base}}, \qquad V_{pu} = \frac{V}{V_{base}}, \qquad S_{pu} = V_{pu}\,I_{pu}^*

Why transformers vanish

Refer an impedance across an ideal transformer and it scales by the turns ratio squared; but the impedance BASE also scales by the voltage-base ratio squared. If voltage bases follow the transformer ratio, the two effects cancel exactly: a device's per-unit impedance is the same from either side. The whole network becomes one flat impedance diagram — the reason fault studies and power flows are always done in per-unit.

Change of base

Manufacturers state impedances on the equipment's own ratings (e.g., 'Z = 6% on 25 MVA, 69 kV'). To use them on the study's system base, convert — impedance scales directly with the MVA base and inversely with the voltage base squared:

Z_{pu}^{new} = Z_{pu}^{old}\left(\frac{V_{base}^{old}}{V_{base}^{new}}\right)^{2}\left(\frac{S_{base}^{new}}{S_{base}^{old}}\right)

Working a per-unit problem

The standard recipe: (1) choose

S_{base}

and one zone's

V_{base}

; (2) propagate

V_{base}

through every transformer; (3) convert all impedances to the common base with the change-of-base formula; (4) solve the simple circuit in pu (Ohm's law, dividers — no

\sqrt{3}

s); (5) convert answers back: multiply by

I_{base}

,

V_{base}

, or

S_{base}

of the zone where the answer lives.

Exam strategy

Per-unit questions are usually one of three: compute a base quantity (

Z_{base}

,

I_{base}

), execute one change of base, or run a small fault/voltage calculation in pu. The change-of-base voltage ratio trips people: it is

(V_{old}/V_{new})^2

— old on top — because a device looks RELATIVELY bigger on a smaller voltage base. When a transformer's nameplate voltage matches the system base on both sides, the ratio is 1 and only the MVA scaling matters.

Key equations

Base current (3-φ)

I_{base} = \dfrac{S_{base}}{\sqrt{3}\,V_{LL,base}}

S in VA, V in volts → A (or MVA and kV → kA).

Base impedance

Z_{base} = \dfrac{V_{LL,base}^2}{S_{base}} = \dfrac{(\text{kV})^2}{\text{MVA}}\ \Omega

The kV²/MVA shortcut works directly in those units.

Per-unit value

X_{pu} = X_{actual}/X_{base}

Dimensionless; ×100 gives percent.

Change of base

Z_{pu}^{new} = Z_{pu}^{old}\left(\dfrac{V_{old}}{V_{new}}\right)^2 \dfrac{S_{new}}{S_{old}}

Old voltage base in the numerator; impedance grows with a larger MVA base.

Per-unit Ohm's law

V_{pu} = I_{pu} Z_{pu}, \qquad S_{pu} = V_{pu} I_{pu}^*

No √3 or 3 factors in per-unit.

Three-phase fault current

I_{f,pu} = \dfrac{V_{pu}}{Z_{pu}} \;\Rightarrow\; I_f = I_{f,pu} \times I_{base}

The most common per-unit application; see the fault-analysis concept.

Worked examples

Base quantities

Problem. Choose $S_{base} = 100$ MVA at 13.8 kV. Find $Z_{base}$ and $I_{base}$ .

Solution.

Z_{base} = (13.8)^2/100 = 190.44/100 = 1.90\,\Omega

.

I_{base} = 100{,}000/(\sqrt{3} \times 13.8) = 100{,}000/23.9 = 4184\,\text{A} = 4.18\,\text{kA}

.

Z_{base} = \frac{(13.8\,\text{kV})^2}{100\,\text{MVA}} = 1.90\ \Omega

Change of base

Problem. A transformer is rated 25 MVA, 69/13.8 kV with $Z = 7\%$ on its own base. Express Z on a 100 MVA system base (system voltage bases are 69 kV and 13.8 kV, matching the nameplate).

Solution. Voltage bases match the nameplate, so the voltage ratio is 1.

Z_{new} = 0.07 \times (100/25) = 0.28\,\text{pu}

(28%). Same physical transformer, bigger MVA yardstick → larger per-unit number.

Round trip through a transformer

Problem. On a 10 MVA base, a source feeds a 10 MVA, 34.5/4.16 kV transformer ( $X = 0.08$ pu) supplying a load drawing 1.0 pu current at the 4.16 kV bus. With the source at $1.0\angle 0^\circ$ pu, find the 4.16 kV bus voltage in pu and in kV (neglect source impedance; load at 0.9 lagging).

Solution. Current angle:

-\cos^{-1}0.9 = -25.84^\circ

, so

I = 1.0\angle{-25.84^\circ}

pu.

V_{bus} = 1.0\angle 0^\circ - (j0.08)(1.0\angle{-25.84^\circ}) = 1.0 - 0.08\angle 64.16^\circ = 1.0 - (0.0349 + j0.0720)

= 0.9651 - j0.0720 = 0.968\angle{-4.27^\circ}

pu. Actual voltage:

0.968 \times 4.16 = 4.03\,\text{kV}

. The transformer never required referring impedances — that's the per-unit payoff.

Common pitfalls

•Inverting the change-of-base voltage ratio — it is $(V_{old}/V_{new})^2$ with the OLD base on top. When nameplate and system voltages match, skip it entirely.
•Forgetting to propagate the voltage base through transformers: each zone has its own $V_{base}$ and $Z_{base}$ ; using one zone's base in another zone corrupts everything downstream.
•Reinserting $\sqrt{3}$ into per-unit equations. With three-phase bases, $V_{pu} = I_{pu}Z_{pu}$ has no $\sqrt{3}$ — the bases already contain it.
•Mixing single-phase and three-phase bases: $S_{base,1\phi} = S_{base,3\phi}/3$ and $V_{base,LN} = V_{base,LL}/\sqrt{3}$ give the SAME $Z_{base}$ — but only if used consistently.
•Treating percent and per-unit interchangeably mid-calculation: 7% = 0.07 pu; dropping the /100 inflates answers by 100×.
•Using equipment-rated voltage as the base when the system base differs (e.g., a 13.2 kV motor on a 13.8 kV bus) — the voltage-ratio term then matters.
•Converting an answer back with the wrong zone's base current — $I_{base}$ differs on each side of every transformer.

References

NCEES PE Power Reference Handbook §3.1.3 — Per Unit System
Grainger & Stevenson, Power System Analysis — ch. 1; per-unit conventions
IEEE Std 399 (Brown Book) — study practices using per-unit

D. Phasor diagrams

Phasors and Complex Power

Phasor representation of sinusoids, complex-number arithmetic, leading vs lagging, and reading R-L-C behavior from a diagram.

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E. Single-phase circuits

Single-Phase AC Circuits

RMS and average values, instantaneous and average power, series/parallel impedance, and resonance in single-phase AC circuits.

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F. Direct current circuits

DC Circuit Analysis

Ohm's and Kirchhoff's laws, dividers, Thevenin/Norton equivalents, maximum power transfer, and RC/RL transients.

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— Single-line diagrams

Single-Line Diagrams and ANSI Device Numbers

Reading one-line diagrams fluently: standard symbols, ANSI/IEEE device numbers and suffixes, and how a system's protection is encoded.

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