IN THE EE SYLLABUS

Balanced three-phase circuits

A balanced three-phase circuit is three equal sinusoidal sources $12 0^{\circ}$ apart feeding identical impedances in every phase, so it can be solved one phase at a time: the other two phases follow by rotating the result through $\pm 12 0^{\circ}$ , and the total power is three times the per-phase power. The topic covers the star and delta line/phase relations with their $3$ and $3 0^{\circ}$ factors, per-phase analysis with star-delta conversion, three-phase active, reactive and apparent power, the two-wattmeter method including its negative reading, and three-phase power-factor correction.

11 previous-year questions · 11 solved · 5 practice questions

Conventions, phase sequence and phasors

A balanced three-phase source is three sinusoidal voltages of the same frequency and the same rms magnitude, displaced by $12 0^{\circ}$ from each other. A balanced three-phase circuit has a balanced source and identical impedances in all three phases (the load, and the line impedances if any). Everything that makes three-phase problems short, namely per-phase analysis, the $3$ relations and zero neutral current, depends on this balance.

Conventions used throughout these notes. Phasors are rms (so the power in one phase is $V I cos ϕ$ with no factor of 1/2). The phase sequence is positive (abc) with $V_{an}$ as the reference unless stated otherwise. The per-phase load impedance is $Z = ∣ Z ∣∠ ϕ$ : $ϕ > 0$ is inductive, i.e. a lagging power factor, and $ϕ$ is the angle by which each phase current lags its own phase voltage. Power is absorbed by the load (passive sign convention). A supply described as "400 V, three-phase" means 400 V rms line-to-line unless the question says otherwise.

BALANCED POSITIVE-SEQUENCE (ABC) SET

V_{an} = V_{p} ∠ 0^{\circ}, V_{bn} = V_{p} ∠ - 12 0^{\circ}, V_{c n} = V_{p} ∠ + 12 0^{\circ}

Phase b lags phase a by $12 0^{\circ}$ and phase c lags phase b by $12 0^{\circ}$ . For negative sequence (acb) the angles of $V_{bn}$ and $V_{c n}$ are interchanged: $V_{bn} = V_{p} ∠ + 12 0^{\circ}$ , $V_{c n} = V_{p} ∠ - 12 0^{\circ}$ .

OPERATOR A AND THE ZERO SUM OF A BALANCED SET

a = 1∠12 0^{\circ} = - \frac{1}{2} + j \frac{3}{2}, 1 + a + a^{2} = 0

For abc: $V_{bn} = a^{2} V_{an}$ and $V_{c n} = a V_{an}$ . Hence any balanced set sums to zero, $V_{an} + V_{bn} + V_{c n} = 0$ , and the same holds for a balanced set of currents.

The phase sequence is the order in which the three phases reach their positive peaks. Phasors rotate counter-clockwise, so in the abc phasor diagram $V_{bn}$ is drawn $12 0^{\circ}$ clockwise from $V_{an}$ , and $V_{c n}$ a further $12 0^{\circ}$ clockwise (which is $12 0^{\circ}$ counter-clockwise of $V_{an}$ ). Interchanging any two supply lines reverses the sequence, and with it the direction of rotation of a three-phase induction motor. The sequence decides the sign of the $3 0^{\circ}$ shifts below and which wattmeter reads more in the two-wattmeter method; it does not change any voltage or current magnitude in a balanced circuit.

EXAM TIP

In a balanced star load the three line currents also form a balanced set and sum to zero, so the neutral wire carries no current, the load neutral sits at the source-neutral potential, and removing the neutral changes nothing. That is exactly why a balanced system can be analysed one phase at a time.

Star and delta: line and phase quantities

In a star (wye) connection each phase sits between a line and the neutral, so the phase current is the line current, while each line voltage is the difference of two phase voltages: $V_{ab} = V_{an} - V_{bn}$ . With the abc set above, $V_{ab} = V_{p} (1 - 1∠ - 12 0^{\circ}) = V_{p} (\frac{3}{2} + j \frac{3}{2}) = 3 V_{p} ∠3 0^{\circ}$ .

In a delta connection each phase sits between two lines, so the phase voltage is the line voltage, while each line current is the difference of two phase currents (KCL at a corner of the delta): $I_{a} = I_{ab} - I_{c a}$ , with $I_{ab}$ taken from line a towards line b, and so on round the delta. For a balanced abc set this gives $I_{a} = I_{ab} (1 - 1∠12 0^{\circ}) = 3 I_{ab} ∠ - 3 0^{\circ}$ .

STAR: LINE AND PHASE RELATIONS (ABC)

V_{ab} = 3 V_{an} ∠ + 3 0^{\circ}, V_{L} = 3 V_{p h}, I_{L} = I_{p h}

Each line voltage leads the phase voltage of its first subscript by $3 0^{\circ}$ ( $V_{b c}$ leads $V_{bn}$ , $V_{c a}$ leads $V_{c n}$ ). For negative sequence the magnitudes are unchanged but $V_{ab} = 3 V_{an} ∠ - 3 0^{\circ}$ .

DELTA: LINE AND PHASE RELATIONS (ABC)

I_{a} = 3 I_{ab} ∠ - 3 0^{\circ}, I_{L} = 3 I_{p h}, V_{L} = V_{p h}

Phase currents $I_{ab}$ , $I_{b c}$ , $I_{c a}$ are taken in the direction of their subscripts; each line current lags, by $3 0^{\circ}$ , the phase current leaving that line's corner ( $I_{b}$ lags $I_{b c}$ , $I_{c}$ lags $I_{c a}$ ). For negative sequence, $I_{a} = 3 I_{ab} ∠ + 3 0^{\circ}$ .

Quantity	Star (Y)	Delta ( $Δ$ )
Phase voltage	$V_{p h} = V_{L} / 3$	$V_{p h} = V_{L}$
Phase current	$I_{p h} = I_{L}$	$I_{p h} = I_{L} / 3$
$3 0^{\circ}$ shift (abc)	$V_{ab}$ leads $V_{an}$ by $3 0^{\circ}$	$I_{a}$ lags $I_{ab}$ by $3 0^{\circ}$
Neutral	Available: three- or four-wire	None: three-wire only
Total active power	$3 V_{L} I_{L} cos ϕ$	$3 V_{L} I_{L} cos ϕ$
Star equivalent for per-phase work	Itself	$Z_{Y} = Z_{Δ} /3$

EXAM TIP

The $3$ factors never change; only the $\pm 3 0^{\circ}$ shifts flip sign with the phase sequence. If a question gives the sequence as acb (negative), redo the angles, not the magnitudes.

Per-phase equivalent and star-delta conversion

Because a balanced circuit is symmetric, the three phases carry equal current and voltage magnitudes displaced by $12 0^{\circ}$ . So you solve one phase of an equivalent star-star circuit, with the source and load neutrals joined by an ideal wire (it carries no current anyway), and write the other two phases by rotating the result through $- 12 0^{\circ}$ and $+ 12 0^{\circ}$ (abc). Any delta, source or load, is first replaced by its star equivalent.

BALANCED STAR-DELTA CONVERSION

Z_{Y} = \frac{Z _{Δ}}{3}, Z_{Δ} = 3 Z_{Y}

Identical impedances only. The division is by a real number, so the impedance angle $ϕ$ , and hence the power factor, is unchanged.

DELTA SOURCE TO EQUIVALENT STAR (ABC)

V_{an} = \frac{V _{ab}}{3} ∠ - 3 0^{\circ}

Replaces a balanced delta-connected source by a star source for per-phase analysis. For negative sequence use $+ 3 0^{\circ}$ .

Per-phase procedure

Convert every balanced delta, source or load, to star: $Z_{Y} = Z_{Δ} /3$ and $V_{an} = (V_{ab} / 3) ∠ - 3 0^{\circ}$ .
Draw phase a only: $V_{an}$ in series with the line impedance $Z_{line}$ and the load $Z_{Y}$ . Leave out any neutral-wire impedance; it carries no current in a balanced system.
Solve $I_{a} = V_{an} / (Z_{line} + Z_{Y})$ . Then $I_{b} = I_{a} ∠ - 12 0^{\circ}$ and $I_{c} = I_{a} ∠ + 12 0^{\circ}$ .
Load voltages: $V_{A N} = I_{a} Z_{Y}$ (star-equivalent phase voltage at the load) and $V_{A B} = 3 V_{A N} ∠3 0^{\circ}$ (line voltage at the load).
If the actual load is delta, its phase current is $I_{A B} = (I_{a} / 3) ∠ + 3 0^{\circ}$ , which also equals $V_{A B} / Z_{Δ}$ .
Multiply per-phase powers by 3 to get totals.

WORKED EXAMPLE · DELTA LOAD FED THROUGH LINE IMPEDANCE

A balanced star-connected source of positive sequence, with $V_{an} = 200∠ 0^{\circ}$ V (rms), feeds a balanced delta-connected load of $Z_{Δ} = (30 + j 39) Ω$ per phase through three lines, each of impedance $(2 + j 3) Ω$ . Find the line current, the phase current of the delta load, the line voltage at the load, the active power absorbed by the load and the power factor at the source.

Solution

Star equivalent of the load: $Z_{Y} = (30 + j 39) /3 = (10 + j 13) Ω = 16.40∠52.4 3^{\circ} Ω$ .
Per-phase series impedance: $Z = (2 + j 3) + (10 + j 13) = (12 + j 16) Ω = 20∠53.1 3^{\circ} Ω$ .
Line current: $I_{a} = \frac{200∠ 0 ^{\circ}}{20∠53.1 3 ^{\circ}} = 10∠ - 53.1 3^{\circ}$ A, so $I_{L} = 10$ A.
Delta phase current: $I_{A B} = (I_{a} / 3) ∠ + 3 0^{\circ} = 5.774∠ - 23.1 3^{\circ}$ A.
Load voltage: $V_{A N} = I_{a} Z_{Y} = (10∠ - 53.1 3^{\circ}) (16.40∠52.4 3^{\circ}) = 164.0∠ - 0.7 0^{\circ}$ V, so the load line voltage is $3 \times 164.0 = 284.1$ V. Check: $∣ I_{A B} ∣∣ Z_{Δ} ∣ = 5.774 \times 49.20 = 284.1$ V.
Load power: $P = 3 I_{L}^{2} R_{Y} = 3 \times 1 0^{2} \times 10 = 3000$ W (equivalently $3 \times 5.77 4^{2} \times 30 = 3000$ W). The lines dissipate $3 \times 1 0^{2} \times 2 = 600$ W.
At the source the per-phase impedance angle is $53.1 3^{\circ}$ , so pf $= cos 53.1 3^{\circ} = 0.6$ lagging. Check: source $P = 3 \times 200 \times 10 \times 0.6 = 3600$ W $= 3000 + 600$ W.

Answer: $I_{L} = 10$ A; delta phase current $= 5.77$ A; load line voltage $\approx 284$ V; load power $= 3.0$ kW; source power factor $= 0.6$ lagging (the load's own power factor is $cos 52.4 3^{\circ} = 0.61$ lagging).

EXAM TIP

Line impedance changes both the magnitude and the angle of the load voltage, so the load line voltage differs from the source line voltage: for an inductive load fed through an inductive line it is smaller, but with a capacitive load it can be larger. The load power factor and the source power factor therefore differ in general. Take each from its own impedance angle: the angle of $Z_{Y}$ for the load, the angle of $Z_{line} + Z_{Y}$ at the source (they coincide only when $Z_{line}$ and $Z_{Y}$ have the same angle). An impedance in the neutral wire of a balanced star-star system has no effect at all.

Three-phase power, and why it is constant

Total power is the sum of the three phase powers. In a balanced load the three are equal, so $P = 3 V_{p h} I_{p h} cos ϕ$ . Substituting the star or the delta relations gives the same line-quantity form, $3 V_{L} I_{L} cos ϕ$ , for both connections. The angle $ϕ$ is always the angle between a phase voltage and its own phase current, i.e. the angle of the per-phase impedance, never the angle between a line voltage and a line current.

THREE-PHASE ACTIVE POWER

P = 3 V_{p h} I_{p h} cos ϕ = 3 V_{L} I_{L} cos ϕ = 3 I_{p h}^{2} R_{p h}

Balanced load, star or delta; rms values; $ϕ$ = angle of the per-phase impedance. Pair the delta resistance with the delta phase current, or the star-equivalent resistance with the line current.

THREE-PHASE REACTIVE POWER

Q = 3 V_{p h} I_{p h} sin ϕ = 3 V_{L} I_{L} sin ϕ = 3 I_{p h}^{2} X_{p h}

$Q > 0$ (absorbed) for an inductive, lagging load; $Q < 0$ for a capacitive, leading load.

COMPLEX AND APPARENT POWER, POWER FACTOR

S = 3 V_{p h} I_{p h}^{*} = P + j Q, ∣ S ∣ = 3 V_{L} I_{L} = P^{2} + Q^{2}, pf = cos ϕ = \frac{P}{∣ S ∣}

$V_{p h}$ and $I_{p h}$ are the phasors of the same phase and $^{*}$ is the complex conjugate. $∣ S ∣$ is in VA (or kVA); a three-phase kVA figure means this total for all three phases, not the per-phase value.

INSTANTANEOUS POWER OF A BALANCED LOAD

p (t) = v_{a} i_{a} + v_{b} i_{b} + v_{c} i_{c} = 3 V_{p h} I_{p h} cos ϕ (constant)

Balanced sinusoidal voltages and a balanced load; $V_{p h}$ , $I_{p h}$ rms. The total does not vary with time and equals the average power $P$ .

Why it is constant: let $V$ and $I$ be the rms phase voltage and current, with $v_{a} = 2 V cos ω t$ and $i_{a} = 2 I cos (ω t - ϕ)$ . Then $v_{a} i_{a} = V I cos ϕ + V I cos (2 ω t - ϕ)$ . Phases b and c give the same constant term plus $V I cos (2 ω t - ϕ - 24 0^{\circ})$ and $V I cos (2 ω t - ϕ + 24 0^{\circ})$ . The three double-frequency terms form a balanced set at $2 ω$ and add to zero, leaving $3 V I cos ϕ$ . Each phase on its own still pulsates at $2 ω$ ; only the total is steady. That is why a balanced three-phase motor can deliver a steady, non-pulsating torque, whereas the power into a single-phase load always pulsates at $2 ω$ .

EXAM TIP

Same three impedances, star versus delta on the same supply: in delta each impedance sees $3$ times the voltage, so it draws $3$ times the phase current, and the line current is $3 \times 3 = 3$ times larger. Hence $P_{Δ} = 3 P_{Y}$ and $I_{L, Δ} = 3 I_{L, Y}$ .

EXAM TIP

One line or one phase open (three identical impedances $Z$ on a balanced three-wire supply). Star, one line open: the other two impedances are in series across one line voltage, so the total power falls to 1/2 of its balanced value. Delta, one line open: one impedance stays across the full line voltage and the other two are in series across that same line voltage, again 1/2. Delta, one phase (one impedance) open: the other two still see the full line voltage, so 2/3. These ratios hold for any identical $Z$ , not only resistors.

EXAM TIP

Trap: in a balanced load (abc, lagging) the angle between $V_{ab}$ and $I_{a}$ is $3 0^{\circ} + ϕ$ , and the angle between $V_{c b}$ and $I_{c}$ is $3 0^{\circ} - ϕ$ . The cosine of either is not the power factor. They are, however, exactly the two wattmeter angles in the next section.

Power measurement: the two-wattmeter method

A wattmeter reads the average of (voltage across its pressure coil) × (current through its current coil): $W = ∣ V ∣∣ I ∣ cos θ$ , where $θ$ is the angle between those two phasors. In a three-wire system the total power can be measured with two wattmeters: current coils in two lines (say a and c), and both pressure coils connected from those lines to the third line (b).

Why it works for any three-wire load: the instantaneous power delivered through the lines is $p = v_{an} i_{a} + v_{bn} i_{b} + v_{c n} i_{c}$ . With no neutral wire, KCL gives $i_{b} = - i_{a} - i_{c}$ , so $p = (v_{an} - v_{bn}) i_{a} + (v_{c n} - v_{bn}) i_{c} = v_{ab} i_{a} + v_{c b} i_{c}$ , which is exactly what the two meters average. The neutral drops out, so nothing here assumes balance, sinusoidal waveforms or an accessible neutral. In general the power of an $n$ -wire system can be measured with $n - 1$ wattmeters by returning all the pressure coils to the one remaining line, so an unbalanced four-wire system needs three.

TWO-WATTMETER SUM (ANY THREE-WIRE LOAD)

P = W_{1} + W_{2}

Readings are taken with their signs. Valid for balanced and unbalanced three-wire loads.

WATTMETER READINGS FOR A BALANCED LOAD

W_{1} = V_{L} I_{L} cos (3 0^{\circ} - ϕ), W_{2} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)

Balanced sinusoidal system. The current coils, in lines a and c, carry the line currents towards the load; the pressure coils measure $V_{ab}$ and $V_{c b}$ (each coil's $\pm$ terminal on its own current-coil line, the other end on line b). For abc sequence the meter in line c reads $W_{1}$ and the meter in line a reads $W_{2}$ . The formulas hold for either sign of $ϕ$ ( $ϕ > 0$ lagging, $ϕ < 0$ leading), so a leading load, or a reversed sequence, swaps which meter reads more.

READINGS IN TERMS OF P AND Q (BALANCED LOAD)

W_{1} = \frac{P}{2} + \frac{Q}{2 3}, W_{2} = \frac{P}{2} - \frac{Q}{2 3}

The same pair written with $P = 3 V_{L} I_{L} cos ϕ$ and $Q = 3 V_{L} I_{L} sin ϕ$ ; handy for predicting the readings when $P$ and $Q$ (or $P$ and the power factor) are given. $W_{2} < 0$ exactly when $Q > 3 P$ , i.e. $ϕ > 6 0^{\circ}$ .

POWER-FACTOR ANGLE AND REACTIVE POWER FROM THE READINGS

tan ϕ = \frac{3 ( W _{1} - W _{2} )}{W _{1} + W _{2}}, Q = 3 (W_{1} - W_{2})

Balanced load only. Follows from $W_{1} + W_{2} = 3 V_{L} I_{L} cos ϕ$ and $W_{1} - W_{2} = V_{L} I_{L} sin ϕ$ . The two readings alone fix $∣ ϕ ∣$ ; whether the load lags or leads needs either the nature of the load or knowledge of which physical meter reads $W_{1}$ , which depends on the connection and the phase sequence.

POWER FACTOR DIRECTLY FROM THE READINGS

cos ϕ = \frac{W _{1} + W _{2}}{2 W _{1}^{2} - W _{1} W _{2} + W _{2}^{2}}

Balanced load; the same content as the $tan ϕ$ form. Insert a negative reading with its sign.

Load pf (lagging)	$ϕ$	$W_{1}$	$W_{2}$	What you see
1	$0^{\circ}$	$P /2$	$P /2$	Equal readings
0.866	$3 0^{\circ}$	$V_{L} I_{L}$	$V_{L} I_{L} /2$	$W_{1} = 2 W_{2}$
0.5	$6 0^{\circ}$	$P$	0	One meter reads zero
Between 0.5 and 0	$6 0^{\circ}$ to $9 0^{\circ}$	Positive	Negative	$P = W_{1} - ∣ W_{2} ∣$
0	$9 0^{\circ}$	$V_{L} I_{L} /2$	$- V_{L} I_{L} /2$	Equal and opposite; $P = 0$

EXAM TIP

The negative reading. Below 0.5 power factor, $W_{2} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)$ is negative and that meter tries to deflect backwards. Its current-coil (or pressure-coil) connections are reversed to get an on-scale reading, and that reading must then be entered as negative: $P = W_{1} + W_{2} = W_{1} - ∣ W_{2} ∣$ . Adding it as positive is the standard trap.

THREE-WATTMETER AND ONE-WATTMETER METHODS

P = W_{a} + W_{b} + W_{c} (four-wire, any load), P = 3 W (balanced load, one meter)

Three-wattmeter method: one meter per line, each pressure coil from its line to the neutral, so each meter reads one phase power; valid for balanced and unbalanced loads. One-wattmeter method: for a balanced load, a single meter with its current coil in one line and its pressure coil from that line to the neutral (or to an artificial neutral formed by three equal impedances, one of them the pressure coil) reads one-third of the total.

REACTIVE POWER WITH ONE WATTMETER (BALANCED LOAD)

Q = 3 W, W = V_{L} I_{L} sin ϕ

Balanced load, abc sequence. Current coil in line a; pressure coil across lines b and c with its $\pm$ terminal on line b, so it measures $V_{b c}$ . Since $V_{b c}$ lags $V_{an}$ by $9 0^{\circ}$ , the meter reads $V_{L} I_{L} cos (9 0^{\circ} - ϕ) = V_{L} I_{L} sin ϕ$ rather than a power; the reading is negative for a leading load.

WORKED EXAMPLE · TWO-WATTMETER READINGS, THEN CORRECTION TO UNITY POWER FACTOR

A balanced three-phase, three-wire inductive load on a 400 V (line, rms), 50 Hz supply is metered by the two-wattmeter method; the readings are 8 kW and 2 kW. Find (a) the total active power, the power factor and the reactive power, (b) the line current, and (c) the capacitance per phase of a delta-connected capacitor bank, connected at the load terminals downstream of the meters, that raises the power factor to unity, together with the wattmeter readings after correction.

Solution

$P = W_{1} + W_{2} = 8 + 2 = 10$ kW.
$tan ϕ = \frac{3 ( 8 - 2 )}{8 + 2} = 1.0392$ , so $ϕ = 46.1 0^{\circ}$ and pf $= cos ϕ = 0.6934$ , lagging because the load is stated to be inductive. Check with the direct form: $\frac{10}{2 64 - 16 + 4} = \frac{10}{14.42} = 0.6934$ .
$Q = 3 (8 - 2) = 10.39$ kvar; $∣ S ∣ = P / cos ϕ = 14.42$ kVA.
$I_{L} = \frac{P}{3 V _{L} cos ϕ} = \frac{10 000}{3 \times 400 \times 0.6934} = 20.82$ A.
For unity pf the bank must supply $Q_{C} = 10.39$ kvar, i.e. $3.464$ kvar per capacitor. Each delta capacitor sees $V_{L} = 400$ V: $C_{Δ} = \frac{3464}{2 π \times 50 \times 40 0 ^{2}} = 68.9 μ F$ . (A star bank would need $3 \times 68.9 = 206.7 μ F$ per phase.)
After correction $P$ is unchanged (ideal capacitors absorb no active power) and $ϕ = 0$ , so the meters read $V_{L} I_{L} cos 3 0^{\circ}$ each, i.e. $P /2 = 5$ kW each; the line current falls to $10 000/ (3 \times 400) = 14.43$ A.

Answer: $P = 10$ kW, pf $= 0.693$ lagging, $Q = 10.39$ kvar; $I_{L} = 20.8$ A; $C_{Δ} = 68.9 μ F$ per phase; after correction both wattmeters read 5 kW.

Power-factor correction in three-phase

Shunt capacitors at the load supply part of the lagging reactive power locally. The active power $P$ is unchanged (ideal capacitors absorb none), while the kVA and the line current fall. The three capacitors of a balanced bank may be connected in star or in delta; the kvar needed is the same, the capacitance is not.

CAPACITOR KVAR REQUIRED

Q_{C} = P (tan ϕ_{1} - tan ϕ_{2})

Total for all three phases; $ϕ_{1}$ is the original angle and $ϕ_{2}$ the target angle, both lagging. For a unity target, $Q_{C}$ equals the original $Q$ .

CAPACITANCE PER PHASE OF THE BANK

C_{Y} = \frac{Q _{C}}{ω V _{L}^{2}}, C_{Δ} = \frac{Q _{C}}{3 ω V _{L}^{2}} = \frac{C _{Y}}{3}

$V_{L}$ = line voltage (rms), $ω = 2 π f$ . Each star capacitor sees $V_{L} / 3$ and supplies $Q_{C} /3 = ω C_{Y} V_{L}^{2} /3$ ; each delta capacitor sees the full $V_{L}$ and supplies $Q_{C} /3 = ω C_{Δ} V_{L}^{2}$ .

EXAM TIP

A delta bank needs one-third the capacitance of a star bank for the same kvar, but each of its capacitors must be rated for the full line voltage instead of $V_{L} / 3$ . Watch whether a question asks for total kvar, kvar per phase, or capacitance per phase.

Unbalanced loads (in brief)

If the three phase impedances differ, the $12 0^{\circ}$ symmetry of the currents is lost and per-phase analysis no longer applies; each phase has to be solved on its own. The source is normally still balanced.

How each case is solved

Star, four-wire (ideal neutral, no line impedance): each phase is independent, $I_{a} = V_{an} / Z_{a}$ , $I_{b} = V_{bn} / Z_{b}$ , $I_{c} = V_{c n} / Z_{c}$ , and the neutral carries $I_{n} = I_{a} + I_{b} + I_{c}$ , which is in general not zero. (A line impedance simply adds in series to its own phase impedance.)
Star, three-wire (no neutral): the load neutral N' shifts away from the source neutral N by $V_{N^{'} N}$ (formula below). Then $I_{a} = (V_{an} - V_{N^{'} N}) / Z_{a}$ , and similarly for b and c; the line currents still add to zero.
Delta with no line impedance: each phase sees a known line voltage, so $I_{A B} = V_{A B} / Z_{A B}$ and so on directly; the line currents follow from KCL, e.g. $I_{a} = I_{A B} - I_{C A}$ .
Power: add the three phase powers. The two-wattmeter sum $W_{1} + W_{2}$ still gives the total active power of a three-wire load, but the $tan ϕ$ and $Q = 3 (W_{1} - W_{2})$ results do not apply.

NEUTRAL-SHIFT VOLTAGE (UNBALANCED STAR, NO NEUTRAL)

V_{N^{'} N} = \frac{Y _{a} V _{an} + Y _{b} V _{bn} + Y _{c} V _{c n}}{Y _{a} + Y _{b} + Y _{c}}, Y_{k} = \frac{1}{Z _{k}}

Nodal equation at N', with the source phase voltages measured from N; any line impedance is included in $Z_{k}$ . It is zero when $Y_{a} = Y_{b} = Y_{c}$ , which recovers the balanced case. If N' and N are joined through a neutral impedance $Z_{n}$ , add $Y_{n} = 1/ Z_{n}$ to the denominator.

EXAM TIP

If the neutral of an unbalanced star load is lost, the load neutral in general shifts ( $V_{N^{'} N} \neq = 0$ ). The load phase voltages $V_{an} - V_{N^{'} N}$ , $V_{bn} - V_{N^{'} N}$ , $V_{c n} - V_{N^{'} N}$ are then unequal, and at least one of them exceeds the normal value $V_{p} = V_{L} / 3$ : because the source phase voltages sum to zero, the squares of the three load phase voltage magnitudes add up to $3 V_{p}^{2} + 3∣ V_{N^{'} N} ∣^{2}$ , which is more than $3 V_{p}^{2}$ .

PREVIOUS-YEAR QUESTIONS

11 questions from the EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2018 · GATE IN 2025 · GATE IN 2022 · GATE IN 2020

GATE 2025 IN Q.44

GATE 2025 · INQ.44 · 2 marks · MCQ

Power consumed by a three-phase balanced load is measured using two-wattmeter method. The per-phase average power drawn by the load is 30 kW at $\frac{3}{2}$ lagging power factor. The readings of the wattmeters will be

(A)
15 kW and 15 kW
(B)
22.5 kW and 7.5 kW
(C)
60 kW and 30 kW
(D)
45 kW and 45 kW

Answer and solution

GATE 2025 IN Q.44: answer and solution

See the question in the paper

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GATE 2025 IN Q.44: answer and solution

GATE 2024 EE Q.50

GATE 2024 · EEQ.50 · 2 marks · NAT

In the circuit shown, $Z_{1} = 50∠ - 9 0^{\circ} Ω$ and $Z_{2} = 200∠ - 3 0^{\circ} Ω$ . It is supplied by a three phase 400 V source with the phase sequence being R-Y-B. Assume the watt meters $W_{1}$ and $W_{2}$ to be ideal. The magnitude of the difference between the readings of $W_{1}$ and $W_{2}$ in watts is ________ (rounded off to 2 decimal places).

GATE 2024 EE question 50: Three supply terminals on the left, top to bottom: R, B, Y (B is drawn in the middle, Y at the bottom). — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

GATE 2024 EE Q.50: answer and solution

GATE 2023 EE Q.39

GATE 2023 · EEQ.39 · 2 marks · MCQ

A 3-phase, star-connected, balanced load is supplied from a 3-phase, 400 V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is $- 400 W$ and the line current is $I_{R} = 2 A$ (rms), then the power factor of the load per phase is

GATE 2023 EE question 39: Left: a box labelled '3 Phase 400 V Balanced Source' with three output lines labelled R (top), Y (middle) and B (bottom). — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
Unity
(B)
0.5 leading
(C)
0.866 leading
(D)
0.707 lagging

Answer and solution

GATE 2023 EE Q.39: answer and solution

GATE 2023 EE Q.64

GATE 2023 · EEQ.64 · 2 marks · NAT

A balanced delta connected load consisting of the series connection of one resistor ( $R = 15 Ω$ ) and a capacitor ( $C = 212.21 μ F$ ) in each phase is connected to three-phase, 50 Hz, 415 V supply terminals through a line having an inductance of $L = 31.83 mH$ per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals $V_{A B}$ in Volts is ________ (Round off to the nearest integer).

GATE 2023 EE question 64: Three supply terminals on the left, labelled '3-phase, 50 Hz, 415 V (r.m.s.) supply terminals'. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

GATE 2023 EE Q.64: answer and solution

GATE 2022 EE Q.21

GATE 2022 · EEQ.21 · 1 mark · MCQ

In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of $1003 V$ with phase sequence $A B C$ . The star connected load has $Z_{A} = 10 Ω$ and $Z_{B} = 20∠6 0^{\circ} Ω$ . The value of $Z_{C}$ in $Ω$ , for which the voltage difference across the nodes $n$ and $n^{'}$ is zero, is

GATE 2022 EE question 21: Left: a star-connected source with common node n (filled dot). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$20∠ - 3 0^{\circ}$
(B)
$20∠3 0^{\circ}$
(C)
$20∠ - 6 0^{\circ}$
(D)
$20∠6 0^{\circ}$

Answer and solution

GATE 2022 EE Q.21: answer and solution

GATE 2022 EE Q.34

GATE 2022 · EEQ.34 · 1 mark · NAT

Two balanced three-phase loads, as shown in the figure, are connected to a $1003$ V, three-phase, 50 Hz main supply. Given $Z_{1} = (18 + j 24) Ω$ and $Z_{2} = (6 + j 8) Ω$ . The ammeter reading, in amperes, is ________. (round off to nearest integer)

GATE 2022 EE question 34: Three supply lines R (top), Y (middle) and B (bottom); — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

GATE 2022 EE Q.34: answer and solution

GATE 2022 IN Q.55

GATE 2022 · INQ.55 · 2 marks · NAT

In the balanced three-phase circuit shown, $C_{0} = 8.2 μ F$ and the line-to-line r.m.s. voltage is $440 V$ at $50 Hz$ . The reading on the wattmeter (in watts) is _______ (round off to two decimal places)

GATE 2022 IN question 55: Three supply lines labelled R (top), Y (middle) and B (bottom) enter from the left. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

GATE 2022 IN Q.55: answer and solution

GATE 2020 IN Q.18

GATE 2020 · INQ.18 · 1 mark · NAT

Three 400 $Ω$ resistors are connected in delta and powered by a 400 V (rms), 50 Hz, balanced, symmetrical R-Y-B sequence, three-phase three-wire mains. The rms value of the line current (in amperes, rounded off to one decimal place) is ______.

Answer and solution

GATE 2020 IN Q.18: answer and solution

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GATE 2020 IN Q.18: answer and solution

GATE 2020 IN Q.34

GATE 2020 · INQ.34 · 2 marks · NAT

In the circuit shown, the rms value of the voltage across the 100 $Ω$ resistor (in volts) is ____.

GATE 2020 IN question 34: On the left, a box labelled "400V Balanced Symmetrical RYB Mains" with three output lines on its right side, labelled R… — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

GATE 2020 IN Q.34: answer and solution

GATE 2020 IN Q.42

GATE 2020 · INQ.42 · 2 marks · NAT

The real power drawn by a balanced load connected to a 400 V, 50 Hz, balanced, symmetrical 3-phase, 3-wire, RYB sequence mains is measured using the two-wattmeter method. Wattmeter $W_{1}$ is connected in the R line and wattmeter $W_{2}$ is connected in the B line. The line current is measured as $\frac{1}{3}$ A. If the wattmeter $W_{1}$ reads zero, the reading on $W_{2}$ (in watts) is ____

Answer and solution

GATE 2020 IN Q.42: answer and solution

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GATE 2020 IN Q.42: answer and solution

GATE 2018 EE Q.47

GATE 2018 · EEQ.47 · 2 marks · NAT

A three-phase load is connected to a three-phase balanced supply as shown in the figure. If $V_{an} = 100∠ 0^{\circ} V$ , $V_{bn} = 100∠ - 12 0^{\circ} V$ and $V_{c n} = 100∠ - 24 0^{\circ} V$ (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is ________ $Ω$ (up to 2 decimal places).

GATE 2018 EE question 47: Four supply terminals on the left, top to bottom: a, n, c, b. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.47: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · MCQ

A balanced star-connected three-phase source has positive (abc) phase sequence and $V_{an} = 230∠ 0^{\circ}$ V (rms). The line voltage $V_{b c}$ is

(A)
$398.4∠ - 9 0^{\circ}$ V
(B)
$398.4∠ + 9 0^{\circ}$ V
(C)
$398.4∠ - 15 0^{\circ}$ V
(D)
$230∠ - 9 0^{\circ}$ V

Answer and solution

Answer: (A)

For abc, $V_{bn} = 230∠ - 12 0^{\circ}$ V and $V_{c n} = 230∠ + 12 0^{\circ}$ V.
$V_{b c} = 3 V_{bn} ∠ + 3 0^{\circ} = 398.4∠ - 9 0^{\circ}$ V.
Direct check: $V_{bn} - V_{c n} = 230 (1∠ - 12 0^{\circ} - 1∠12 0^{\circ}) = 230 (- j 3) = 398.4∠ - 9 0^{\circ}$ V.

PRACTICE 21 mark · MCQ

Three identical impedances connected in star across a balanced three-phase supply absorb a total active power of 1.5 kW. If the same three impedances are reconnected in delta across the same supply, the total active power absorbed is

(A)
0.5 kW
(B)
1.5 kW
(C)
2.6 kW
(D)
4.5 kW

Answer and solution

Answer: (D)

In star each impedance sees $V_{L} / 3$ ; in delta it sees $V_{L}$ .
Power in a fixed impedance is proportional to the square of its voltage, so each impedance absorbs $(3)^{2} = 3$ times more.
$P_{Δ} = 3 \times 1.5 = 4.5$ kW.

PRACTICE 31 mark · NAT

The total power of a balanced three-phase, three-wire load, supplied from a balanced sinusoidal three-phase source, is measured by the two-wattmeter method. One wattmeter reads 4 kW. The other deflects backwards, and after its current-coil connections are reversed it reads 1 kW. The power factor of the load is ______ (round off to three decimal places).

Answer and solution

Answer: $0.327$ (accepted range $\pm 0.005$ )

The reversed meter's reading is negative: $W_{1} = 4$ kW, $W_{2} = - 1$ kW.
$P = W_{1} + W_{2} = 3$ kW and $W_{1} - W_{2} = 5$ kW.
$tan ϕ = \frac{3 \times 5}{3} = 2.887$ , so $ϕ = 70.8 9^{\circ}$ .
pf $= cos 70.8 9^{\circ} = 0.327$ (below 0.5, consistent with one negative reading). Check with the direct form: $\frac{3}{2 16 + 4 + 1} = \frac{3}{2 21} = 0.327$ .

PRACTICE 42 marks · NAT

A balanced star-connected source of positive sequence, with phase voltage 230 V (rms), supplies a balanced delta-connected load of $(27 + j 36) Ω$ per phase through three lines, each of impedance $(3 + j 4) Ω$ . The total active power absorbed by the delta load, in W, is ______ (round off to the nearest integer).

Answer and solution

Answer: $3571$ (accepted range $\pm 5$ )

Star equivalent of the load: $Z_{Y} = (27 + j 36) /3 = (9 + j 12) Ω$ .
Per-phase series impedance: $(3 + j 4) + (9 + j 12) = (12 + j 16) Ω$ , of magnitude $20 Ω$ .
Line current: $I_{L} = 230/20 = 11.5$ A.
Load power: $P = 3 I_{L}^{2} R_{Y} = 3 \times 11. 5^{2} \times 9 = 3570.75$ W $\approx 3571$ W. Check with the delta phase current $11.5/ 3 = 6.640$ A: $3 \times 6.64 0^{2} \times 27 = 3571$ W. The line resistance's $3 \times 11. 5^{2} \times 3 = 1190$ W is not part of the load power.

PRACTICE 52 marks · MSQ

The two-wattmeter method is used in a three-phase, three-wire system with balanced sinusoidal supply voltages. Which of the following statements is/are correct?

(A)
For a balanced load of power factor 0.5 lagging, one wattmeter reads zero.
(B)
For a balanced load of unity power factor, each wattmeter reads half the total active power.
(C)
For a balanced, purely inductive load, both wattmeters read zero.
(D)
The algebraic sum of the two readings equals the total active power only when the load is balanced.

Answer and solution

Answer: (A), (B)

For a balanced load, $W_{1} = V_{L} I_{L} cos (3 0^{\circ} - ϕ)$ and $W_{2} = V_{L} I_{L} cos (3 0^{\circ} + ϕ)$ .
A: $ϕ = 6 0^{\circ}$ gives $W_{2} = V_{L} I_{L} cos 9 0^{\circ} = 0$ . Correct.
B: $ϕ = 0$ gives $W_{1} = W_{2} = V_{L} I_{L} cos 3 0^{\circ}$ , and their sum is $3 V_{L} I_{L} = P$ , so each is $P /2$ . Correct.
C: $ϕ = 9 0^{\circ}$ gives $W_{1} = V_{L} I_{L} cos (- 6 0^{\circ}) = + V_{L} I_{L} /2$ and $W_{2} = V_{L} I_{L} cos 12 0^{\circ} = - V_{L} I_{L} /2$ : equal and opposite, not zero (only their sum is zero). Incorrect.
D: $W_{1} + W_{2} = P$ follows from KCL, $i_{a} + i_{b} + i_{c} = 0$ , alone, so it holds for unbalanced three-wire loads too. Incorrect.

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