Three-Phase Power | Power Analysis

Tutorial 19 — Three-Phase Systems & Circuits

These problems work through balanced and unbalanced three-phase circuits in both star (Y) and delta (Δ) connections. They cover the line–phase relationships for voltage and current, per-phase and total power, the active–reactive–apparent power triangle, neutral current in a four-wire system, and recovering load \(R\) and \(L\) from power and power-factor data.

Power Analysis · Electric Circuit Analysis · 9 solved problems

Demonstrative VideoWalkthrough

ReferenceStar vs Delta

Comparison of the defining relationships for star- and delta-connected balanced three-phase loads.

	Star (Y) Connection	Delta (Δ) Connection
1	\(I_{L}=I_{ph}\)	\(I_{L}=\sqrt{3}\,I_{ph}\)
2	\(V_{L}=\sqrt{3}\,V_{ph}\)	\(V_{L}=V_{ph}\)
3	Total power \(=\sqrt{3}\,V_{L}I_{L}\cos\phi\)	\(=\sqrt{3}\,V_{L}I_{L}\cos\phi\)
4	Per-phase power \(=V_{ph}I_{ph}\cos\phi\)	\(=V_{ph}I_{ph}\cos\phi\)
5	\(3\phi\) 3/4-wire systems are possible	\(3\phi\) 3-wire system is possible
6	\(V_{L}\) leads the respective \(V_{ph}\) by \(30^{\circ}\)	\(I_{L}\) lags the respective \(I_{ph}\) by \(30^{\circ}\)

Problem 1Delta Load — Currents & Impedance

A three-phase system with a line voltage of 400 V is supplying a delta-connected load of 1500 W at 0.8 pf lagging. Determine the phase and line currents and also the phase impedance.

Solution

The total load is shared equally by the three phases, so the per-phase power is:

\[ P_{ph} = \frac{1500}{3} = 500\ \text{W} \]

For a delta connection the phase voltage equals the line voltage:

\[ V_{ph} = V_L = 400\ \text{V} \]

From the per-phase power, solve for the phase current:

\[ P_{ph} = V_{ph}\,I_{ph}\cos\phi \;\Rightarrow\; 500 = 400 \times I_{ph}\times 0.8 \;\Rightarrow\; I_{ph} = 1.56\ \text{A} \]

For a delta connection the line current is \(\sqrt{3}\) times the phase current:

\[ I_L = \sqrt{3}\,I_{ph} = 1.732 \times 1.56 = 2.71\ \text{A} \]

The phase impedance is the phase voltage divided by the phase current (with angle \(\cos^{-1}0.8 = 36.9^{\circ}\) lagging):

\[ Z_{ph} = \frac{V_{ph}}{I_{ph}} = \frac{400\angle 0^{\circ}}{1.56\angle -36.9^{\circ}} = 256\angle 36.9^{\circ}\ \Omega \]

Answer\(I_{ph} = 1.56\ \text{A},\quad I_L = 2.71\ \text{A},\quad Z_{ph} = 256\angle 36.9^{\circ}\ \Omega\)

Problem 2Star Load — Currents & Impedance

A three-phase system with a line voltage of 400 V supplies 1200 W to a star-connected load at 0.8 pf lagging. Determine the line and phase current and the phase impedance.

Solution

Per-phase power:

\[ P_{ph} = \frac{1200}{3} = 400\ \text{W} \]

For a star connection the phase voltage is the line voltage divided by \(\sqrt{3}\):

\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} = 231\ \text{V} \]

Solve for the phase current (in star, \(I_L = I_{ph}\)):

\[ P_{ph} = V_{ph}\,I_{ph}\cos\phi \;\Rightarrow\; 400 = 231 \times I_{ph}\times 0.8 \;\Rightarrow\; I_{ph} = I_L = 2.16\ \text{A} \]

The impedance angle and the phase impedance:

\[ \theta = \cos^{-1}(0.8) = 36.9^{\circ}\ (\text{lag}),\qquad Z_{ph} = \frac{V_{ph}}{I_{ph}} = \frac{231}{2.16\angle -36.9^{\circ}} = 106.9\angle 36.9^{\circ}\ \Omega \]

Answer\(I_{ph} = I_L = 2.16\ \text{A},\quad Z_{ph} = 106.9\angle 36.9^{\circ}\ \Omega\)

Problem 3Delta R+jX — P, Q, S

A 400 V, three-phase, 50 Hz power supply is applied across the three terminals of a delta-connected three-phase load. The resistance and reactance of each phase are 6 Ω and 8 Ω, respectively. Determine:

the line current and phase current;
the active, reactive, and apparent power of the circuit.

Solution

The load is delta-connected, so the phase voltage equals the line voltage:

\[ V_{ph} = V_L = 400\ \text{V} \]

The per-phase impedance in polar form:

\[ Z_{ph} = R + jX = 6 + j8 = \sqrt{6^2+8^2}\,\angle \tan^{-1}\!\frac{8}{6} = 10\angle 53^{\circ}\ \Omega \]

aPhase current, then line current:

\[ I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{400\angle 0^{\circ}}{10\angle 53^{\circ}} = 40\angle -53^{\circ}\ \text{A},\qquad I_L = \sqrt{3}\,I_{ph} = 1.732 \times 40 = 69.28\ \text{A} \]

The power factor is set by the impedance angle:

\[ \cos\phi = \frac{R}{Z} = \frac{6}{10} = 0.6\ \text{lagging} \;\Rightarrow\; \sin\phi = 0.8 \]

bActive power:

\[ P = \sqrt{3}\,V_L I_L\cos\phi = 1.732 \times 400 \times 69.28 \times 0.6 = 28798\ \text{W} \simeq 28.8\ \text{kW} \]

Reactive power:

\[ Q = \sqrt{3}\,V_L I_L\sin\phi = 1.732 \times 400 \times 69.28 \times 0.8 = 38397\ \text{VAR} \simeq 38.4\ \text{kVAR} \]

Apparent power (and the check via the power triangle):

\[ S = \sqrt{3}\,V_L I_L = 1.732 \times 400 \times 69.28 = 47997\ \text{VA} \simeq 48\ \text{kVA},\qquad S = \sqrt{P^2+Q^2} = \sqrt{28.8^2+38.4^2} = 48\ \text{kVA} \]

Answer\(I_{ph} = 40\ \text{A},\ I_L = 69.28\ \text{A},\ P = 28.8\ \text{kW},\ Q = 38.4\ \text{kVAR},\ S = 48\ \text{kVA}\)

Problem 4Unbalanced 4-Wire — Neutral Current

A three-phase, four-wire supply system has a line voltage of 400 V. Three non-inductive loads of 16 kW, 8 kW and 12 kW are connected between the R, Y and B phases and the neutral, respectively. Calculate the neutral current.

Solution

For a star (four-wire) connection the phase voltage is:

\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} = 231\ \text{V} \]

Taking the three phase voltages 120° apart:

\[ V_R = 231\angle 0^{\circ},\quad V_Y = 231\angle -120^{\circ},\quad V_B = 231\angle -240^{\circ} \]

Each load is purely resistive, so its current is in phase with its own phase voltage. The magnitude is \(P/V_{ph}\) and the angle matches the corresponding voltage:

\[ \begin{aligned} I_R &= \frac{16\times10^3}{231}\angle 0^{\circ} = 69.3\angle 0^{\circ}\ \text{A}\\[2pt] I_Y &= \frac{8\times10^3}{231}\angle -120^{\circ} = 34.6\angle -120^{\circ}\ \text{A}\\[2pt] I_B &= \frac{12\times10^3}{231}\angle -240^{\circ} = 51.95\angle -240^{\circ}\ \text{A} \end{aligned} \]

The neutral carries the phasor sum of the three line currents. Resolving into rectangular components:

\[ \begin{aligned} I_N &= I_R + I_Y + I_B\\ &= 69.3\angle 0^{\circ} + 34.6\angle -120^{\circ} + 51.95\angle 120^{\circ}\\ &= (69.3 - 17.3 - 25.98) + j(0 - 29.97 + 45.0)\\ &= 26.0 + j15.0\ \text{A} \end{aligned} \]

The magnitude of the neutral current:

\[ |I_N| = \sqrt{26.0^2 + 15.0^2} = \sqrt{900} = 30.0\ \text{A} \]

Note: because the load is resistive, each branch current sits at the same angle as its phase voltage (\(V_B = 231\angle -240^{\circ} = 231\angle +120^{\circ}\)). Using the conjugate angle by mistake (e.g. dividing a scalar power by a phasor voltage) flips the sign of the reactive components and inflates \(|I_N|\).

Answer\(I_N = 26.0 + j15.0\ \text{A},\quad |I_N| \approx 30.0\ \text{A}\)

Problem 5Star R = X — Full Analysis

Each phase of a three-phase load has a resistance and a reactance of 6 Ω each and is connected in star. A 400 V, 50 Hz, three-phase supply is connected across the load. Calculate the phase voltage, phase current, power factor, power consumed per phase and the total power consumed by the load.

Solution

Star connection — phase voltage:

\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} = 231\ \text{V} \]

Per-phase impedance:

\[ Z_{ph} = 6 + j6 = 8.48\angle 45^{\circ}\ \Omega \]

Phase current (equal to line current in star):

\[ I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{231\angle 0^{\circ}}{8.48\angle 45^{\circ}} = 27.2\angle -45^{\circ}\ \text{A},\qquad I_L = I_{ph} = 27.2\ \text{A} \]

Power factor:

\[ \text{p.f.} = \cos 45^{\circ} = 0.707\ \text{lagging} \]

Power consumed per phase — computed from \(I^2R\) (equivalently \(V_{ph}I_{ph}\cos\phi\)):

\[ P_{ph} = I_{ph}^2\,R = (27.2)^2 \times 6 = 4439\ \text{W} \approx 4.44\ \text{kW} \]

Total power consumed by the load:

\[ P_{\text{total}} = 3 \times 4439 = 13317\ \text{W} \approx 13.3\ \text{kW} \]

Answer\(V_{ph}=231\ \text{V},\ I_{ph}=27.2\ \text{A},\ \text{p.f.}=0.707,\ P_{ph}\approx4.44\ \text{kW},\ P_{\text{total}}\approx13.3\ \text{kW}\)

Problem 6Star (8+j6) — Line Current & Power

A balanced star-connected load of \((8+j6)\ \Omega\) per phase is connected to a balanced three-phase, 400 V supply. Find the line current, power factor, power and total volt-amperes.

Solution

Phase voltage:

\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} = 231\ \text{V} \]

Impedance magnitude per phase:

\[ Z_{ph} = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = 10\ \Omega \]

Phase current, and line current (equal in star):

\[ I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{231}{10} = 23.1\ \text{A},\qquad I_L = I_{ph} = 23.1\ \text{A} \]

Power factor:

\[ \cos\phi = \frac{R}{Z} = \frac{8}{10} = 0.8\ \text{(lagging)} \]

Total active power and total apparent power:

\[ P = \sqrt{3}\,V_L I_L\cos\phi = \sqrt{3}\times 400\times 23.1\times 0.8 = 12{,}800\ \text{W},\qquad S = \sqrt{3}\,V_L I_L = \sqrt{3}\times 400\times 23.1 = 16{,}000\ \text{VA} \]

Answer\(I_L = 23.1\ \text{A},\ \text{p.f.}=0.8,\ P = 12.8\ \text{kW},\ S = 16\ \text{kVA}\)

Problem 7Recover R and L per Phase

A three-phase star-connected load consumes a total of 12 kW at a power factor of 0.8 lagging when connected to a 400 V, three-phase, 50 Hz power supply. Calculate the resistance and inductance of the load per phase.

Solution

Total and per-phase power:

\[ P_{\text{total}} = 12\ \text{kW},\qquad P_{ph} = \frac{12}{3} = 4\ \text{kW} \]

Phase current from the per-phase power (\(V_{ph} = 400/\sqrt{3} = 231\ \text{V}\)):

\[ V_{ph} I_{ph}\cos\phi = 4000 \;\Rightarrow\; I_{ph} = \frac{4000}{V_{ph}\cos\phi} = \frac{4000}{(400/\sqrt{3})\times 0.8} = 21.6\ \text{A} \]

Per-phase impedance magnitude:

\[ Z_{ph} = \frac{V_{ph}}{I_{ph}} = \frac{231}{21.6} = 10.7\ \Omega \]

With \(\cos\phi = 0.8 \Rightarrow \sin\phi = 0.6\), resolve into resistance and reactance:

\[ R = Z_{ph}\cos\phi = 10.7\times 0.8 = 8.56\ \Omega,\qquad X = Z_{ph}\sin\phi = 10.7\times 0.6 = 6.42\ \Omega \]

Convert the reactance to an inductance at 50 Hz:

\[ X = 2\pi f L \;\Rightarrow\; L = \frac{6.42}{2\pi \times 50} = 20.4\times 10^{-3}\ \text{H} = 20.4\ \text{mH} \]

Answer\(R = 8.56\ \Omega,\quad L = 20.4\ \text{mH}\) per phase

Additional Practice Problems

Two further worked examples with schematics: a balanced delta load given its complex per-phase impedance, and a side-by-side comparison showing how the same impedance behaves when reconnected from star to delta.

Problem 8Delta Load — P and S

A balanced delta-connected load of \((12 + j9)\ \Omega\) per phase is supplied from a 415 V, 50 Hz three-phase system. Find the phase current, line current, power factor, total active power and total apparent power.

Solution

Delta connection — phase voltage equals line voltage:

\[ V_{ph} = V_L = 415\ \text{V} \]

Per-phase impedance in polar form:

\[ Z_{ph} = 12 + j9 = \sqrt{12^2+9^2}\,\angle\tan^{-1}\!\frac{9}{12} = 15\angle 36.9^{\circ}\ \Omega \]

Phase current, then line current:

\[ I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{415}{15} = 27.67\ \text{A},\qquad I_L = \sqrt{3}\,I_{ph} = 1.732 \times 27.67 = 47.9\ \text{A} \]

Power factor from the impedance:

\[ \cos\phi = \frac{R}{Z} = \frac{12}{15} = 0.8\ \text{lagging} \]

Total active power and total apparent power:

\[ P = \sqrt{3}\,V_L I_L\cos\phi = 1.732 \times 415 \times 47.9 \times 0.8 \approx 27.6\ \text{kW},\qquad S = \sqrt{3}\,V_L I_L = 1.732 \times 415 \times 47.9 \approx 34.4\ \text{kVA} \]

Answer\(I_{ph}=27.67\ \text{A},\ I_L=47.9\ \text{A},\ \text{p.f.}=0.8,\ P\approx27.6\ \text{kW},\ S\approx34.4\ \text{kVA}\)

Problem 9Star vs Delta Comparison

A balanced three-phase load has a per-phase impedance of \(20\angle 30^{\circ}\ \Omega\) and is supplied from a 400 V three-phase system. Compare the line current and the total power drawn when the load is connected (a) in star and (b) in delta.

Solution

The impedance magnitude is 20 Ω with \(\cos\phi = \cos 30^{\circ} = 0.866\) in both cases.

aStar: phase voltage \(V_{ph} = 400/\sqrt{3} = 231\ \text{V}\), and \(I_L = I_{ph}\):

\[ I_L = \frac{231}{20} = 11.55\ \text{A},\qquad P_Y = \sqrt{3}\,V_L I_L\cos\phi = 1.732\times 400\times 11.55\times 0.866 \approx 6.93\ \text{kW} \]

bDelta: phase voltage \(V_{ph} = V_L = 400\ \text{V}\), and \(I_L = \sqrt{3}\,I_{ph}\):

\[ I_{ph} = \frac{400}{20} = 20\ \text{A},\quad I_L = \sqrt{3}\times 20 = 34.64\ \text{A},\qquad P_\Delta = \sqrt{3}\,V_L I_L\cos\phi = 1.732\times 400\times 34.64\times 0.866 \approx 20.8\ \text{kW} \]

Comparing the two connections for the same impedance and line voltage:

\[ \frac{I_{L,\Delta}}{I_{L,Y}} = \frac{34.64}{11.55} = 3,\qquad \frac{P_\Delta}{P_Y} = \frac{20.8}{6.93} = 3 \]

Key result: reconnecting the same load from star to delta on the same line voltage triples both the line current and the total power, because each phase now sees \(\sqrt{3}\) times the voltage and carries \(\sqrt{3}\) times the current.

Answer\(\text{Star: } I_L=11.55\ \text{A},\ P=6.93\ \text{kW};\quad \text{Delta: } I_L=34.64\ \text{A},\ P=20.8\ \text{kW}\ (3\times)\)