Basic Electrical & Electronics Engineering (BT-104) - Unit 2 Short Notes

How unit 2 is examined

This unit covers sinusoidal AC quantities, single-phase RLC circuits, power, and balanced three-phase star and delta systems; the marks sit in RLC circuit numericals, line-phase relations, and RMS/form/peak factor definitions.

Generation of sinusoidal AC voltage

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Definition. A coil rotating at constant speed in a uniform magnetic field cuts flux and produces the sinusoidal emf $e = E_m \sin\omega t$.

Key points.

1. Emf is induced by Faraday's law, $e = -N\,d\phi/dt$, and with $\phi=\phi_m\cos\omega t$ it becomes $e = N\omega\phi_m \sin\omega t$.
2. The peak value is $E_m = 2\pi f N \phi_m = NBA\omega$ and the angular speed is $\omega = 2\pi f$ rad/s.
3. One revolution of the coil gives one complete cycle, so frequency is $f = 1/T$ and $f = PN_s/120$ for a P-pole machine.
4. The emf is zero when the coil sides move parallel to the flux and maximum when they cut it at right angles.

Average, RMS, form factor and peak factor

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Definition. <mark>The RMS value of an alternating quantity is that steady DC value which produces the same heat in a given resistance in the same time as the AC does.</mark>

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u2-01" viewBox="0 0 596 424" width="596" height="424" role="img" aria-label="One cycle of v = Vm sin(wt): peak Vm at P, zero at O, Q and S, negative peak at R; RMS = 0.707 Vm, average (half cycle) = 0.637 Vm"><style>#dsfig-u2-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u2-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u2-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u2-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u2-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u2-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u2-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u2-01 .t{fill:#16181D;font-weight:500}#dsfig-u2-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u2-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u2-01 .dot{fill:#16181D}#dsfig-u2-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u2-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u2-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u2-01 .ah{fill:#454C5A}#dsfig-u2-01 .ah.hi{fill:#2340B8}#dsfig-u2-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u2-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u2-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u2-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u2-01 .e{stroke:#B1B7C3}html.dark #dsfig-u2-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u2-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u2-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u2-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u2-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u2-01 .t{fill:#E6E8ED}html.dark #dsfig-u2-01 .t.inv{fill:#0F1115}html.dark #dsfig-u2-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u2-01 .dot{fill:#E6E8ED}html.dark #dsfig-u2-01 .ann{fill:#8FA3FF}html.dark #dsfig-u2-01 .lbl{fill:#858D9C}html.dark #dsfig-u2-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u2-01 .ah{fill:#B1B7C3}html.dark #dsfig-u2-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u2-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u2-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u2-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah5" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh5" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M51.4,196.8 L156.4,56.8" marker-end="url(#ah5)"/><path class="e" d="M180.4,55.2 L285.4,195.2" marker-end="url(#ah5)"/><path class="e" d="M309.4,227.2 L414.4,367.2" marker-end="url(#ah5)"/><path class="e" d="M438.4,368.8 L543.4,228.8" marker-end="url(#ah5)"/><g class="wl"><rect x="94.9" y="117" width="19.2" height="18" rx="9"/><text class="t" x="104.5" y="126" dy=".35em" text-anchor="middle">0</text></g><g class="wl"><rect x="223.9" y="117" width="19.2" height="18" rx="9"/><text class="t" x="233.5" y="126" dy=".35em" text-anchor="middle">0</text></g><g class="wl"><rect x="352.9" y="289" width="19.2" height="18" rx="9"/><text class="t" x="362.5" y="298" dy=".35em" text-anchor="middle">0</text></g><g class="wl"><rect x="481.9" y="289" width="19.2" height="18" rx="9"/><text class="t" x="491.5" y="298" dy=".35em" text-anchor="middle">0</text></g><circle class="n" cx="40" cy="212" r="18"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">O</text><circle class="n" cx="169" cy="40" r="18"/><text class="t" x="169" y="40" dy=".35em" text-anchor="middle">P</text><circle class="n" cx="298" cy="212" r="18"/><text class="t" x="298" y="212" dy=".35em" text-anchor="middle">Q</text><circle class="n" cx="427" cy="384" r="18"/><text class="t" x="427" y="384" dy=".35em" text-anchor="middle">R</text><circle class="n" cx="556" cy="212" r="18"/><text class="t" x="556" y="212" dy=".35em" text-anchor="middle">S</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">One cycle of v = Vm sin(wt): peak Vm at P, zero at O, Q and S, negative peak at R; RMS = 0.707 Vm, average (half cycle) = 0.637 Vm</figcaption></figure>

Formula. $$V_{rms}=\sqrt{\frac{1}{T}\int_0^T v^2\,dt}=\frac{V_m}{\sqrt2}=0.707V_m,\qquad V_{av}=\frac{2}{T}\int_0^{T/2}v\,dt=\frac{2V_m}{\pi}=0.637V_m$$

Key points.

1. Instantaneous value is the value of the quantity at any instant, $v=V_m\sin\omega t$; peak (maximum) value $V_m$ is its greatest value in a cycle.
2. Time period $T$ is the time for one complete cycle, and frequency $f=1/T$ is cycles per second.
3. Average value is the mean of all instantaneous values; over a full sine cycle it is zero, so it is taken over one half cycle.
4. RMS value is the square root of the mean of the squared values and gives the effective (heating) value; meters and the 230 V supply quote RMS.
5. Form factor $=V_{rms}/V_{av}=0.707V_m/0.637V_m=1.11$ for a sine wave.
6. Peak (crest) factor $=V_m/V_{rms}=1.414$ for a sine wave.
7. Alternating quantity is one whose magnitude and direction vary periodically with time.

Answer frame. Open with the heating-effect definition of RMS; draw one sine cycle marking $V_m$, $V_{rms}$, $V_{av}$; then define time period, average, RMS, form factor, peak factor in that order with the integral and sine-wave value each; close with 1.11 and 1.414.

Asked: [7 marks] (Jun 2022, Dec 2023, Jun 2023, Dec 2024, Jun 2025) Define RMS value, form factor, peak factor (with diagram); also time period, average value, alternating quantity, peak value, instantaneous value. Pitfall: Average value of a full sine cycle is zero; always state that $0.637V_m$ is over a half cycle.

Concept of phasor

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Definition. A phasor is a rotating line whose length is the RMS (or peak) value of a sinusoid and whose angle is its phase, so that AC quantities can be added like vectors.

Key points.

1. A sinusoid $V_m\sin(\omega t+\phi)$ is drawn as a line of length $V$ at angle $\phi$ from the reference axis, all rotating counter-clockwise at $\omega$.
2. Only quantities of the same frequency can be shown on one phasor diagram.
3. Rectangular form $V=a+jb$ and polar form $V\angle\phi$ let phasors be added (rectangular) and multiplied or divided (polar).
4. A leading phasor is ahead of the reference in the anticlockwise direction and a lagging phasor is behind it.

Concept of power factor

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Definition. Power factor is the cosine of the phase angle between voltage and current, $\cos\phi = P/S = R/Z$.

Key points.

1. It lies between 0 and 1 and shows what fraction of the apparent power is really used.
2. It is lagging for an inductive load (current behind voltage) and leading for a capacitive load.
3. A low power factor needs more current for the same power, raising $I^2R$ loss and cable size, so it is corrected by parallel capacitors.
4. Pure resistance gives unity; a pure L or C gives zero.

Concept of impedance and admittance

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Definition. Impedance $Z$ is the total opposition to AC current, $Z=V/I=R+jX$ ohm; admittance $Y=1/Z=G+jB$ siemens.

Key points.

1. $|Z|=\sqrt{R^2+X^2}$ and $\phi=\tan^{-1}(X/R)$, where $X=X_L-X_C$ is the net reactance.
2. $X_L=\omega L=2\pi fL$ and $X_C=1/(\omega C)=1/(2\pi fC)$.
3. Conductance $G=R/Z^2$ and susceptance $B=X/Z^2$ give $|Y|=\sqrt{G^2+B^2}$.
4. Impedances in series add; admittances in parallel add.

Active, reactive and apparent power

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Definition. ==Active (real) power $P=VI\cos\phi$ watt is the power actually converted to work or heat; reactive power $Q=VI\sin\phi$ VAR is the power exchanged back and forth with the source; apparent power $S=VI$ VA is their vector sum.==

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u2-02" viewBox="0 0 424 338" width="424" height="338" role="img" aria-label="Power triangle, angle phi at A: P horizontal (active, W), Q vertical (reactive, VAR), S hypotenuse (apparent, VA)"><style>#dsfig-u2-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u2-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u2-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u2-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u2-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u2-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u2-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u2-02 .t{fill:#16181D;font-weight:500}#dsfig-u2-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u2-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u2-02 .dot{fill:#16181D}#dsfig-u2-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u2-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u2-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u2-02 .ah{fill:#454C5A}#dsfig-u2-02 .ah.hi{fill:#2340B8}#dsfig-u2-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u2-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u2-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u2-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u2-02 .e{stroke:#B1B7C3}html.dark #dsfig-u2-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u2-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u2-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u2-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u2-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u2-02 .t{fill:#E6E8ED}html.dark #dsfig-u2-02 .t.inv{fill:#0F1115}html.dark #dsfig-u2-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u2-02 .dot{fill:#E6E8ED}html.dark #dsfig-u2-02 .ann{fill:#8FA3FF}html.dark #dsfig-u2-02 .lbl{fill:#858D9C}html.dark #dsfig-u2-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u2-02 .ah{fill:#B1B7C3}html.dark #dsfig-u2-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u2-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u2-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u2-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah6" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh6" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,298 L365,298"/><path class="e" d="M55.2,286.6 L368.8,51.4"/><path class="e" d="M384,279 L384,59"/><g class="wl"><rect x="181.3" y="289" width="61.5" height="18" rx="9"/><text class="t" x="212" y="298" dy=".35em" text-anchor="middle">P=VIcos</text></g><g class="wl"><rect x="191.6" y="160" width="40.8" height="18" rx="9"/><text class="t" x="212" y="169" dy=".35em" text-anchor="middle">S=VI</text></g><g class="wl"><rect x="353.3" y="160" width="61.5" height="18" rx="9"/><text class="t" x="384" y="169" dy=".35em" text-anchor="middle">Q=VIsin</text></g><circle class="n" cx="40" cy="298" r="18"/><text class="t" x="40" y="298" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="384" cy="298" r="18"/><text class="t" x="384" y="298" dy=".35em" text-anchor="middle">B</text><circle class="n" cx="384" cy="40" r="18"/><text class="t" x="384" y="40" dy=".35em" text-anchor="middle">C</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Power triangle, angle phi at A: P horizontal (active, W), Q vertical (reactive, VAR), S hypotenuse (apparent, VA)</figcaption></figure>

Key points.

1. Active power is dissipated only in resistance, $P=I^2R=VI\cos\phi$, and is measured by a wattmeter.
2. Reactive power is stored and returned by L and C each cycle, $Q=I^2X=VI\sin\phi$; it is taken positive for an inductive load and negative for a capacitive one.
3. Apparent power $S=VI=I^2Z$ is the product of RMS voltage and current and rates generators and transformers in kVA.
4. They form a right triangle, $S^2=P^2+Q^2$, with $\cos\phi=P/S$ as the power factor.
5. In a series RLC circuit $P=I^2R$, $Q=I^2(X_L-X_C)$ and $S=I^2Z$.

Answer frame. Open by naming the three powers and $\phi$; draw the power triangle; define P, Q, S with formula and unit each, then power factor $=P/S$; close with $S^2=P^2+Q^2$.

Asked: [7 marks] (Jun 2022, Dec 2023) Discuss real power, reactive power, apparent power and power factor in an AC circuit (also in a series RLC circuit).

Analysis of R-L, R-C, R-L-C series and parallel circuits

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Definition. ==In a series RLC circuit the same current flows through all elements, and $Z=\sqrt{R^2+(X_L-X_C)^2}$.==

Diagram. Phasor diagram with $I$ as horizontal reference (labels: $V_R$ along $I$, $V_L$ up 90 deg, $V_C$ down 90 deg):

Derivation (series). Let $i=I_m\sin\omega t$. Then $V_R=IR$ (in phase), $V_L=IX_L$ (leads 90 deg), $V_C=IX_C$ (lags 90 deg). Adding phasors: $$V=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}\Rightarrow Z=\sqrt{R^2+(X_L-X_C)^2}$$ $$I=\frac{V}{Z},\quad \phi=\tan^{-1}\frac{X_L-X_C}{R},\quad \cos\phi=\frac{R}{Z},\quad P=VI\cos\phi=I^2R$$

Key points.

1. R-L series: $Z=\sqrt{R^2+X_L^2}$, current lags voltage by $\phi=\tan^{-1}(X_L/R)$, and $\cos\phi=R/Z$.
2. R-C series: $Z=\sqrt{R^2+X_C^2}$ and current leads voltage by $\tan^{-1}(X_C/R)$.
3. Resistance opposes current and turns energy into heat ($P=I^2R$); inductance opposes change of current and stores $\tfrac12LI^2$ in a magnetic field ($v=L\,di/dt$); capacitance opposes change of voltage and stores $\tfrac12CV^2$ in an electric field ($i=C\,dv/dt$).
4. In an inductor current lags voltage by 90 deg; in a capacitor current leads by 90 deg; in a resistor they are in phase.
5. Series resonance occurs at $X_L=X_C$, $f_r=\dfrac{1}{2\pi\sqrt{LC}}$, where $Z=R$ is minimum and current is maximum.
6. Parallel circuit: same voltage across each branch, $I=\sqrt{I_R^2+(I_C-I_L)^2}$, $Y=\sqrt{G^2+(B_C-B_L)^2}$.
7. Power factor is $\cos\phi=R/Z$ in series and $\cos\phi=I_R/I$ in parallel.

Example (Dec 2023, Dec 2024). $R=10\ \Omega$, $L=0.05$ H, $C=100\ \mu$F, 200 V, 50 Hz, series.

Z = 18.97 ohm, I = 10.54 A, pf = 0.527 leading, P = 1111 W.

Example (Jun 2023). $R=10$, $L=0.2$ H, $C=100\ \mu$F, 230 V, 50 Hz: $X_L=62.83$, $X_C=31.83$, $Z=\sqrt{10^2+31^2}=32.57\ \Omega$, $\phi=72.1^\circ$ lagging, $I=7.06$ A. Active component $I\cos\phi=2.17$ A; reactive component $I\sin\phi=6.72$ A. Coil voltage $=I\sqrt{R^2+X_L^2}=7.06\times63.62=449.2$ V. Answer: 2.17 A, 6.72 A, 449.2 V. Phasor diagram: $I$ reference, $V_R=70.6$ V along it, $V_L=443.6$ V up, $V_C=224.7$ V down, $V=230$ V at $72.1^\circ$ ahead.

Example (Jun 2025). $R=4$, $L=10$ mH, 100 V, 50 Hz: $X_L=3.14\ \Omega$, $Z=5.086\ \Omega$, $I=19.66$ A, $V_R=78.6$ V, $V_L=61.8$ V, $\cos\phi=0.786$ lagging, $P=I^2R=1546$ W.

Example (Nov 2022). Resonance: $C=\dfrac{1}{4\pi^2f_r^2L}=\dfrac{1}{4\pi^2(100)^2(0.05)}=50.66\ \mu$F.

Answer frame. Numerical: write the given data, $X_L$, $X_C$, $Z$, $I$, then $\cos\phi=R/Z$ and $P=I^2R$, and state lead or lag. Derivation: draw the series circuit and phasor diagram first, then the $Z$ expression, then $I$, $\phi$, pf and $P$. Explain-type: one paragraph per element with equation, energy stored and phase.

Asked: [7 marks] (Dec 2023, Dec 2024) Choke coil 10 ohm, 0.05 H in series with 100 uF on 200 V, 50 Hz: find impedance, current, power factor, real power (Dec 2024 variant: coil 100 ohm, 0.1 H, 150 uF, 200 V, 50 Hz; find X_L, X_C, Z, I, pf, coil voltage, capacitor voltage). Asked: [7 marks] (Nov 2022) Draw phasor diagrams of series RLC for X_L > X_C, X_L < X_C, X_L = X_C. Asked: [7 marks] (Nov 2022) 220 V, 100 Hz series RLC with coil of 50 mH: find capacitor at resonance at 100 Hz. Asked: [7 marks] (Dec 2023) Derive expressions for impedance, current, power factor and power in a series RLC circuit. Asked: [7 marks] (Jun 2023) Coil 10 ohm, 0.2 H in series with 100 uF on 230 V, 50 Hz: active and reactive components of current, voltage across coil, phasor diagram. Asked: [7 marks] (Jun 2025) 4 ohm resistor with 10 mH inductor on 100 V, 50 Hz: current, voltage across R and L, power factor, real power. Asked: [7 marks] (Jun 2025) Explain the role of resistance, capacitance and inductance in an electric circuit.

Necessity and advantages of three phase systems

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Definition. A three-phase system uses three sinusoidal voltages of equal magnitude, 120 deg apart in phase.

Key points.

1. For the same power, a three-phase machine is smaller, lighter and cheaper than a single-phase one.
2. Three-phase transmission needs less conductor material (75% of the single-phase copper) for the same power and loss.
3. It gives a constant, non-pulsating power and hence smooth torque, and three-phase induction motors are self-starting.
4. It offers two voltage levels (line and phase) and higher efficiency.

Meaning of phase sequence

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Definition. Phase sequence is the order in which the three phase voltages reach their positive maximum, normally R-Y-B.

Key points.

1. In R-Y-B, $V_R$ leads $V_Y$ by 120 deg and $V_Y$ leads $V_B$ by 120 deg.
2. Reversing any two lines changes the sequence to R-B-Y.
3. Sequence decides the direction of rotation of a 3-phase motor, and it must match when alternators are paralleled.

Balanced and unbalanced supply and loads

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Definition. A balanced supply has three equal-magnitude voltages 120 deg apart, and a balanced load has equal impedance in every phase.

Key points.

1. In a balanced system the three line currents are equal, 120 deg apart, and sum to zero, so the star neutral carries no current.
2. In an unbalanced system the phase impedances (or voltages) differ, so currents are unequal and the neutral current is $I_R+I_Y+I_B\ne0$.
3. A balanced load is solved for one phase only; an unbalanced load needs loop or nodal analysis.

Line and phase values, balanced star and delta

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Definition. <mark>Phase values are measured across one winding or load branch; line values are measured between the lines (line voltage) or in the supply wire (line current).</mark>

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u2-03" viewBox="0 0 424 381" width="424" height="381" role="img" aria-label="Star: R, Y, B line terminals, N neutral; phase voltages VR, VY, VB are 120 deg apart, line voltages VRY, VYB, VBR (each a difference of two phase voltages) lead them by 30 deg"><style>#dsfig-u2-03 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u2-03 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u2-03 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u2-03 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u2-03 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u2-03 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u2-03 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u2-03 .t{fill:#16181D;font-weight:500}#dsfig-u2-03 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u2-03 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u2-03 .dot{fill:#16181D}#dsfig-u2-03 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u2-03 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u2-03 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u2-03 .ah{fill:#454C5A}#dsfig-u2-03 .ah.hi{fill:#2340B8}#dsfig-u2-03 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u2-03 .wl .t{font-size:12px;font-weight:700}#dsfig-u2-03 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u2-03 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u2-03 .e{stroke:#B1B7C3}html.dark #dsfig-u2-03 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u2-03 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u2-03 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u2-03 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u2-03 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u2-03 .t{fill:#E6E8ED}html.dark #dsfig-u2-03 .t.inv{fill:#0F1115}html.dark #dsfig-u2-03 .kd{stroke:#E6E8ED}html.dark #dsfig-u2-03 .dot{fill:#E6E8ED}html.dark #dsfig-u2-03 .ann{fill:#8FA3FF}html.dark #dsfig-u2-03 .lbl{fill:#858D9C}html.dark #dsfig-u2-03 .ptr{fill:#8FA3FF}html.dark #dsfig-u2-03 .ah{fill:#B1B7C3}html.dark #dsfig-u2-03 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u2-03 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u2-03 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u2-03 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah7" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh7" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M212,193 L212,59"/><path class="e" d="M196.8,223.4 L55.2,329.6"/><path class="e" d="M227.2,223.4 L368.8,329.6"/><path class="e" d="M202.6,56.5 L49.4,324.5"/><path class="e" d="M59,341 L365,341"/><path class="e" d="M374.6,324.5 L221.4,56.5"/><g class="wl"><rect x="198.8" y="117" width="26.4" height="18" rx="9"/><text class="t" x="212" y="126" dy=".35em" text-anchor="middle">VR</text></g><g class="wl"><rect x="112.8" y="267.5" width="26.4" height="18" rx="9"/><text class="t" x="126" y="276.5" dy=".35em" text-anchor="middle">VY</text></g><g class="wl"><rect x="284.8" y="267.5" width="26.4" height="18" rx="9"/><text class="t" x="298" y="276.5" dy=".35em" text-anchor="middle">VB</text></g><g class="wl"><rect x="109.2" y="181.5" width="33.6" height="18" rx="9"/><text class="t" x="126" y="190.5" dy=".35em" text-anchor="middle">VRY</text></g><g class="wl"><rect x="195.2" y="332" width="33.6" height="18" rx="9"/><text class="t" x="212" y="341" dy=".35em" text-anchor="middle">VYB</text></g><g class="wl"><rect x="281.2" y="181.5" width="33.6" height="18" rx="9"/><text class="t" x="298" y="190.5" dy=".35em" text-anchor="middle">VBR</text></g><circle class="n" cx="212" cy="212" r="18"/><text class="t" x="212" y="212" dy=".35em" text-anchor="middle">N</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">R</text><circle class="n" cx="40" cy="341" r="18"/><text class="t" x="40" y="341" dy=".35em" text-anchor="middle">Y</text><circle class="n" cx="384" cy="341" r="18"/><text class="t" x="384" y="341" dy=".35em" text-anchor="middle">B</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Star: R, Y, B line terminals, N neutral; phase voltages VR, VY, VB are 120 deg apart, line voltages VRY, VYB, VBR (each a difference of two phase voltages) lead them by 30 deg</figcaption></figure>

Derivation (star). Line current equals phase current, $I_L=I_{ph}$, since each line is in series with one phase. Line voltage $V_{RY}=V_R-V_Y$. The angle between $V_R$ and $-V_Y$ is 60 deg, so $$V_{RY}=2V_{ph}\cos30^\circ=\sqrt3V_{ph}$$ so $V_L=\sqrt3V_{ph}$, and $V_L$ leads the corresponding phase voltage by 30 deg.

Derivation (delta). Each phase is connected directly across two lines, so $V_L=V_{ph}$. Applying KCL at node R: $I_R=I_{RY}-I_{BR}$. The angle between $I_{RY}$ and $-I_{BR}$ is 60 deg, so $$I_R=2I_{ph}\cos30^\circ=\sqrt3I_{ph}$$ so $I_L=\sqrt3I_{ph}$, and each line current lags its phase current by 30 deg.

Key points.

1. Phase angle $\phi$ is the angle between phase voltage and phase current, not line quantities.
2. For the same line voltage and load, delta draws three times the line current and three times the power of star, $P_Y=\tfrac13P_\Delta$.
3. Phasor diagram (delta, inductive): phase currents lag their phase voltages by $\phi$, and line currents lag phase currents by 30 deg.

Example (Nov 2022, Dec 2024). Delta, $V_L=110$ V, $Z_{ph}=3.54+j3.54=5\angle45^\circ\ \Omega$. $V_{ph}=110$ V, $I_{ph}=110/5=22$ A (lagging $V_{ph}$ by 45 deg), $I_L=\sqrt3\times22=38.1$ A. Line currents = 38.1 A each, lagging the phase currents by 30 deg.

Example (Jun 2022, capacitive delta). $Z_{ph}=\dfrac{2(-j5)}{2-j5}=1.857\angle-21.8^\circ\ \Omega$, $V_{ph}=440$ V. $I_{ph}=440/1.857=236.9$ A (leading voltage by $21.8^\circ$), $I_L=\sqrt3\times236.9=410.4$ A. I_ph = 236.9 A, I_L = 410.4 A. Phase currents lead their phase voltages, line currents lag phase currents by 30 deg.

Example (Jun 2022, star vs delta). The question gives no power, so assume $Z_{ph}=10\angle36.87^\circ\ \Omega$ (pf 0.8) on 220 V.

Power in star is one third of delta ($P_Y=P_\Delta/3$), whatever the impedance.

Answer frame. Derivations: draw the star (or delta) connection, then the phasor diagram with the 30 deg angle, derive the current relation, then the voltage relation using $2\cos30^\circ$, and close with the boxed result. Numericals: state which quantity equals the phase value, get $|Z_{ph}|$, then $I_{ph}$, then $I_L$, and draw the phasor diagram.

Asked: [7 marks] (Nov 2022, Dec 2024) 3-phase balanced system supplies 110 V to a delta load of $3.54+j3.54$ ohm per phase: find line currents and draw phasor diagram. Asked: [7 marks] (Jun 2022) Power consumed by a 3-phase, 220 V, 50 Hz delta load; find power if connected in star, and phase and line quantities in both cases, pf 0.8 lagging. Asked: [7 marks] (Jun 2022) Draw phasor diagram of a 3-phase star load and find the relations between phase and line voltages and currents. Asked: [7 marks] (Jun 2022) Delta load with 2 ohm in parallel with $-j5$ ohm per phase on 440 V: find phase and line currents and phasor diagram. Asked: [7 marks] (Nov 2022) Draw phasor diagram of a 3-phase delta load and find the relations between phase and line voltages and currents. Pitfall: In a delta, do not divide the line voltage by $\sqrt3$; $V_{ph}=V_L$, and the $\sqrt3$ applies to current.

Power in balanced and unbalanced three-phase systems and measurement

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Definition. Total three-phase power is the sum of the three phase powers, $P=3V_{ph}I_{ph}\cos\phi=\sqrt3V_LI_L\cos\phi$ for a balanced load.

Key points.

1. $Q=\sqrt3V_LI_L\sin\phi$ VAR and $S=\sqrt3V_LI_L$ VA hold for both star and delta.
2. For an unbalanced load, the power of each phase is found separately and added.
3. Power is measured with the two-wattmeter method, $P=W_1+W_2$ and $\tan\phi=\sqrt3(W_1-W_2)/(W_1+W_2)$; a balanced load can use one wattmeter, $P=3W$.

Example (Jun 2023). Star load $8+j6\ \Omega$, 440 V: $V_{ph}=440/\sqrt3=254$ V, $|Z|=10\ \Omega$, $\cos\phi=0.8$. $I_L=I_{ph}=25.4$ A; $P=\sqrt3\cdot440\cdot25.4\cdot0.8=15.49$ kW ($=3I^2R$); $Q=\sqrt3\cdot440\cdot25.4\cdot0.6=11.61$ kVAR. I_L = 25.4 A, P = 15.5 kW, Q = 11.6 kVAR.

Asked: [7 marks] (Jun 2023) Balanced star load $8+6j$ ohm on 3-phase, 50 Hz, 440 V: line current, power absorbed, reactive volt-amperes.

Last-minute revision

• RMS $=0.707V_m$, average $=0.637V_m$, form factor 1.11, peak factor 1.414 (sine).
• $X_L=2\pi fL$, $X_C=1/(2\pi fC)$, $Z=\sqrt{R^2+(X_L-X_C)^2}$, $\cos\phi=R/Z$.
• $X_L>X_C$ gives lagging pf; $X_L<X_C$ gives leading pf; $X_L=X_C$ is resonance, $f_r=1/(2\pi\sqrt{LC})$.
• $P=VI\cos\phi$ W, $Q=VI\sin\phi$ VAR, $S=VI$ VA, $S^2=P^2+Q^2$.
• Star: $V_L=\sqrt3V_{ph}$, $I_L=I_{ph}$; delta: $V_L=V_{ph}$, $I_L=\sqrt3I_{ph}$.
• 3-phase power $=\sqrt3V_LI_L\cos\phi$; $P_Y=\tfrac13P_\Delta$ for the same load and line voltage.
• Phase sequence R-Y-B; phases are 120 deg apart.
• Two-wattmeter method: $P=W_1+W_2$.
• Answers: 10 ohm/0.05 H/100 uF/200 V gives Z = 18.97 ohm, I = 10.54 A, pf 0.527 leading; 50 mH at 100 Hz needs 50.66 uF.

Memory hooks

• "CIVIL": in C, I leads V; in L, V leads I.
• RMS heats, average rectifies: form factor is RMS over average, 1.11.
• Star line voltage is root-3 times phase (voltage up, current same); delta is the opposite.
• ELI the ICE man: inductor voltage before current, capacitor current before voltage.
• Power triangle: P adjacent, Q opposite, S hypotenuse, pf = adjacent over hypotenuse.

Coverage checklist

• Generation of sinusoidal AC voltage: covered (no past questions).
• definition of average value, R.M.S. value, form factor and peak factor of AC quantity: Jun 2022, Dec 2023, Jun 2023, Dec 2024, Jun 2025 definition questions.
• Concept of phasor: covered (no past questions).
• Concept of Power factor: covered (no past questions).
• Concept of impedance and admittance: covered (no past questions).
• Active, reactive and apparent power: Jun 2022, Dec 2023 discussion.
• analysis of R-L, R-C, R-L-C series & parallel circuit: Q Nov 2022 (phasors, resonance), Dec 2023, Dec 2024, Jun 2023, Jun 2025 (two).
• Necessity and advantages of three phase systems: covered (no past questions).
• Meaning of Phase sequence: covered (no past questions).
• balanced and unbalanced supply and loads: covered (no past questions).
• Relationship between line and phase values for balanced star and delta connections: Nov 2022, Dec 2024, Jun 2022 (three).
• Power in balanced & unbalanced three-phase system and their measurements: Jun 2023.