AC Fundamentals¶

Possible Exam Questions¶

Exam Questions and Answer Map

These are pattern-based predictions, not claimed past questions. For each one, rehearse the answer plan closed-book, then use the links to check the complete answer in this chapter.

Define RMS value, average value, form factor and peak factor of an AC waveform. [5] — [likely]
Answer plan: Write sinusoidal voltage expression \(v(t) = V_m\sin(\omega t + \phi)\) → define and compute each: \(V_{rms} = V_m/\sqrt{2}\), \(V_{avg} = 2V_m/\pi\), form factor = 1.11, peak factor = 1.414 → state that mains 230 V is RMS.
Model answer: RMS, Average, Form Factor and Peak Factor
Explain the behaviour of R, L and C in an AC circuit; define impedance and power factor. [5–10] — [likely]
Answer plan: State V–I phase for each (R in-phase, L lags 90°, C leads 90° — CIVIL) → write impedances \(R\), \(j\omega L\), \(1/j\omega C\) → derive series RLC \(Z = R + j(X_L - X_C)\) → define power factor \(\cos\theta = P/S = R/|Z|\) → draw power triangle (\(P\), \(Q\), \(S\)).
Model answer: R, L and C Behaviour, Impedance and Power Factor
Explain single-phase and three-phase circuits with their benefits and applications. [6] — [PYQ 2082]
Answer plan: Define 1-φ system (230 V, 50 Hz, 2 conductors) → define 3-φ system (three voltages 120° apart, 400 V L-L) → list 6 benefits of 3-φ (constant power, self-starting motors, conductor saving, etc.) → compare in table → give applications.
Model answer: Single-Phase and Three-Phase Circuits
Write the formulas for star-to-delta and delta-to-star conversion. [4] — [PYQ 2082]
Answer plan: Draw both networks → for \(Y\to\Delta\), divide the sum of pairwise star products by the opposite star arm → for \(\Delta\to Y\), divide the product of the two adjacent delta arms by the sum of all delta arms → state balanced shortcuts \(R_\Delta=3R_Y\) and \(R_Y=R_\Delta/3\).
Model answer: Star-Delta Resistance Conversion
Explain series and parallel resonance in AC circuits. [5] — [likely]
Answer plan: State resonance condition \(X_L = X_C\) → derive \(f_0 = 1/(2\pi\sqrt{LC})\) → compare series (min Z, max I, acceptor) vs parallel (max Z, min I, rejector) → define Q factor \(Q = (1/R)\sqrt{L/C} = f_0/BW\) → state applications (tuning, filtering).
Model answer: Series and Parallel Resonance

Model Answer — RMS, Average, Form Factor and Peak Factor [5 marks]¶

Exam-ready answer

For a sinusoidal voltage

\[ v(t)=V_m\sin(\omega t+\phi), \qquad \omega=2\pi f\ \text{rad/s}, \qquad T=\frac{1}{f}\ \text{s}, \]

\(V_m\) is the maximum or peak voltage in volts (V), \(f\) is frequency in hertz (Hz), and \(T\) is period in seconds (s).

RMS value is the DC value that produces the same heat in a resistor as the AC waveform. Its general definition and sinusoidal result are

\[ V_{rms}=\sqrt{\frac{1}{T}\int_0^T v^2(t)\,dt} =V_m\sqrt{\frac{1}{T}\int_0^T\sin^2\omega t\,dt} =\boxed{\frac{V_m}{\sqrt2}}=0.707V_m. \]

The algebraic average of a symmetrical sine wave over a complete cycle is zero. The conventional average value of AC means the mean of the rectified wave, equivalently one positive half-cycle:

\[ V_{avg}=\frac{1}{\pi}\int_0^\pi V_m\sin\theta\,d\theta =\boxed{\frac{2V_m}{\pi}}=0.637V_m. \]

The form factor compares heating value with rectified average, while the peak (crest) factor compares insulation/current stress with RMS value:

\[ \boxed{k_f=\frac{V_{rms}}{V_{avg}}=\frac{\pi}{2\sqrt2}=1.11}, \qquad \boxed{k_p=\frac{V_m}{V_{rms}}=\sqrt2=1.414}. \]

Fig: Sinusoidal cycle with peak, period, RMS level, rectified average and defining integrals

Quantity	Meaning	Sinusoidal value
RMS	Equal-heating DC equivalent	\(0.707V_m\)
Rectified average	Mean magnitude over a cycle	\(0.637V_m\)
Form factor	Wave-shape/heating ratio	\(1.11\)
Peak factor	Peak stress relative to rating	\(1.414\)

Numerical check: a stated \(230\,\text{V}\) mains supply is RMS. Therefore \(V_m=\sqrt2(230)=325.3\,\text{V}\) and \(V_{avg}=2(325.3)/\pi=207.1\,\text{V}\). The checks \(230/207.1=1.11\) and \(325.3/230=1.414\) reproduce the two factors.

Thus RMS is used for equipment voltage, current and power ratings; average value is important in rectifier measurements; and peak factor determines insulation and device peak-voltage requirements.

Practice target: 8–9 minutes; derive both integrals and clearly state that full-cycle algebraic average is zero.

Model Answer — R, L and C Behaviour, Impedance and Power Factor [10 marks]¶

Exam-ready answer

In sinusoidal steady state, a phasor represents the magnitude and phase of a sinusoid at angular frequency \(\omega=2\pi f\,\text{rad/s}\). Impedance is the complex ratio

\[ \boxed{Z=\frac{\underline V}{\underline I}}\ \Omega, \]

so it combines resistance, which consumes real power, and reactance, which alternately stores and returns energy.

Pure elements¶

For a resistor, \(v=Ri\). If \(i=I_m\sin\omega t\), then \(v=RI_m\sin\omega t\); voltage and current are in phase and

\[ \boxed{Z_R=R}\ \Omega. \]

For an inductor, \(v=L\,di/dt=\omega LI_m\cos\omega t\), so voltage leads current by \(90^\circ\) and current lags voltage by \(90^\circ\):

\[ \boxed{Z_L=j\omega L=jX_L}, \qquad X_L=2\pi fL\ \Omega. \]

For a capacitor, \(i=C\,dv/dt\). Therefore current leads voltage by \(90^\circ\):

\[ \boxed{Z_C=\frac{1}{j\omega C}=-\frac{j}{\omega C}=-jX_C}, \qquad X_C=\frac{1}{2\pi fC}\ \Omega. \]

Fig: R, L and C circuits with voltage-current phasors and series-RLC voltage and impedance triangles

Element	Phase relation	Average power	Energy behaviour
\(R\)	\(V\) and \(I\) in phase	\(P=VI=I^2R\) W	Dissipates heat
Ideal \(L\)	\(I\) lags \(V\) by \(90^\circ\)	\(P=0\) W, \(Q=+I^2X_L\) VAR	Stores magnetic energy \(Li^2/2\) J
Ideal \(C\)	\(I\) leads \(V\) by \(90^\circ\)	\(P=0\) W, \(Q=-I^2X_C\) VAR	Stores electric energy \(Cv^2/2\) J

The memory rule is CIVIL: in a Capacitor, I leads V; V leads I in an Inductor.

Series RLC derivation¶

Using current as the reference phasor, \(\underline V_R=IR\), \(\underline V_L=jIX_L\), and \(\underline V_C=-jIX_C\). KVL gives

\[ \underline V=\underline V_R+\underline V_L+\underline V_C =\underline I\,[R+j(X_L-X_C)], \]

hence

\[ \boxed{Z=R+j(X_L-X_C)}, \quad \boxed{|Z|=\sqrt{R^2+(X_L-X_C)^2}}, \quad \boxed{\theta=\tan^{-1}\!\frac{X_L-X_C}{R}}. \]

The circuit is inductive and current lags when \(X_L>X_C\); it is capacitive and current leads when \(X_C>X_L\).

Power factor is the cosine of the phase angle between RMS voltage and current:

\[ \boxed{\mathrm{pf}=\cos\theta=\frac{P}{S}=\frac{R}{|Z|}}. \]

Real, reactive and apparent power are

\[ \boxed{P=VI\cos\theta}\ \text{W},\qquad \boxed{Q=VI\sin\theta}\ \text{VAR},\qquad \boxed{S=VI}\ \text{VA}, \]

with \(S^2=P^2+Q^2\). Positive \(Q\) is lagging/inductive and negative \(Q\) is leading/capacitive.

Fig: Lagging and leading power triangles with P, Q, S, phase angle and capacitor correction

Worked example: let \(R=30\,\Omega\), \(L=0.1\,\text{H}\) and \(C=100\,\mu\text{F}\) be connected to \(230\,\text{V}\), \(50\,\text{Hz}\). Then

\[ X_L=2\pi(50)(0.1)=31.42\,\Omega, \qquad X_C=\frac{1}{2\pi(50)(100\times10^{-6})}=31.83\,\Omega. \]

Thus \(Z=30-j0.41\,\Omega\), \(|Z|\approx30.003\,\Omega\), and \(I=230/30.003=7.67\,\text{A}\). The angle is about \(-0.79^\circ\), so pf \(\approx0.9999\) leading. As a check, \(P=I^2R\approx1.763\,\text{kW}\), \(Q=I^2(X_L-X_C)\approx-24\,\text{VAR}\), and \(S=VI\approx1.764\,\text{kVA}\).

Low lagging power factor raises line current and \(I^2R\) loss for the same useful power; shunt capacitor banks supply leading VAR locally. RLC impedance and power-factor calculations therefore control cable, transformer, filter and compensation design.

Practice target: 16–18 minutes; derive each impedance, draw both triangles, and complete one signed-power calculation.

Model Answer — Single-Phase and Three-Phase Circuits [6 marks, NTC 2082]¶

Exam-ready answer

A single-phase system supplies one sinusoidal voltage, normally through phase and neutral conductors. Nepalese low-voltage supply is nominally \(230\,\text{V}\) RMS at \(50\,\text{Hz}\). A balanced three-phase system supplies three equal sinusoidal phase voltages displaced by \(120^\circ\) electrical:

\[ v_R=V_m\sin\omega t, \quad v_Y=V_m\sin(\omega t-120^\circ), \quad v_B=V_m\sin(\omega t-240^\circ). \]

Their instantaneous sum is zero. A three-winding alternator produces these voltages; when they feed a balanced load, the total instantaneous power is constant and creates smooth motor torque.

Fig: Single-phase and balanced three-phase waveforms, 120-degree phasors and power comparison

For a balanced load,

\[ \boxed{P_{1\phi}=VI\cos\theta}\ \text{W}, \qquad \boxed{P_{3\phi}=\sqrt3V_LI_L\cos\theta}\ \text{W}. \]

In star, \(V_L=\sqrt3V_{ph}\) and \(I_L=I_{ph}\); in delta, \(V_L=V_{ph}\) and \(I_L=\sqrt3I_{ph}\). Thus a \(400\,\text{V}\) line-to-line star supply provides about \(400/\sqrt3=230\,\text{V}\) line-to-neutral.

Feature	Single phase	Three phase
Wave sets	One	Three, \(120^\circ\) apart
Conductors	Usually phase + neutral	Three wires, or three phases + neutral
Instantaneous power	Pulsates at twice supply frequency	Constant for a balanced load
Motors	Need auxiliary starting means	Naturally produces rotating field; self-starting
Economy	More conductor for a given transmitted power	More power per unit conductor; smaller machines
Typical use	Homes, lighting, small appliances	Generation, transmission, industry, pumps and HVAC

Benefits of three phase: smoother torque, self-starting induction motors, higher machine efficiency and power density, approximately 25% conductor saving for comparable transmission conditions, better voltage regulation, and availability of both \(400\,\text{V}\) three-phase and \(230\,\text{V}\) single-phase loads from a four-wire system.

Numerical check: a balanced \(400\,\text{V}\) motor drawing \(10\,\text{A}\) at 0.8 lagging power factor takes

\[ P=\sqrt3(400)(10)(0.8)=5.54\,\text{kW}. \]

Supplying \(5.54\,\text{kW}\) from \(230\,\text{V}\) single phase at the same power factor would require \(I=5540/(230\times0.8)=30.1\,\text{A}\), illustrating the larger current burden. Therefore single phase is economical for small domestic loads, while three phase is preferred for bulk power and telecom-site machinery.

Practice target: 10–11 minutes; include the 120-degree equations, line relations, six benefits, and one power calculation.

Model Answer — Star-Delta Resistance Conversion [4 marks, NTC 2082]¶

Exam-ready answer

In a star (Y) network, three resistances \(R_A,R_B,R_C\) connect their external terminals A, B and C to one common point. In a delta (\(\Delta\)) network, \(R_{AB},R_{BC},R_{CA}\) form a closed triangle. Equivalent conversion keeps the resistance seen between every pair of terminals unchanged.

Fig: Corresponding star and delta networks with terminals, common point and opposite arms labeled

Define

\[ S_Y=R_AR_B+R_BR_C+R_CR_A\ \Omega^2. \]

For star to delta, divide the sum of pair-products by the star arm opposite the required delta side:

\[ \boxed{R_{AB}=\frac{S_Y}{R_C}},\qquad \boxed{R_{BC}=\frac{S_Y}{R_A}},\qquad \boxed{R_{CA}=\frac{S_Y}{R_B}}. \]

For delta to star, let \(S_\Delta=R_{AB}+R_{BC}+R_{CA}\,\Omega\) and divide the product of adjacent delta arms by the total:

\[ \boxed{R_A=\frac{R_{AB}R_{CA}}{S_\Delta}},\qquad \boxed{R_B=\frac{R_{AB}R_{BC}}{S_\Delta}},\qquad \boxed{R_C=\frac{R_{BC}R_{CA}}{S_\Delta}}. \]

Item	Star	Delta
Physical form	Common neutral point	Closed three-side loop
Balanced conversion	\(R_Y=R_\Delta/3\)	\(R_\Delta=3R_Y\)
Three-phase line relation	\(V_L=\sqrt3V_{ph}\), \(I_L=I_{ph}\)	\(V_L=V_{ph}\), \(I_L=\sqrt3I_{ph}\)

The line relations describe an energised balanced three-phase connection; they are not substitutes for the resistance-conversion formulas.

Numerical check: if \(R_A=2\,\Omega\), \(R_B=3\,\Omega\), and \(R_C=6\,\Omega\), then \(S_Y=36\,\Omega^2\). Therefore \(R_{AB}=36/6=6\,\Omega\), \(R_{BC}=36/2=18\,\Omega\), and \(R_{CA}=36/3=12\,\Omega\). Back-conversion gives \(S_\Delta=36\,\Omega\) and returns \(R_A=(6)(12)/36=2\,\Omega\), confirming equivalence.

The same formulas apply to complex impedances at one frequency and are used to simplify bridge networks and to compare balanced star and delta loads.

Practice target: 6–7 minutes; draw labels first, then write both conversions and the balanced shortcut.

Model Answer — Series and Parallel Resonance [5 marks]¶

Exam-ready answer

Resonance occurs when the net reactive part of a circuit's impedance or admittance is zero, so inductive and capacitive reactive effects cancel. For ideal \(L\) and \(C\),

\[ X_L=X_C, \qquad \omega_0L=\frac{1}{\omega_0C}, \qquad \boxed{f_0=\frac{1}{2\pi\sqrt{LC}}}\ \text{Hz}. \]

For a series RLC circuit,

\[ Z=R+j\left(\omega L-\frac{1}{\omega C}\right). \]

At \(f_0\), \(Z=R\) is minimum, current \(I=V/R\) is maximum, power factor is unity, and \(V_L\) and \(V_C\) are equal and opposite. It is an acceptor circuit. Its quality factor and half-power bandwidth are

\[ \boxed{Q_s=\frac{\omega_0L}{R}=\frac{1}{\omega_0CR}}, \qquad \boxed{BW=f_2-f_1=\frac{R}{2\pi L}=\frac{f_0}{Q_s}}\ \text{Hz}. \]

For an ideal parallel RLC, admittance is

\[ Y=\frac{1}{R}+j\left(\omega C-\frac{1}{\omega L}\right). \]

At resonance, input susceptance is zero, impedance is maximum, supply current is minimum, and large opposing currents may circulate in \(L\) and \(C\); it is a rejector or tank circuit. If a practical coil has series resistance \(r\) and is in parallel with \(C\),

\[ f_r=\frac{1}{2\pi}\sqrt{\frac{1}{LC}-\frac{r^2}{L^2}}, \qquad R_{dynamic}\approx\frac{L}{Cr} \]

for a high-\(Q\) coil.

Fig: Practical series and parallel resonant circuits with extrema, half-power points, bandwidth and Q

Property	Series resonance	Parallel resonance
Input impedance	Minimum	Maximum
Supply current	Maximum	Minimum
Magnification	Voltage across \(L\) or \(C\)	Circulating branch current
Function	Acceptor/band-pass action	Rejector/tank action

Design check: for a series circuit with \(L=0.1\,\text{H}\), \(C=100\,\mu\text{F}\) and \(R=10\,\Omega\),

\[ f_0=\frac{1}{2\pi\sqrt{0.1(100\times10^{-6})}}=50.33\,\text{Hz}, \]

\(Q_s=2\pi(50.33)(0.1)/10=3.16\) and \(BW=f_0/Q_s=15.9\,\text{Hz}\). At \(100\,\text{V}\) resonance, \(I=10\,\text{A}\) and each reactive voltage is about \(QV=316\,\text{V}\), showing why component ratings matter.

Resonance and \(Q\) set the selectivity of radio tuning, oscillators, filters and impedance-matching networks.

Practice target: 8–9 minutes; derive \(f_0\), compare the extrema, and calculate both Q and bandwidth.

Syllabus Focus¶

Single-phase and three-phase systems (asked NTC 2082 [6])
Power factor
RLC circuits
Resonance

1. AC Quantities¶

Likely Exam Question (5 marks)

"Define RMS value, average value, form factor and peak factor of a sinusoidal wave."

A sinusoidal voltage:

\[ \boxed{v(t) = V_m\sin(\omega t + \phi)}, \qquad \omega = 2\pi f, \qquad T = \frac{1}{f} \]

Quantity	Definition	Sinusoid Value
Peak value \(V_m\)	Maximum instantaneous value	\(V_m\)
Average value	Mean over half cycle	\(V_{avg} = \frac{2V_m}{\pi} = 0.637V_m\)
RMS value	DC equivalent producing same heating	\(V_{rms} = \frac{V_m}{\sqrt{2}} = 0.707V_m\)
Form factor	\(V_{rms}/V_{avg}\)	1.11
Peak factor	\(V_m/V_{rms}\)	1.414

Fig: One sinusoidal cycle with peak and period, full-cycle RMS definition, rectified positive-half-cycle area and average, RMS and average levels, and correct form-factor and peak-factor formulas

Mains supply "230 V, 50 Hz" is the RMS value; \(V_m = 325\,\text{V}\).

2. AC Through R, L, C¶

Element	Impedance	Phase Relation	Power
R	\(Z = R\)	V and I in phase	Dissipates \(P = I^2R\)
L	\(Z = jX_L\), \(X_L = \omega L = 2\pi fL\)	I lags V by 90°	Zero average power
C	\(Z = -jX_C\), \(X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}\)	I leads V by 90°	Zero average power

Memory aid: CIVIL — in C, I leads V; V leads I in L.

Series RLC Impedance¶

\[ \boxed{Z = R + j(X_L - X_C)}, \qquad |Z| = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \theta = \tan^{-1}\frac{X_L - X_C}{R} \]

\(X_L > X_C\): inductive circuit, current lags
\(X_C > X_L\): capacitive circuit, current leads

Fig: Schemdraw R, L, and C AC circuits with voltage-current phasors, plus series-RLC voltage and impedance triangles for inductive lagging and capacitive leading cases

3. AC Power and Power Factor¶

Likely Exam Question (5 marks)

"Define active, reactive and apparent power. What is power factor and why is its improvement important?"

The Power Triangle¶

\[ \boxed{P = VI\cos\theta \;\text{(W, active/real)}} \qquad \boxed{Q = VI\sin\theta \;\text{(VAR, reactive)}} \qquad \boxed{S = VI \;\text{(VA, apparent)}} \]

\[ S^2 = P^2 + Q^2 \]

Power Factor¶

\[ \boxed{\text{pf} = \cos\theta = \frac{P}{S} = \frac{R}{|Z|}} \]

Unity for pure R; zero (lagging) for pure L; zero (leading) for pure C.
Low pf means more current for the same real power → larger \(I^2R\) line losses, bigger conductors/transformers, voltage drop, utility penalties.

Power Factor Improvement¶

Connect capacitor banks (or synchronous condensers) in parallel with inductive loads (motors, transformers) to supply the reactive power locally:

\[ Q_C = P(\tan\theta_1 - \tan\theta_2) \]

where \(\theta_1, \theta_2\) are angles before and after correction.

Fig: Lagging and leading AC power triangles with signed reactive power, real and apparent power units, phase angle, power factor, and an inductive-load correction inset

4. Resonance (Series and Parallel)¶

At resonance \(X_L = X_C\):

\[ \boxed{f_0 = \frac{1}{2\pi\sqrt{LC}}} \]

Property	Series Resonance	Parallel Resonance
Impedance	Minimum (\(=R\))	Maximum (\(=L/CR\))
Current	Maximum	Minimum
pf	Unity	Unity
Magnification	Voltage (\(V_L = QV\))	Current (\(I_C = QI\))
Called	Acceptor circuit	Rejector circuit

\[ Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{f_0}{BW} \]

Fig: Practical series and parallel RLC resonance circuits with loss assumptions, current and impedance extrema, half-power frequencies, bandwidth, quality factor, and resonance condition

(Full treatment with derivations: see Series and Parallel Resonant Circuits.)

5. Three-Phase Systems¶

Likely Exam Question (NTC 2082, 6 marks)

"Explain single-phase and three-phase circuits with their benefits and applications."

Single-Phase System¶

One alternating voltage delivered over two conductors (phase + neutral). Standard Nepal LV supply: 230 V, 50 Hz.

Applications: residential lighting, small appliances, small motors (< ~3 kW).

Limitations: pulsating instantaneous power (twice per cycle), single-phase motors are not self-starting, lower transmission efficiency.

Three-Phase System¶

Three sinusoidal voltages of equal magnitude, displaced by 120°, generated by a three-winding alternator. Standard LV: 400 V line-to-line / 230 V line-to-neutral.

\[ v_R = V_m\sin\omega t,\quad v_Y = V_m\sin(\omega t - 120°),\quad v_B = V_m\sin(\omega t - 240°) \]

Fig: Single-phase and balanced three-phase AC: three equal sinusoidal voltages displaced by 120 degrees, their phasor representation, and the contrast between pulsating single-phase power and constant total three-phase power

Benefits of Three-Phase over Single-Phase¶

Constant instantaneous power — smooth torque in motors, less vibration.
Self-starting motors — rotating magnetic field is produced naturally.
More power per conductor material — ~75% conductor material of equivalent 1-φ for same power; more economical transmission.
Higher efficiency and smaller machines for the same rating.
Both 400 V (power) and 230 V (lighting) available from one system.
Better voltage regulation and less pulsating flux.

Applications: power generation/transmission/distribution, industrial motors, telecom exchange power plants, large HVAC, data-center feeds.

Star and Delta Connections¶

Star (Y): line-neutral available.

\[ \boxed{V_L = \sqrt{3}\,V_{ph}, \qquad I_L = I_{ph}} \]

Delta (Δ): no neutral.

\[ \boxed{V_L = V_{ph}, \qquad I_L = \sqrt{3}\,I_{ph}} \]

Fig: Star and delta three-resistor networks showing corresponding terminals, the common star node, and opposite delta arms for resistance conversion

For star arms \(R_A,R_B,R_C\) and delta arms \(R_{AB},R_{BC},R_{CA}\), define

\[ S_Y=R_AR_B+R_BR_C+R_CR_A \]

Star to delta:

\[ \boxed{R_{AB}=\frac{S_Y}{R_C},\qquad R_{BC}=\frac{S_Y}{R_A},\qquad R_{CA}=\frac{S_Y}{R_B}} \]

Delta to star, with \(S_\Delta=R_{AB}+R_{BC}+R_{CA}\):

\[ \boxed{R_A=\frac{R_{AB}R_{CA}}{S_\Delta},\qquad R_B=\frac{R_{AB}R_{BC}}{S_\Delta},\qquad R_C=\frac{R_{BC}R_{CA}}{S_\Delta}} \]

For balanced networks, \(\boxed{R_\Delta=3R_Y}\) and \(\boxed{R_Y=R_\Delta/3}\).

Power (both connections, balanced):

\[ \boxed{P = \sqrt{3}\,V_L I_L\cos\theta, \qquad Q = \sqrt{3}\,V_L I_L\sin\theta, \qquad S = \sqrt{3}\,V_L I_L} \]

Single-Phase vs Three-Phase — Comparison Table¶

Feature	Single-Phase	Three-Phase
Conductors	2 (P + N)	3 or 4 (3P + N)
Voltage (Nepal)	230 V	400 V line / 230 V phase
Instantaneous power	Pulsating	Constant
Motor starting	Needs auxiliary winding/capacitor	Self-starting
Power capacity	Low	High
Transmission economy	Poorer	~25% conductor saving
Use	Homes, small loads	Industry, distribution, big loads

6. Solved Examples¶

Example 1 — RLC Series¶

Q. \(R = 30\,\Omega\), \(L = 0.1\,\text{H}\), \(C = 100\,\mu\text{F}\) across 230 V, 50 Hz. Find current and pf.

Solution:

\(X_L = 2\pi(50)(0.1) = 31.4\,\Omega\); \(X_C = \frac{1}{2\pi(50)(100\times10^{-6})} = 31.8\,\Omega\)

\(|Z| = \sqrt{30^2 + (31.4-31.8)^2} \approx 30\,\Omega\) → \(I = 230/30 = 7.67\,\text{A}\), pf \(\approx 1\) (near resonance).

Example 2 — Three-Phase Power¶

Q. A balanced 400 V, 3-φ motor draws 10 A at pf 0.8. Find input power.

Solution:

\[ P = \sqrt{3} \times 400 \times 10 \times 0.8 = 5.54\,\text{kW} \]

Example 3 — PF Correction¶

Q. A 10 kW load at pf 0.6 lagging is corrected to 0.9. Find capacitor kVAR.

Solution:

\(\tan\theta_1 = 1.333\), \(\tan\theta_2 = 0.484\)

\[ Q_C = 10(1.333 - 0.484) = 8.49\,\text{kVAR} \]

7. Quick Revision Table¶

Topic	Key Result
RMS of sine	\(V_m/\sqrt{2}\)
\(X_L\), \(X_C\)	\(2\pi fL\), \(1/(2\pi fC)\)
Series impedance	\(Z = R + j(X_L - X_C)\)
Power triangle	\(S^2 = P^2 + Q^2\); pf \(=\cos\theta = P/S\)
Resonance	\(f_0 = 1/(2\pi\sqrt{LC})\); series → min Z, parallel → max Z
Star	\(V_L = \sqrt{3}V_{ph}\), \(I_L = I_{ph}\)
Delta	\(V_L = V_{ph}\), \(I_L = \sqrt{3}I_{ph}\)
3-φ power	\(\sqrt{3}V_LI_L\cos\theta\)
3-φ benefits	Constant power, self-start motors, conductor saving

Key Exam Points - AC Fundamentals

1-φ vs 3-φ comparison was directly asked (NTC 2082) — memorize the 6 benefits + comparison table.
\(\sqrt{3}\) goes with voltage in star and current in delta; power formula is the same: \(\sqrt{3}V_LI_L\cos\theta\).
Power factor improvement = parallel capacitors; know \(Q_C = P(\tan\theta_1 - \tan\theta_2)\).
CIVIL: capacitor I leads V; inductor I lags V.