Single-Phase and Three-Phase Circuits

Tip

Exam Focus

Directly asked in NTC 2082 [6] (1$\phi$ vs 3$\phi$ with benefits). Nail: RMS $=V_m/\sqrt2$, average $=2V_m/\pi$, form factor $1.11$, peak factor $1.414$; CIVIL (in C, I leads V; in L, V leads I); $Z=R+j(X_L-X_C)$ and $\mathrm{pf}=\cos\varphi=R/|Z|$; the power triangle $S^2=P^2+Q^2$; resonance $f_0=1/(2\pi\sqrt{LC})$; and the star/delta line relations with $P=\sqrt3\,V_LI_L\cos\varphi$.

AC Quantities

Caution

Key Formula — Sinusoid and its characteristic values

$$v(t)=V_m\sin(\omega t+\phi),\quad \omega=2\pi f,\quad T=\tfrac1f;$$$$V_{rms}=\frac{V_m}{\sqrt2}=0.707V_m,\quad V_{avg}=\frac{2V_m}{\pi}=0.637V_m,\quad k_f=\frac{V_{rms}}{V_{avg}}=1.11,\quad k_p=\frac{V_m}{V_{rms}}=1.414.$$

One cycle of a sinusoid with the RMS (equal-heating) and rectified average levels.

Mains “230 V, 50 Hz” is the RMS value, so $V_m=\sqrt2\times230\approx325\,V$.

R, L, C in AC Circuits

Element	Impedance	Phase	Avg. power
$R$	$Z=R$	$V,I$ in phase	$P=I^2R$
$L$	$Z=jX_L,\ X_L=2\pi fL$	$I$ lags $V$ by $90^\circ$	$0$
$C$	$Z=-jX_C,\ X_C=\dfrac{1}{2\pi fC}$	$I$ leads $V$ by $90^\circ$	$0$

Impedance and phase of the three elements. Mnemonic: CIVIL.

Voltage–current phasor relations for the three elements.

Caution

Key Formula — Series RLC impedance

$$Z=R+j(X_L-X_C),\quad |Z|=\sqrt{R^2+(X_L-X_C)^2},\quad \varphi=\tan^{-1}\frac{X_L-X_C}{R}.$$

AC Power and Power Factor

Caution

Key Formula — Real, reactive, apparent power

$$P=VI\cos\varphi\ \text{(W)},\quad Q=VI\sin\varphi\ \text{(VAR)},\quad S=VI\ \text{(VA)},\quad S^2=P^2+Q^2.$$

Power factor $\mathrm{pf}=\cos\varphi=P/S=R/|Z|$.

Power triangle (lagging load): $S^2=P^2+Q^2$, $\cos\varphi=P/S$.

Note

Key Point

Low power factor draws more current for the same real power, raising $I^2R$ losses and voltage drop. It is corrected by connecting shunt capacitors across inductive loads: $\;Q_C=P(\tan\varphi_1-\tan\varphi_2)$.

Resonance

Caution

Key Formula — Resonant frequency, quality factor

At $X_L=X_C$: $f_0=\dfrac{1}{2\pi\sqrt{LC}}$, $Q=\dfrac1R\sqrt{\dfrac{L}{C}}=\dfrac{f_0}{\mathrm{BW}}$.

Property	Series resonance	Parallel resonance
Impedance	Minimum ($=R$)	Maximum ($=L/CR$)
Current	Maximum	Minimum
Called	Acceptor circuit	Rejector (tank) circuit

Series vs parallel resonance.

Three-Phase Systems

A balanced three-phase source gives three equal voltages $120^\circ$ apart:

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

whose instantaneous sum is zero, giving constant total power.

Balanced three-phase voltage phasors, $120^\circ$ apart.

Caution

Key Formula — Line/phase relations and three-phase power

$$\text{Star (Y):}\ V_L=\sqrt3\,V_{ph},\ I_L=I_{ph};\qquad \text{Delta ($\Delta$):}\ V_L=V_{ph},\ I_L=\sqrt3\,I_{ph};$$$$P=\sqrt3\,V_LI_L\cos\varphi,\quad Q=\sqrt3\,V_LI_L\sin\varphi,\quad S=\sqrt3\,V_LI_L.$$

Three-phase star and delta connections.

Feature	Single-phase	Three-phase
Conductors	2 (P + N)	3 or 4
Voltage (Nepal)	230 V	400 V line / 230 V phase
Instantaneous power	Pulsating	Constant (balanced)
Motor starting	Needs aux. winding	Self-starting
Transmission economy	Poorer	Approximately 25% conductor saving

Single-phase vs three-phase. Benefits of 3$\phi$: constant torque, self-starting motors, conductor saving, smaller machines, both 400/230 V available.

Note

Example — Three-phase power

A balanced 400 V motor draws 10 A at $\mathrm{pf}=0.8$: $P=\sqrt3\times400\times10\times0.8=5.54\,kW$.