Module 3 Full Notes PDF

# MODULE Ⅲ ## 3.1 AC Circuits - Phasor representation of sinusoidal quantities: 1 - Trigonometric, Rectangular, Polar and complex forms. - Analysis of simple AC circuits: Purely resistive, inductive & capacitive circuits: 2 - Inductive and capacitive reactance, concept of impedance. - Average Po...

# MODULE Ⅲ ## 3.1 AC Circuits - Phasor representation of sinusoidal quantities: 1 - Trigonometric, Rectangular, Polar and complex forms. - Analysis of simple AC circuits: Purely resistive, inductive & capacitive circuits: 2 - Inductive and capacitive reactance, concept of impedance. - Average Power, Power factor. - Analysis of RL, RC and RLC series circuits-active, reactive and apparent power: 1 - Simple numerical problems. ## 3.2 Three phase AC systems - Generation of three phase voltages - Advantages of three phase systems, star and delta connections - Balanced only - relation between line and phase voltages, line and phase currents - Numerical problems. ## Phase The phase of a particular value of an alternating quantity is a fractional part of time period or cycle through which the quantity has advanced from a selected zero position. ## Phase Difference When two alternating quantities of same frequency have different zero points, they are said to have a phase difference. It is represented by $\phi$ and measured in radians or degree. - In phase: $\phi = 0$ - lagging: $\phi = -\phi$ - leading: $\phi = +\phi$ ## Representation of AC Quantity using Phasor Sinusoidal alternating quantity is represented by a line of definite length rotating in counter-clockwise direction with an angular velocity $\omega$ rad/sec. Such a rotating line is called a phasor. ## Phasor representation of in-phase, lagging, leading quantities - **In phase** - $v = Vm sin \omega t$ - $i = Im sin \omega t$ - $\phi = 0$ - **lagging** - $v = Vm sin \omega t$ - $i = Im sin (\omega t - \phi)$ - **leading** - $v = Vm sin \omega t$ - $i = Im sin (\omega t + \phi)$ ## Mathematical Representation of phasors Phasors can be mathematically represented in 4 forms: 1. **Rectangular form:** 2. **Polar form:** 3. **Exponential form:** 4. **Trigonometric form:** ## 1 Rectangular form Any phasor can be resolved into x-component and y component. - $v = x + yj$ - $|v| = \sqrt{x^2 + y^2}$ - $\theta = tan^{-1} \left(\frac{y}{x}\right)$ ## 2 Polar form Any phasor can be represented using its magnitude and angle from reference axis. - $v = V \angle \pm \theta$ - Rectangular form can be converted into polar form by: - $v= x + yj$ - $v = |V|\angle \pm \theta$ - where $|V| = \sqrt{x^2 + y^2}$, $\theta = tan^{-1} \left(\frac{y}{x}\right)$ - Or using calculator - Rect $\rightarrow$ Polar Calculator : Shift Pol(a, b)= |V|$\angle$ $\theta$ - Polar $\rightarrow$ Rect Calculator : Shift Rec(V, $\theta$) = a + jb ## 3 Exponential form The phasor can be represented using its magnitude and angle. - $v = Ve^{j\theta}$ - $v = Ve^{-j\theta}$ ## 4 Trignometric form The phasor can be represented using its magnitude and angle by substituting $e^{j\theta}= cos\theta \pm jsine$ - $v = V(cos\theta + jsine)$ where V is the magnitude & $\theta$ is the angle. - $v = V(cos\theta - jsine)$ where V is the magnitude & $\theta$ is the angle. - Generally, it can be represented as $v = V cos\theta \pm j sin\theta$. ## Phasor Addition & Subtraction Phasor addition & subtraction rectangular form is used. - $v_1 = x_1 + jy_1$, $v_2 = x_2 + jy_2$ - $v_1 + v_2 = (x_1 + x_2) + j(y_1 + y_2)$ - $v_1 - v_2 = (x_1 - x_2) + j(y_1 - y_2)$ ## Phasor multiplication & Division. Phasor multiplication & division polar form is used. - $v_1 = V_1 \angle \theta_1$, $v_2 = V_2 \angle \theta_2$ - $v_1 \times v_2 = V_1 V_2 \angle \theta_1 + \theta_2$ - $v_1 / v_2 = \frac{V_1}{V_2} \angle \theta_1 - \theta_2$ ## Rectangular form of representation of impedance - $z= R + jX_L$ - $z = 0 + jX_L$ - $z = 0 - jXc$ - $z = R + jX_L = jXc$ - $z = R + j(X_L - Xc)$ - $z = R - jXc$ - $z = R + j(X_L - Xc)$ ## 3.1 Perform the following operations & express the result in polar form. - $5 \angle 130^o (2 - j4)$ - $(3 + j3) \times (2 \angle -15^o )$ **Equation breakdown:** - ** Step 1: Convert everything to polar form (if it isn't already).** - $5 \angle130^o = 5 \angle 130^o$ - $(2 - j4) = 4.47 \angle -63.43^o$ - $(3 + j3) = 3 \sqrt{2} \angle 45^o = 3 \sqrt{2} \angle 45^o$ - $2 \angle -15^o = 2 \angle -15^o$ - ** Step 2: Apply multiplication and division rules** - $5 \angle 130^o \times 4.47 \angle -63.43^o = 22.35 \angle (130 - 63.43) = 22.35 \angle 66.57^o$ - $3 \sqrt{2} \angle 45^o \times 2 \angle -15^o = 6 \sqrt{2} \angle (45 - 15) = 6 \sqrt{2} \angle 30^o$ **Answer:** $3 \angle 53.40^o$ ## 3.2 An alternating voltage of (80+j60) V is applied to a circuit and current flowing is (~4+jlo) A - Find the impedance of the circuit. - **Equation:** - Impedance = Voltage/Current - **Step 1: Convert both voltage and current into their polar forms.** - $(80 + j60) = 100 \angle 36.86^o$ - $(-4 + j10) = 10.77 \angle 111.80^o$ - **Step 2: Apply the impedance equation** $Z = \frac{100 \angle 36.86^o}{10.77 \angle 111.80^o} = 9.28 \angle -74.94^o \Omega$ ## Behavior of AC in different circuits We are analysing 6 different circuits. 1. **Resistive R (Purely resistive)** - $v = Vm sin \omega t$ - $i_R = i = Im sin \omega t$ - $v_R = v = Vm sin \omega t$ - $i_R = \frac{V_R}{R}$ (Ohm's law) - $i_R = Im sin \omega t$ where $Im = \frac{Vm}{R}$ - **Waveform & phasor: v & i are in phase.** - **Phase difference: $\phi = 0$ (between v & i)** - **Instantaneous power: P = v * i** - **Power factor: cos$\phi = cos 0 = 1$** - **Power: P = $Vm * Im sin^2 \omega t$** - **Average Power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T VmIm sin^2 \omega t dt$ - $= \frac{VmIm}{T} \int_0^T sin^2 \omega t dt$ - $= \frac{VmIm}{T} \int_0^T \frac{1 - cos2\omega t }{2} dt$ - $= \frac{VmIm}{T} \left[\frac{t}{2} - sin \frac{2\omega t}{2\omega} \right]_0^T$ - $ = \frac{VmIm}{T} [\frac{T}{2} - sin(\omega T)-0-sin(0)]$ - $= \frac{VmIm}{T} [\frac{T}{2} - sin(2\pi)-0-sin(0)]$ - $= \frac{VmIm}{T} [\frac{T}{2} - 0 - 0 - 0]$ - $= \frac{VmIm}{2}$ - $Pav = \frac{VmIm}{2} = Vrms Irms$ - **Apparent Power: S= Vrms Irms** - **Active Power: P = Vrms Irms cos$\phi$** - $P = Vrms Irms$ as cos$\phi=1$ - **Reactive Power: Q = Vrms Irms sin$\phi$** - $ Q= 0$ as $sin \phi = sin 0 = 0$ 2. **Inductive L (Purely Inductive)** - $v_L = Vm sin \omega t$ - **Current: We know that for inductor: $v_L = L \frac{di}{dt}$ or $v_L = L di$** - **Equation:** $L \frac{di}{dt} = v_L = Vm sin \omega t$. - **Solve for i:** $di = \frac{Vm sin \omega t dt}{L}$ - **Integrate both sides:** $i = \int \frac{Vm sin \omega t dt}{L}$ - $i = \frac{Vm}{L} \int sin\omega t dt$ - $i = \frac{Vm}{L} [\frac{-cos\omega t}{\omega}]$ - $i = \frac{-Vm}{L\omega} [cos \omega t]$ - **Solve for i:** $i = Im sin(\omega t - \frac{\pi}{2})$: - $I_m = \frac{Vm}{L\omega} = \frac{Vm}{XL}$ - **Phase dilexience: $i$ lags voltage by $\frac{\pi}{2}$** - **Power factor: cos$\phi = cos \frac{\pi}{2} = 0$** - **Instantaneous power: P:** - $P = vi$ - $P = Vm sin\omega t \times Im sin(\omega t - \frac{\pi}{2})$ - $P = VmIm sin\omega t(cos\omega t)$ - **Average Power/Active Power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T VmIm sin\omega t cos \omega t dt$ - $= \frac{VmIm}{T} \int_0^T \frac{sin2\omega t}{2} dt$ - $= \frac{VmIm}{2T} [-\frac{cos2\omega t}{\omega}]_0^T$ - $= - \frac{VmIm}{2T} [\frac{cos2\omega t}{\omega}]_0^T$ - $ = \frac{VmIm}{2\omega T} [-cos(2\pi) + cos(0)]$ - $ =\frac{VmIm }{2\omega T} [-1 +1]$ - $Pav = 0$ - **Apparent Power: S= Vrms Irms** - **Active Power: P= Vrms Irms cos$\phi$ = 0** - **Reactive Power: Q = Vrms Irms sin$\phi$** - $ Q = $ Vrms Irms. 3. **Capacitive C (Purely Capacitive)** - $v_C = Vm sin \omega t$ - **Current:** - **We know that for capacitor: $i = C \frac{dv}{dt}$ or $i = CdV$ ** - **Equation:** $i = C \frac{d}{dt} (Vm sin\omega t)$ - **Solve for i:** $i = C Vm \frac{d}{dt} (sin\omega t)$ - $i = C Vm \omega cos \omega t $ - $i = CVm \omega cos \omega t $ - **Solve for i:** - $i = Im sin(\omega t + \frac{\pi}{2})$: - where $I_m = CVm\omega = \frac{Vm}{Xc}$, $Xc = \frac{1}{C \omega}$. - **Capacitive reactance:** $X_c = \frac{1}{\omega C}$ - **Waveform & phasor diagram:** - $i$ leads voltage by - **Phase difference: $\phi = + \frac{\pi}{2}$** - **Power factor: cos$\phi = cos\frac{\pi}{2} = 0$** - **Instantaneous Power: P:** - $P = v_C \times i$ - $P = Vm sin\omega t \times Im sin(\omega t + \frac{\pi}{2}) = VmIm sin\omega t cos \omega t$ - **Average Power/Active Power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T VmIm sin2\omega t dt$ - $= \frac{VmIm}{T} \int_0^T sin2\omega t dt$ - $= \frac{VmIm }{2T} [-\frac{cos2\omega t}{\omega}]_0^T$ - $= \frac{VmIm}{2\omega T} [-cos(2\pi) + cos(0)]$ - $= \frac{VmIm }{2\omega T} [-1 +1]$ - $Pav = 0$ - **Apparent Power: S= Vrms Irms** - **Active Power: P= Vrms Irms cos$\phi$ =0** - **Reactive Power: Q = Vrms Irms sin$\phi$ ** - $Q = $ Vrms Irms 4. **RL - Series circuit** - **Current:** $i = Im sin \omega t$, same in both R & L. - $z = z_R + z_L $ (phasor addition) - **Voltage:** $V = Vm sin(\omega t + \phi)$ - **Waveform:** - **Phase difference: $\phi$ = lagging** - **Power factor: cos$\phi$** - **Instantaneous power: p:** - $P = v \times i$ - $P = Vm sin(\omega t + \phi) Im sin\omega t$ - $P = \frac{VmIm}{2} [cos \phi - cos(2\omega t + \phi)]$ - **Average power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T \frac{VmIm}{2} [cos \phi - cos(2\omega t + \phi)] dt$ - $= \frac{VmIm}{2T} \int_0^T cos \phi dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t + \phi) dt$ - $= \frac{VmIm}{2T} cos\phi \int_0^T dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t + \phi) dt$ - $= \frac{VmIm}{2T} cos \phi [t ]_0^T - \frac{VmIm}{2T} \int_0^T cos(2\omega t + \phi) dt$ - $= \frac{VmIm}{2T} cos\phi [T - 0] - \frac{VmIm}{4\omega T} [sin(2\omega t + \phi)]_0^T $ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [sin(2\pi + \phi) - sin(0 + \phi)]$ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [sin \phi - sin \phi]$ - $= \frac{VmIm}{2} cos\phi$ - $Pav = \frac{VmIm}{2} cos \phi = Vrms Irms cos \phi $ - **Apparent Power: S= Vrms Irms ** - **Active Power: P = Vrms Irms cos$\phi$** - **Reactive Power: Q = Vrms Irms sin $\phi$** - **Voltage Triangle: From phasor diagram:** - $Z = \sqrt{R^2 + X_L^2}$ - $\phi = tan^{-1}\left(\frac{X_L}{R} \right)$ - cos$\phi = \frac{R}{Z}$ - **Power Triangle: $S = \sqrt{P^2 + Q^2}$** - $\phi = tan^{-1}\left(\frac{Q}{P} \right)$ - cos$\phi = \frac{P}{S}$ 5. **RC - series circuit** - **Current:** $i = Im sin \omega t$, same in both R & C. - **Voltage:** $V = Vm sin(\omega t - \phi)$ - **Waveform:** - **Phase difference: $\phi$ = leading** - **Power factor: cos $\phi$** - **Instantaneous power p:** - $p = v \times i$ - $p = Vm sin(\omega t - \phi) Im sin \omega t$ - $p = \frac{VmIm}{2}[cos \phi - cos(2\omega t - \phi)]$ - **Average power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T \frac{VmIm}{2} [cos \phi - cos(2\omega t - \phi)] dt$ - $= \frac{VmIm}{2T} \int_0^T cos \phi dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t - \phi) dt$ - $= \frac{VmIm}{2T} cos\phi \int_0^T dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t - \phi) dt$ - $= \frac{VmIm}{2T} cos \phi [t ]_0^T - \frac{VmIm}{2T} \int_0^T cos(2\omega t - \phi) dt$ - $= \frac{VmIm}{2T} cos\phi [T - 0] - \frac{VmIm}{4\omega T} [sin(2\omega t - \phi)]_0^T $ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [sin(2\pi - \phi) - sin(0 - \phi)]$ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [-sin \phi + sin \phi]$ - $ = \frac{VmIm}{2} cos\phi$ - $Pav = \frac{VmIm}{2} cos \phi = Vrms Irms cos \phi $ - **Apparent Power: S= Vrms Irms ** - **Active Power: P = Vrms Irms cos$\phi$** - **Reactive Power: Q = Vrms Irms sin $\phi$** - **Voltage Triangle: From phasor diagram:** - $Z = \sqrt{R^2 + Xc^2}$ - $\phi = tan^{-1}\left(\frac{Xc}{R} \right)$ - cos$\phi = \frac{R}{Z}$ - **Power Triangle: $S = \sqrt{P^2 + Q^2}$** - $\phi = tan^{-1}\left(\frac{Q}{P} \right)$ - cos$\phi = \frac{P}{S}$ 6. **RLC - series circuit** - **Current: $i = Im sin \omega t$, same in all (R, L, C)** - **Voltage:** $V = Vm sin(\omega t + \phi)$ - **Phase difference: $\phi$ = lagging** - **power factor: cos $\phi$** - **Instantaneous power p:** - $p = v \times i$ - $p = Vm sin(\omega t + \phi) Im sin \omega t$ - $p = \frac{VmIm}{2}[cos \phi - cos(2\omega t + \phi)]$ - **Average power:** - $Pav = \frac{1}{T} \int_0^T p dt$ - $= \frac{1}{T} \int_0^T \frac{VmIm}{2} [cos \phi - cos(2\omega t + \phi)] dt$ - $= \frac{VmIm}{2T} \int_0^T cos \phi dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t + \phi) dt$ - $= \frac{VmIm}{2T} cos\phi \int_0^T dt - \frac{VmIm}{2T} \int_0^T cos(2\omega t + \phi) dt$ - $= \frac{VmIm}{2T} cos \phi [t ]_0^T - \frac{VmIm}{2T} \int_0^T cos(2\omega t - \phi) dt$ - $= \frac{VmIm}{2T} cos\phi [T - 0] - \frac{VmIm}{4\omega T} [sin(2\omega t + \phi)]_0^T $ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [sin(2\pi + \phi) - sin(0 + \phi)]$ - $= \frac{VmIm}{2} cos\phi - \frac{VmIm}{4\omega T} [sin \phi - sin \phi]$ - $ = \frac{VmIm}{2} cos\phi$ - $Pav = \frac{VmIm}{2} cos \phi = Vrms Irms cos \phi $ - **Apparent Power: S= Vrms Irms ** - **Active Power: P = Vrms Irms cos$\phi$** - **Reactive Power: Q = Vrms Irms sin $\phi$** - **Voltage Triangle: From phasor diagram:** - $Z = \sqrt{R^2 + (X_L - Xc)^2}$ - $\phi = tan^{-1}\left(\frac{X_L - Xc }{R} \right)$ - cos$\phi = \frac{R}{Z}$ - **Power Triangle: $S = \sqrt{P^2 + Q^2}$** - $\phi = tan^{-1}\left(\frac{Q}{P} \right)$ - cos$\phi = \frac{P}{S}$ ## Problems **Equations:** - $Vrms = \frac{Vm}{\sqrt{2}}$ - $Irms = \frac{Im}{\sqrt{2}}$ - $z = \frac{Vrms}{Irms}$ - $z = \sqrt{R^2 + (X_L - Xc)^2}$ - $\phi = tan^{-1}\left(\frac{X_L-Xc}{R}\right)$ - $\phi$ = (Voltage phase angle) - (Current phase angle) - $cos \phi = \frac{R}{Z}$ - $cos \phi = \frac{V_R}{V_m}$ - $S = \frac{P}{cos \phi}$ - $Q = \sqrt{S^2 - P^2}$ or $ Q = \frac{P}{tan \phi}$ - $P = Vrms Irms cos \phi$ - $P = I^2R$ - $P = \frac{V^2}{R}$ - $Q = Vrms Irms sin \phi$ - $S = Vrms Irms$ - $X_L = 2\pi fL$ - $X_c = \frac{1}{2 \pi fC}$ ## 3.3 A series RC circuit takes a power of 7000 W when connected to a 200 V, 50 Hz supply. The voltage across the resistor is 130 V. Calculate: (i) Resistance (ii) Current (iii) Power factor (iv) Capacitance (v) Impedance (vi) Equations for instantaneous value of voltage and current. **Solution:** - **Given:** - P= 7000 W - Vrms = 200V - f = 50 Hz - VR = 130 V - **(i) Resistance:** - We know that P = I^2R and P = \frac{V_R^2}{R} - $ 7000 = \frac{130^2}{R}$ Solving for R: - $R = \frac{130^2}{7000} = 2.414 \Omega$ - **(ii) Current:** - $P = I^2R$ - $7000 = I^2 \times 2.414$ - $I^2 = \frac{7000}{2.414} = 2899.75 $ - $I = \sqrt{2899.75} = 53.84 A$ - **(iii) Power factor cos$\phi$: ** - $cos \phi = \frac{V_R}{V} = \frac{130}{200} = 0.65$ leading. - **(iv) Capacitance** - $V = V_R^2 + V_C^2$ - $V_C = \sqrt{V^2 - V_R^2} = \sqrt{200^2 - 130^2} = 151.98 V$ - $V_C = I Xc$ - $151.98 = 53.84 \times Xc$ - $Xc = \frac{151.98}{53.84} = 2.822\Omega$ - $X_c = \frac{1}{2\pi fC} $ - $C = \frac{1}{2 \pi fX_c} = \frac{1}{2 \pi \times 50 \times 2.822} = 0.00112 F = 1.12 mF$ - **(v) Impedance** - $Z = \sqrt{R^2 + X_C^2} = \sqrt{2.414^2 + 2.822^2} = 3.713 \Omega $ - **(vi) v and i equations:** - $v = Vm sin \omega t$ - $i = Im sin(\omega t + \phi)$ - $Vm = \sqrt{2} \times Vrms = \sqrt{2} \times 200 = 282.84 V$ - $Im = \sqrt{2} \times Irms = \sqrt{2} \times 53.84 = 76.14 A$ - $\phi = cos^{-1}(0.65) = 49.45^o = 0.863 rad$ - $\omega=2 \pi f = 2 \times 50 = 314.15 rad/sec$ - $v = 282.84 sin(314.15t)$ - $i = 76.14 sin(314.15t + 0.863)$ ## 3.4 An alternating voltage of (80+j60) V is applied to an RX circuit and the current flowing through the circuit is (~4+j10) A - Calculate the impedance of the circuit in rectangular and polar form. - Also determine if X is inductive or capacitive. **Solution:** - **Given:** - $V = (80 + j60) V$ - $i = (-4 + j10)A$ - **Z = ? ** - **Equation:** - $Z = \frac{V}{I}= \frac{80+j60}{-4+j10}$ - **Step 1: Convert to rectangular form:** - $Z = 2.413 - j8.96 \Omega$ - **Step 2: Convert to polar form:** - $Z = 9.28 \angle -74.93 \Omega$ - **Is X inductive of capacitive?** - The impedance is in the form $Z = R - jX_c $ which shows that X is **capacitive**. ## 3.5 For an AC circuit, if v(t)= 160 sin(wt+1o), and i(t) = 5sin (wt-20); find the power factor and active power absorbed by the circuit. Draw the phasor diagram. **Solution:** - **Given:** - v(t) = $160sin(\omega t + 10) => Vm = 160$ - i(t) = $5sin(\omega t - 20) => Im = 5$ - **Equation:** - cos$\phi$ = ? - P = ? - **Phasor diagram:** - **$\phi$ = Voltage phase angle - current phase angle = (10) - (-20) = 30°** - **cos$\phi$ = cos 30° = 0.866 lagging** - **P = Vrms Irms cos$\phi$ = $\frac{VmIm}{2} \times 0.866 = \frac{160 \times 5}{2} \times 0.866 = 346.4 W$** - **Phasor diagram:** ## Three Phase Systems ### Generation of 3$\phi$ AC - A 3$\phi$ alternator has 3 identical windings R, Y, B 120° displaced from each other. - The magnet rotates in counter-clockwise direction with an angular velocity $\omega$. The corresponding terminals R, Y, B are kept 120° apart. - So emf of Y lags emf of R by 120°. And emf of B lags emf of R by 240°. - **Equations:** - $V_R = Vm sin \omega t$ - $V_Y = Vm sin(\omega t - 120^o)$ - $V_B = Vm sin(\omega t - 240^o)$ ### Advantages of 3$\phi$ AC over single phase (1$\phi$) AC 1. Greater output - The output of a 3$\phi$ alternator is more than that of a single phase generator of the same size. 2. Constant Power - The instantaneous power in a single phase system is fluctuating; but in 3$\phi$ it is almost steady. 3. Cheaper - 3$\phi$ motors are much smaller and less expensive than single $\phi$ motors. 4. Self Starting - The most commonly used induction motors are self starting. For single phase motors starting winding is required. 5. Power Transmission economics - Transmission of electric power by 3$\phi$ system is cheaper than single phase system. 6. Three phase suctification - Rectified3$\phi$ voltage is smoother than single $\phi$ voltage. 7. Higher efficiency - 3$\phi$ systems have higher efficiency and higher power factor when compared to single $\phi$ machine. 8. Rotating magnetic field - 3$\phi$ system can produce a rotating magnetic field which cannot be done by a single phase system. ## Phase Sequence The order in which max value of each phase appears is called phase sequence. It can be RYB (tve) or RBY (-ve) sequence. ## 3 phase winding connections. There are 2 methods of interphase connection: - **Star connection or Y connection** - Star connection is formed by connecting the starting or finishing ends of all 3 windings towards a single point known as neutral or star point. The remaining 3 ends are taken out for connecting the load. - **Delta or mesh connection** - Delta connection is formed by connecting the finishing end of one winding to the starting end of 2nd winding, finishing end of 2nd coil to starting end of 3rd coil, and similarly forming a closed loop. ## Important terms - **Phase voltage Vph:** - Voltage across a phase winding (voltage across phase & neutral ). - **Line voltage VL:** - Voltage across any 2