11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECRICAL CIRCUITS (EENG312) LECTURER’S NOTES 1 ELECTRICAL CIRCUITS ( EENG 312 ) 2024/2025 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECRICAL CIRCUITS (EENG312) LECTURER’S NOTES 2 METHOD OF ASSESSMENT Attendance Assignments Classwork Tests Final Exam TOTAL 5% 5% 5% 15% 70% 100% 1 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRICAL CIRCUITS (EENG312) LECTURER’S NOTES 3 Lesson 1: WAVEFORMS A waveform is a representation of how an alternating voltage or current varies with time (b) (a) FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 4 Simple Sine Waveform Equations The standard form of an alternating voltage e = Em sin θ, or e = Em sin ω t e = Im Z sin ωt, or e = Em sin 2 π f t or e = Em sin ๐๐ t ๐ป Z = √(R2 + ω2L2) Em = Im √(R2 + ω2L2) and the frequency f = ω/2π Hz. 2 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 5 EC-2020/21 Question 7. (7a) An alternating wave has a maximum voltage of 415 volts and produces a current of 25 A when subjected to a capacitive load. At time t=0, the instantaneous values of the voltage and current are 240 v and 15 A respectively while they are increasing positively. If the two waves are operating at 50 Hz, (i) Write the expression for the voltage and current at time t. (ii) Determine the power consumed by the circuit. [10 Marks] Ans. (i) The general expression for an a.c. voltage is given by: v = Vm sin (๐๐ก + ๐) Where ๐ is the phase difference with respect the point where t = 0. For an instantaneous voltage v = 240 v, and maximum voltage V = 415 v, the above equation would become: 240 = 415 sin ((๐๐ก + ๐) At time t = 0, 240 = 415 sin (๐(0) + ๐) 240 = 415 sin ๐ FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES Sin ๐ = , ๐ = sin-1 0.578 = 35.332o (ii) Power: P = IV cos ∅ Hence equation (i) would become 240 = 415 sin (๐๐ + 35.332o) Similarly, the current equation is i.e. 6 i = Im sin (๐๐ก + ๐) 15 = 25 sin (๐๐ก + ๐) Where I and V are the rms values and ∅ is the phase difference between the voltage and current. I = Im/ 2 and V = Vm/ 2 ∅ = (๐ - ๐) P= ๐ = 36.87o Hence the equation: 15 = 25 sin (๐๐ +36.87o) ∗ cos (36.87 – 35.33) P = 5187.5 cos 1.54 P = 5186 W, P = 5.186 kW 3 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 7 EC-2020/21 (7b) Determine the resultant of the following two currents by adding them as waves and verify your answer as vectors addition. i1 = 12 sin ๐t, and i2 = 15 sin (๐t + 60o) [10 Marks] Ans. The resultant sum of the two currents is ir = i1 + i2 ir = 12 sin ๐๐ก + 15 sin (๐๐ก + 60o) ir = 12 sin ๐๐ก + 15 [sin ๐๐กcos60 +sin 60cos ๐๐ก] ir = 12 sin ๐๐ก + 15 [0.5sin ๐๐ก + 0.866cos ๐๐ก] ir = 12 sin ๐๐ก + 7.5 sin ๐๐ก + 13 cos ๐๐ก Z 13 ๐ผ 19.5 Z = 19.52 + 132 = 23.44 ๐ผ = 33.69o ir = 19.5 sin ๐๐ก + 13 cos ๐๐ก FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 8 ir = 19.5 sin ๐๐ก + 13 cos ๐๐ก . ir = 23.44 [ sin ๐๐ก + . . cos ๐๐ก ] ir = 23.44 [cos ๐ผ sin ๐๐ก + sin ๐ผ cos ๐๐ก ] Where cos ๐ผ = Hence and . . and sin ๐ผ = 13 60o . Z 15 ๐ผ 12 19.5 ir = 23.44 sin (๐๐ก + ๐ผ ) ir = 23.44 sin (๐๐ก + 33.69o ) Vector diagram 4 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 9 EC-2020/21 (6b) An e.m.f. whose instantaneous values at time t is applied to a coil (L), having some internal resistance (R). The e.m.f and the current in the circuit are given by: v = 283 sin (314t + ) volts i = 5.66 sin (314t - ) amps Determine: (i) the frequency of the e.m.f. (ii) the resistance and the inductance of the circuit (iii) the power consumed by the load. (iv) sketch the two sine waves. [10 Marks] Ans. (i) The angular velocity ๐ = 2๐f = 314 f = = 49.975 Hz. f ≅ 50 Hz FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES (ii) Lets consider a time t = 0, v = 283 sin [314(0) + ] The impedance Z= v = 283 sin 45o The rms voltage V= Similarly, = √ √ = 200.11∠ i = 5.66 sin [314(0) - ] The rms current √ = . √ Z= 200.11∠ 45o 4∠ −30o Z = 50 ∠ 75๐ Z = 50 cos 75o + j50 sin 75o Z = (12.94 + j48.30 ) Ω Hence the resistance R = 12.94 Ω and i = 5.66 sin I= 45o , 10 = 4∠ -30o The inductive reactance XL = 48.0 Ω 5 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 11 The inductance L= (iv) . = ( ) V I,V L = 0.154 H I (iii) The power consumed by the circuit 0 75o 180o 360o FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRICAL CIRCUITS (EENG312) LECTURER’S NOTES 12 P = I2R P = 42 * 12.94 P = 207 W t Lesson 1: WAVEFORMS Any waveform that is not sinusoidal is a complex wave 6 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRICAL CIRCUITS (EENG322) LECTURER’S NOTES 13 Lesson 2.4. Complex Waveform Fundamental 3rd Harmonic 5th Harmonic Resulting COMPLEX WAVE ๐ 0 2๐ A complex waveform is the result of combining the instantaneous amplitudes of two (or more) sine waves FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRICAL CIRCUITS (EENG312) LECTURER’S NOTES (a) 14 (b) The first harmonic is also called the FUNDAMENTAL (50 Hz) The 2nd harmonic has a frequency of 2 x 50 Hz = 100 Hz The 3rd harmonic has a frequency of 3 x 50 Hz = 150 Hz The 4th harmonics with frequencies 4 x 50 Hz = 200 Hz 7 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURER’S NOTES ๏ ๏ 15 The two halves of the complex wave are identical in shape. In other words, there is no distortion. This is always the case when only odd harmonic (3rd, 5th, 7th, 9th etc.) are present. Fig (b) above. The two halves of the complex wave are not identical. It is always so when even harmonics (2nd, 4th, 6th etc.) are present. Fig (a) above. (C) ๏ Sometimes, a combination of an alternating and direct current ID, flows simultaneously through a circuit. Fig (c) above. FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURER’S NOTES 16 8 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURER’S NOTES 17 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURER’S NOTES 18 9 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURER’S NOTES 19 Causes of HARMONICS waveforms ๏ Harmonics are the result of nonlinear loads that convert AC line voltage to DC. ๏ Harmonics flow into the electrical system because of nonlinear electronic switching devices, such as variable frequency drives (VFDs), computer power supplies, and energy-efficient lighting. ๏ The second harmonic causes a negative sequence ๏ The third harmonic causes zero sequence leading to an increase in the neutral ๏ The fourth harmonic causes more current flow than normal and hence more heat ๏ The fifth harmonic causes an opposite rotation of motors ๏ 3, 6, and 9 do not cancel each other therefore Reactors are used in power systems as a control measure. FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 20 Complex waveforms The instantaneous voltage waveform maybe represented by: e = Em sin (ω t + ∅) volts, where ∅ ๐๐ ๐กโ๐ ๐โ๐๐ ๐ ๐ โ๐๐๐ก. Consider a complex wave which is made up of the fundamental and few harmonics and of which has its own peak values and phase angles as follows: e1 = E1m sin (ω t + ∅๐) e2 = E2m sin (ω t + ∅๐) e3 = E3m sin (ω t + ∅๐) The equation for the instantaneous complex voltage waveform is given by: e = e1 + e2 + e3 ............... en volts en = Enm sin (ω t + ∅๐ง) 10 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 21 Hence e = E1m sin (ω t + ∅๐) + E2m sin (ω t + ∅๐) + E3m sin (ω t + ∅๐) … … Enm sin (ω t + ∅๐ง) Volts With the knowledge that Enm are the maximum voltage values and ∅๐ง are the individual phase shifts Similarly, the complex current waveform would be: i = I1m sin (ω t + ๐๐) + I2m sin (ω t + ๐๐) + I3m sin (ω t + ๐๐) + ….. Inm sin (ω t + ๐๐) ๐จ๐๐๐ Apparently, (∅๐ - ๐๐) is the phase difference between the harmonic voltage and current for the fundamental, (∅๐ - ๐๐) for the second harmonic, (∅3 - ๐3) for the third harmonic, (∅๐ง - ๐n) for the nth harmonic. FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG322) LECTURE NOTES 22 Harmonics in a single-phase circuits When a complex alternating voltage wave, i.e., one containing harmonics, is applied to a single-phase circuit containing resistance, inductance and/or capacitance (i.e., linear circuit elements), then the resulting current will also be complex and contain harmonics. PURELY RESISTIVE LOADS The impedance of a pure resistance R is independent of frequency and the current and voltage are in phase for each harmonic. Thus the general expression for current i is given by The percentage second harmonic content to the fundamental is given by: ๐ฝ๐๐ ๐น i.e 100% ๐ฝ๐๐ * 100% ๐น * 11 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES Example 1. Determine the instantaneous current flowing in the circuit below. 10Ω i v = 50 sin ๐ + 25 sin 2๐ + 30 sin 3๐ ๐ฃ๐๐๐ก๐ Ans. 1. 50 sin θ … … … … . fundamental voltage 25 sin 2๐ … … … . . 2nd harmonic voltage 30 sin 3๐ …………. 3rd harmonic voltage 23 Using KVL i=Σ , i= i= sin ๐ + sin 2๐ + + + sin 3๐ i = 5 sin ๐ + 2.5 sin 2๐ + 3 sin 3๐ Amps The percentage 3rd harmonic current to the fundamental is: * 100% = 60% FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 24 PURELY INDUCTIVE LOADS ๏ Consider a voltage wave without harmonics as v = Vm sin ๐๐ก. ๏ If such voltage is applied to a pure inductor (L), the current wave will be i = Im sin (๐๐ก - ) ๏ If the applied voltage is complex, given by: v = Vm sin ๐๐ก + V2m sin 2๐๐ก + V3m sin 3๐๐ก + ……………….. ๏ The impedance of a pure inductance L, i.e., inductive reactance XL = ๐L, varies with the harmonic frequency when voltage v is applied to it. ๏ The current i is given by: ๏ Also, for every harmonic term, the current will lag the voltage by 90° or /2 rad. 12 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 25 Example 1. For the fundamental. Determine the instantaneous current flowing in the XL1 = 2๐ ∗ 50 ∗ 15.9 circuit below at 50 Hz frequency. X = 5Ω 7.96 mH i L1 For the 2nd harmonic frequency XL2 = 2๐๐ฟ XL2 = 2 * 5 = 10 Ω For the 3rd harmonic frequency XL3 = 3๐๐ฟ XL3 = 3 * 5 = 15 Ω v = 50 sin ๐ + 25 sin 2๐ + 30 sin 3๐ ๐ฃ๐๐๐ก๐ i = sin (๐ − ) + sin (2๐ − ) + sin (3๐ − ) Ans. The inductive reactance XL = 2๐๐๐ฟ = ๐๐ฟ i = 10 sin (๐ − ) + 2.5 sin (2๐ − ) + 2 sin (3๐ − ) Note. ๐ฝ = ๐๐ FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 26 PURELY CAPACITIVE LOADS ๏ The impedance of a pure capacitance C, i.e., capacitive reactance XC = 1/2๐fC, varies with the harmonic frequency when voltage v is applied to it. ๏ The current i is given by: Hence, ๏ Notice that for each harmonic term the current will lead the voltage by 90° or /2 rad. 13 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES Example 1. Determine the instantaneous current flowing in the circuit below at 50 Hz frequency. 25 ๐F i v = 50 sin ๐ + 25 sin 2๐ + 30 sin 3๐ ๐ฃ๐๐๐ก๐ 27 For the fundamental. ๐๐ 1 = (2๐ ∗ 50 ∗ 0.159 ∗ 10 ๐๐ 1 = 0.05 Ω For the 2nd harmonic frequency ๐๐2 = 2 * ๐๐ 1 ๐๐2 = 2 * 0.05 = 0.10 Ω For the 3rd harmonic frequency ๐๐3 = 3 * ๐๐ 1 ๐๐3 = 3 * 0.05 = 0.15 Ω i = (50*0.05) sin (๐ + ) ) + (25*0.1) sin (2๐ + ) + (30*0.15) sin (3๐ + ) Ans. i = 2.5 sin (๐ + ) + 2.5 sin (2๐ + The capacitive reactance + 4.5 sin (3๐ + ) A Note. ๐ฝ = ๐๐ XC = 1/2๐๐๐ถ = 1/๐๐ถ FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES Example 4. Determine the instantaneous current flowing in the circuit below. 12 Ω ) 16 mH 170 ๐F i v = 250 sin 377t + 50 sin (1885t + π/3) v Ans. The angular displacement ๐ = 377 rad/s 28 The inductive reactance XL = ๐๐ฟ, XL = 377 * 16 * 10-3, XL = 6.032 Ω The capacitive reactance XC = 1/๐๐, XC = 1 / (377 * 170 * 10-6 ) XC = 1 / (0.0641), Xc = 15.6 Ω The impedance, Z = √[๐ 2 + ๐๐ถ − ๐๐ฟ 2] Z = √[122 + 15.6 − 6.03 2] Z = √[122 + 9.57 2] Z = √[144 + 91.585] Z = 235.585 , Z1 = 15.349 Ω ๐1 = tan-1 . = -38.57o ๐1 = -38.57 = -0.673 rad 14 11/1/2024 FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES For the next harmonic frequency ๐5 = tan-1 Hence, its inductive reactance XL5 = 5๐๐ฟ, XL5 = 5 ∗ 6.032, XL5 = 30.16 Ω ๐5 = 66.07o 1885 = 5th 377 Its capacitive reactance XC5 = 1/5๐๐, XC5 = 1 / (5*377 * 170 * 10-6 ) XC5 = 1 / (0.32), XC5 = 3.125 Ω The 5th harmonic impedance Z5 = √[๐ 2 + ๐๐ฟ5 − ๐๐ถ5 2] Z5 = √[122 + 30.16 − 3.125 2] Z5 = √[122 + 27.035 2] Z5 = √[144 + 730.89] Z5 = 874.89 , Z5 = 29.58 Ω 29 . [66.07 = 1.15 rad ] i= i= i= sin(377t + ๐1 ) + . sin(377t + 0.673) + sin(1885t + π/3- ๐5 ) . sin(1885t + 1.05 – 1.15) i = 16.29 sin(377t + 0.673) + 1.69 sin(1885t – 0.1) FOURAH BAY COLLEGE ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT ELECTRIC CIRCUITS (EENG312) LECTURE NOTES 30 Homework 1. Determine the instantaneous current (i) flowing in the circuit below. 20 Ω 35 mH 200 ๐F i To be submitted on Tuesday, Nov. 5 v = 50 sin (1885t + π/6) + 250 sin 377t v Ans. 15
