Principles-and-Applications-of-Electrica - G Rizzoni...

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Unformatted text preview: G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Chapter 1 Instructor Notes Chapter 1 is introductory in nature, establishing some rationale for studying electrical engineering methods, even though the students' primary interest may lie in other areas. The material in this chapter should be included in every syllabus, and can typically be thoroughly covered in a single-day introductory lecture. Oftentimes, reading of this material is left up to the discretion of the student. 1.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Chapter 1 problem solutions treadmill _________________________________ 1.1 Miscellaneous A few examples are: lawn tools Bathroom power tools ventilation fan electric toothbrush _________________________________ hair dryer ________________________________ electric shaver 1.2 Several examples are listed below for each system: electric heater fan Kitchen microwave fan a) microwave turntable A ship Circuit Analysis mixer design of the ship's food processor electrical system blender Electromagnetics coffee grinder radar garbage disposal Solid-State Electronics ceiling fan radio electric clock sonar exhaust fan Electric Machines refrigerator compressor pump dish washer elevator Utility Room Electric Power Systems clothes washer lighting dryer generators air conditioner Digital Logic Circuits furnace blower elevator control pump Computer Systems Family Room navigation VCR drive Communication Systems cassette tape drive radio reel-to-reel tape drive telephone record turntable drive Electro-Optics computer fan Morse light bridge displays 1.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Control Systems Instrumentation compass rudder speed indicator flaps Control Systems c) rudder Household Circuit Analysis HVAC design of the home's electrical system b) A Commercial Passenger Aircraft Electromagnetics Circuit Analysis Design of the plane's microwave oven electrical system stereo speakers Solid-State Electronics Electromagnetics radar television microwave oven stereo VCR Solid-State Electronics Electric Machines radio appliances Electric Machines turbines power tools fans fans Electric Power Systems Electric Power Systems lighting lighting HVAC HVAC receptacles Digital Logic Circuits Digital Logic Circuits seat belts Computer Systems clocks navigation timers Computer Systems Communication Systems radio microwave oven telephone programmable VCR Electro-Optics Communication Systems cockpit displays telephone Instrumentation compass CB radio air speed indicator television inclinometer radio Electro-Optics altimeter 1.3 G. Rizzoni, Principles and Applications of Electrical Engineering digital clocks Instrumentation electric meter Control Systems thermostat _________________________________ _________________________________ 1.3 Some examples are: a) HVAC lighting office equipment typewriter computer copy machine clock stapler shredder elevator b) conveyor punch press lighting ventilation drill press hoist lathe c) power saw drill lighting elevator pump compressor _________________________________ 1.4 Problem solutions, Chapter 1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Chapter 2 Instructor Notes Chapter 2 develops the foundations for the first part of the book. Coverage of the entire Chapter would be typical in an introductory course. The first four sections provide the basic definitions and cover Kirchoff’s Laws and the passive sign convention; the box Focus on Methodology: The Passive Sign Convention (p. 35) and two examples illustrate the latter topic. The sidebars Make The Connection: Mechanical Analog of Voltage Sources (p. 20) and Make The Connection: Hydraulic Analog of Current Sources (p. 22) present the concept of analogies between electrical and other physical domains; these analogies will continue through the first six chapters. Sections 2.5and 2.6 introduce the i-v characteristic and the resistance element. Tables 2.1 and 2.2 on p. 41 summarize the resistivity of common materials and standard resistor values; Table 2.3 on p. 44 provides the resistance of copper wire for various gauges. The sidebar Make The Connection: Electric Circuit Analog of Hydraulic Systems – Fluid Resistance (p. 40) continues the electric-hydraulic system analogy. Finally, Sections 2.7 and 2.8 introduce some basic but important concepts related to ideal and nonideal current sources, and measuring instruments. The Instructor will find that although the material in Chapter 2 is quite basic, it is possible to give an applied flavor to the subject matter by emphasizing a few selected topics in the examples presented in class. In particular, a lecture could be devoted to resistance devices, including the resistive displacement transducer of Focus on Measurements: Resistive throttle position sensor (pp. 52-54), the resistance strain gauges of Focus on Measurements: Resistance strain gauges (pp. 54-55), and Focus on Measurements: The Wheatstone bridge and force measurements (pp. 55-56). The instructor wishing to gain a more in-depth understanding of resistance strain gauges will find a detailed analysis in1. Early motivation for the application of circuit analysis to problems of practical interest to the nonelectrical engineer can be found in the Focus on Measurements: The Wheatstone bridge and force measurements. The Wheatstone bridge material can also serve as an introduction to a laboratory experiment on strain gauges and the measurement of force (see, for example2). Finally, the material on practical measuring instruments in Section2.8b can also motivate a number of useful examples. The homework problems include a variety of practical examples, with emphasis on instrumentation. Problem 2.36 illustrates analysis related to fuses; problems 2.44-47 are related to wire gauges; problem 2.52 discusses the thermistor; problems 2.54 and 2.55 discuss moving coil meters; problems 2.52 and 2.53 illustrate calculations related to temperature sensors; an problems 2.56-66 present a variety of problems related to practical measuring devices. It has been the author's experience that providing the students with an early introduction to practical applications of electrical engineering to their own disciplines can increase the interest level in the course significantly. Learning Objectives 1. Identify the principal elements of electrical circuits: nodes, loops, meshes, branches, and voltage and current sources. 2. Apply Kirchhoff’s Laws to simple electrical circuits and derive the basic circuit equations. 3. Apply the passive sign convention and compute power dissipated by circuit elements. 4. Apply the voltage and current divider laws to calculate unknown variables in simple series, parallel and series-parallel circuits. 5. Understand the rules for connecting electrical measuring instruments to electrical circuits for the measurement of voltage, current, and power. 1 E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990. 2 G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rd Edition, Kendall-Hunt, 1998. 2.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Section 2.1: Definitions Problem 2.1 Solution: Known quantities: Initial Coulombic potential energy, potential energy, Vi = 17kJ / C ; initial velocity, U i = 93M m ; final Coulombic s V f = 6kJ / C . Find: The change in velocity of the electron. Assumptions: ∆PE g << ∆PE c Analysis: Using the first law of thermodynamics, we obtain the final velocity of the electron: Qheat − W = ∆KE + ∆PE c + ∆PE g + ... Heat is not applicable to a single particle. W=0 since no external forces are applied. ∆KE = −∆PE c 1 me (U 2f − U i2 ) = −Qe (V f − Vi ) 2 2Q U 2f = U i2 − e (V f − Vi ) me ( ) m · 2 − 1.6 × 10 −19 C § = ¨ 93 M ¸ − (6kV − 17kV ) 9.11× 10 −37 g s¹ © 2 2 m2 15 m − 3 . 864 × 10 s2 s2 m U f = 6.917 × 10 7 s m m m U f − U i = 93 M − 69.17 M = 23.83 M . s s s = 8.649 × 1015 ________________________________________________________________________ Problem 2.2 Solution: Known quantities: MKSQ units. Find: Equivalent units of volt, ampere and ohm. 2.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: Joule J V= Coulomb C Coulomb C Current = Ampere = a= second s Volt Joule × second Resistance = Ohm = = Ampere Coulomb 2 Voltage = Volt = Conductance = Siemen or Mho = Ω= J ⋅s C2 Ampere C 2 = Volt J ⋅s ________________________________________________________________________ Problem 2.3 Solution: Known quantities: Battery nominal rate of 100 A-h. Find: a) Charge potentially derived from the battery b) Electrons contained in that charge. Assumptions: Battery fully charged. Analysis: a) C· s · § § 100 A × 1hr = ¨100 ¸(1hr )¨ 3600 ¸ s¹ hr ¹ © © = 360000 C b) charge on electron: − 1.602 × 10 −19 C no. of electrons = 360 × 103 C = 224.7 × 10 22 −19 1.602 × 10 C ________________________________________________________________________ Problem 2.4 Solution: Known quantities: Two-rate change charge cycle shown in Figure P2.4. 2.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: a) The charge transferred to the battery b) The energy transferred to the battery. Analysis: a) To find the charge delivered to the battery during the charge cycle, we examine the charge-current relationship: i= dq dt or dq = i ⋅ dt thus: t1 Q = ³ i (t )dt t0 5hrs 10 hrs Q = ³ 50mAdt + ³ 20mAdt 0 5hrs 18000 s 36000 s = ³ 0.05dt + ³ 0.02dt 0 18000 = 900 + 360 = 1260C b) To find the energy transferred to the battery, we examine the energy relationship p= dw dt or dw = p (t )dt t1 t1 t0 t0 w = ³ p (t )dt = ³ v(t )i (t )dt observing that the energy delivered to the battery is the integral of the power over the charge cycle. Thus, 18000 w= ³ 36000 0.05(1 + 0 = (0.05t + 0.25t 0.75t ) dt ) dt + ³ 0.02(1 + 18000 18000 18000 0.25 2 36000 0.75 2 18000 t ) 0 + (0.02t + t ) 18000 36000 36000 w = 1732.5 J ________________________________________________________________________ Problem 2.5 Solution: Known quantities: Rated voltage of the battery; rated capacity of the battery. Find: a) The rated chemical energy stored in the battery b) The total charge that can be supplied at the rated voltage. 2.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: a) ∆PE c ∆Q I= ∆Q ∆t Chemical energy = ∆PE c ∆V ≡ = ∆V ⋅ ∆Q = ∆V ⋅ (I ⋅ ∆t ) = 12 V 350 a hr 3600 s hr = 15.12 MJ . As the battery discharges, the voltage will decrease below the rated voltage. The remaining chemical energy stored in the battery is less useful or not useful. b) ∆Q is the total charge passing through the battery and gaining 12 J/C of electrical energy. ∆Q = I ⋅ ∆t = 350 a hr = 350 C s hr ⋅ 3600 s hr = 1.26 MC. ________________________________________________________________________ Problem 2.6 Solution: Known quantities: Resistance of external circuit. Find: a) Current supplied by an ideal voltage source b) Voltage supplied by an ideal current source. Assumptions: Ideal voltage and current sources. Analysis: a) An ideal voltage source produces a constant voltage at or below its rated current. Current is determined by the power required by the external circuit (modeled as R). I= Vs R P = Vs ⋅ I b) An ideal current source produces a constant current at or below its rated voltage. Voltage is determined by the power demanded by the external circuit (modeled as R). V = Is ⋅ R P = V ⋅ Is A real source will overheat and, perhaps, burn up if its rated power is exceeded. ________________________________________________________________________ 2.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Sections 2.2, 2.3: KCL, KVL Problem 2.7 Solution: Known quantities: Circuit shown in Figure P2.7 with currents I 0 = −2 A, I1 = −4 A, I S = 8 A, and voltage source VS = 12 V. Find: The unknown currents. Analysis: Applying KCL to node (a) and node (b): ­°I 2 = −(I 0 + I1 ) = 6 A ­I 0 + I1 + I 2 = 0 Ÿ ® ® °¯I 3 = I 0 + I S + I1 = 2 A ¯ I 0 + I S + I1 − I 3 = 0 ________________________________________________________________________ 2.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Section 2.4: Sign Convention Focus on Methodology: Passive Sign Convention 1. 2. 3. 4. Choose an arbitrary direction of current flow. Label polarities of all active elements (current and voltage sources). Assign polarities to all passive elements (resistors and other loads); for passive elements, current always flows into the positive terminal. Compute the power dissipated by each element according to the following rule: If positive current flows into the positive terminal of an element, then the power dissipated is positive (i.e., the element absorbs power); if the current leaves the positive terminal of an element, then the power dissipated is negative (i.e., the element delivers power). Problem 2.8 Solution: Known quantities: Direction and magnitude of the current through the elements in Figure P2.8; voltage at the terminals. Find: Class of the components A and B. Analysis: The current enters the negative terminal of element B and leaves the positive terminal: its coulombic potential energy increases. Element B is a power source. It must be either a voltage source or a current source. The reverse is true for element A. The current loses energy as it flows through element A. Element A could be 1. a resistor or 2. a power source through which current is being forced to flows ‘backwards’. ________________________________________________________________________ Problem 2.9 Solution: Known quantities: Current absorbed by the heater; voltage at which the current is supplied; cost of the energy. Find: a) Power consumption b) Energy dissipated in 24 hr. c) Cost of the Energy Assumptions: The heater works for 24 hours continuously. Analysis: a) 2.7 G. Rizzoni, Principles and Applications of Electrical Engineering P = VI = 110 V (23 a ) = 2.53 K Problem solutions, Chapter 2 J a = 2.53 KW a s b) W = Pt = 2.53 K J s 24 hr 3600 = 218.6 MJ hr s c) Cost = ( Rate)W = 6 cents (2.53 KW )(24 hr ) = 364.3 cents = $3.64 KW hr ________________________________________________________________________ Problem 2.10 Solution: Known quantities: Current through elements A, B and C shown in Figure P2.10; voltage across elements A, B and C. Find: Which components are absorbing power, which are supplying power; verify the conservation of power. Analysis: A absorbs (35 V )(15 A) = 525 W B absorbs (15 V )(15 A) = 225 W C supplies (50 V )(15 A) = 750 W Total power supplied = PC = 750 W Total power absorbed = PB + PA = 225 W + 525 W = 750 W Total power supplied = Total power absorbed, so conservation of power is satisfied. ________________________________________________________________________ Problem 2.11 Solution: Known quantities: Vs = 12V ; internal resistance of the source, Rs = 5kΩ ; and resistance of the load, RL = 7kΩ . Circuit shown in Figure P2.11 with voltage source, Find: The terminal voltage of the source; the power supplied to the circuit, the efficiency of the circuit. Assumptions: Assume that the only loss is due to the internal resistance of the source. 2.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: KVL : − V S + I T RS + VT = 0 OL : VT = I T R L ∴ I T = − VS + VT = VT RL VT RS + VT = 0 RL VS 12 V = = 7V 5 KΩ RS 1+ 1+ 7 KΩ RL or VD : VT = VS R L 12 V 7 KΩ = = 7 V. R S + R L 5 KΩ + 7 KΩ VR2 VT2 (7 V ) PL = = = = 7 mW V RL RL K 7 a P P I 2R 7 KΩ RL = 2 T L2 = η = out = = 0.5833 or 58.33% . Pin PRS + PRL I T RS + I T RL 5 KΩ + 7 KΩ 2 ________________________________________________________________________ Problem 2.12 Solution: Known quantities: Circuit shown in Figure P2.12; Current through elements D and E; voltage across elements B, C and E. Find: a) Which components are absorbing power and which are supplying power b) Verify the conservation of power. Analysis: a) By KCL, the current through element B is 5 A, to the right. By KVL, v D = v E = 10 V (positive at the top) v A + 5 − 10 − 10 = 0 Therefore the voltage across element A is v A = 15 V (positive on top) A supplies (15 V )(5 A) = 75 W B supplies (5 V )(5 A) = 25 W C absorbs (10 V )(5 A) = 50 W D absorbs (10 V )(4 A) = 40 W E absorbs (10 V )(1 A) = 10 W b) Total power supplied = PA + PB = 75 W + 25 W = 100 W 2.9 G. Rizzoni, Principles and Applications of Electrical Engineering Total power absorbed Problem solutions, Chapter 2 = PC + PD + PE = 50 W + 40 W + 10 W = 100 W Total power supplied = Total power absorbed, so conservation of power is satisfied. ________________________________________________________________________ Problem 2.13 Solution: Known quantities: Headlights connected in parallel to a 24-V automotive battery; power absorbed by each headlight. Find: Resistance of each headlight; total resistance seen by the battery. Analysis: Headlight no. 1: P = v × i = 100 W = R= v2 or R v 2 576 = = 5.76Ω 100 100 Headlight no. 2: P = v × i = 75 W = R= v2 or R v 2 576 = = 7.68Ω 75 75 The total resistance is given by the parallel combination: 1 RTOTAL = 1 1 or RTOTAL = 3.29 Ω + 5.76Ω 7.68Ω ________________________________________________________________________ Problem 2.14 Solution: Known quantities: Headlights and 24-V automotive battery of problem 2.13 with 2 15-W taillights added in parallel; power absorbed by each headlight; power absorbed by each taillight. Find: Equivalent resistance seen by the battery. Analysis: The resistance corresponding to a 75-W headlight is: R 75W = v 2 576 = = 7.68Ω 75 75 For each 15-W tail light we compute the resistance: 2.10 G. Rizzoni, Principles and Applications of Electrical Engineering R 15W = Problem solutions, Chapter 2 v 2 576 = = 38.4Ω 15 15 Therefore, the total resistance is computed as: 1 RTOTAL = 1 1 1 1 or RTOTAL = 3.2 Ω + + + 7.68Ω 7.68Ω 38.4Ω 38.4Ω ________________________________________________________________________ Problem 2.15 Solution: Known quantities: Circuit shown in Figure P2.15 with voltage source, Vs = 20V ; and resistor, Ro = 5Ω . Find: The power absorbed by variable resistor R (ranging from 0 to 20 Ω). Analysis: The current flowing clockwise in the series circuit is: i= 20 5+ R The voltage across the variable resistor R, positive on the left, is: v R = Ri = 20 R R+5 2 § 20 · Therefore, PR = v R i = ¨ ¸ R ©5+ R ¹ ________________________________________________________________________ 2.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.16 Solution: Known quantities: Circuit shown in Figure P2.16 with source voltage, Vs = 12V ; internal resistance of the source, Rs = 0.3Ω . Current, I T = 0, 5, 10, 20, 30 A. Find: a) b) c) d) The power supplied by the ideal source as a function of current The power dissipated by the nonideal source as a function of current The power supplied by the source to the circuit Plot the terminal voltage and power supplied to the circuit as a function of current Assumptions: There are no other losses except that on Rs. Analysis: a) Ps = power supplied by the source = VS I S = VS I T . b) Rs = equivalent resistance for internal losses Ploss = I T2 RS c) VT = voltage at the battery terminals: VD : VT = VS − RS I T P0 = power supplied to the circuit ( RL in this case) = I T VT . Conservation of energy: PS = Ploss + P0 . I T ( A) PS (W ) Ploss (W ) VT (V ) P0 (W ) 0 2 5 10 20 30 0 30 60 120 240 360 0 1.875 7.5 30 120 270 0 11.4 10.5 9 6 3 0 28.13 52.5 90 120 90 2.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Terminal Voltage 12 11 10 9 t V (V) 8 7 6 5 4 3 0 5 10 15 It (A) 20 25 30 400 350 300 2...
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