AC Circuits: Instructional Module for Electrical Engineering
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Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 College: Engineering Campus: Bambang DEGREE PROGRAM SPECIALIZATION YEAR LEVEL I. Bachelor of Science in Electrical Engineering nd 2 Year COURSE NO. EE 02 COURSE TITLE TIME FRAME Electrical Circuits 2 WK NO. IM NO. 01 UNIT TITLE/CHAPTER TITLE AC Circuit II. LESSON TITLE 1. 2. 3. 4. III. Generation of AC voltage Frequency and Electrical degrees RMS value of an AC AC phase difference LESSON OVERVIEW This module will provide students with an explanations of the specific characteristics of AC waveform and give students the skills necessary to calculate the current and voltage in a simple AC resistive circuit. IV. DESIRED LEARNING OUTCOMES At the end of the lesson, the students should be able to: 1. 2. 3. 4. 5. V. Identify a sinusoidal waveform and measure its characteristics Describe how sine waves are generated Calculate the voltage and current values of sine waves Describe angular relationships of sine waves Apply the basic circuit laws to AC resistive circuits COURSE CONTENT ALTERNATING CURRENTINTRODUCTION AC is either a voltage or current that alternates. The alternating effect is due to the way that it is generated. When AC is graphed out it follows a sinusoidal waveform. The positive and negative sides of the graph represent one direction then the other GENERATION OF ALTERN ATING EMF AC is produced using laws and properties of magnets and their relationship to electricity. Metal rods are rotated in a magnetic field. This is what generates AC. There are different ways to generate AC. One set up is to have permanent magnets with the north and south poles facing each other. The attraction between the two poles creates a magnetic field between the magnets. A loop of wire or armature is NVSU-FR-ICD-05-00 (081220) Page 1 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 rotated inside of the magnetic field. The rotation causes the magnetic flux to change within the wire, this generates electricity. To improve the amount of electricity that is produced, more magnets or loops can be added. A permanent magnet that’s placed center and is rotated, if windings of wire are placed close to the magnet, will produce electricity in the windings. Emf can be developed in a coil of wire in one of three ways: 1. By changing the flux through a coil 2. By moving a coil through a magnetic field 3. By altering the direction of the flux with respect to the coil The induced emf can be calculated as π=π΅ π π± π π where : Nππ· ππ‘ - number of turns in the coil rate at which the flux changes through the coil (unit: Weber) 1 weber= 108 Maxwells If there is a physical motion of coil or magnet, generated voltage is equal to π = π·ππ where βlv- flux density ( unit :Tesla) 1 Tesla = 1 Weber/m2 length of wire (unit: m) velocity of the wire (unit: m/sec) Calculate the average voltage induced in a coil of 300 turns, through which the flux changes from 250, 000 maxwells to 20, 000 Maxwells in 0.15 sec. Solution: π = 300 π‘ππ’ππ ππ = 250, 000 πππ₯π€ππππ − 20, 000 πππ₯π€ππππ = 230, 000 πππ₯π€ππππ ππ‘ = 15 π ππ π=π ππ· ππ‘ = 300 230,000 πππ₯π€ππππ 1 πππππ π₯ 8 = 4.6 ππππ‘π 0.15 π ππ 10 πππ₯π€πππ How many turns of wire are there in a coil in which 35. 7 volts are induced when the flux through it increases uniformly at the rate of 3 x10 6 maxwells per second? NVSU-FR-ICD-05-00 (081220) Page 2 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Solution: π = 35.7 π ππ· = 3π₯106 πππ₯π€ππππ πππ π ππ ππ‘ π= π÷ ππ· ππ‘ = 35.7π ÷ 3π₯106 πππ₯π€πππ 1 πππππ π₯ 8 = 1190π‘π’πππ π ππ 10 πππ₯π€πππ GENERATION OF A SINE WAVE OF A VOLTAGE The voltage developed in a coil of generator changes in magnitude from instant to instant as varying flux are cut per second and in direction as coil sides changes positions under north and south poles, implies that an alternating emf is generated. Thus, an alternating current (AC) follows a sinusoidal waveform. The current/voltage starts at zero and increases to its maximum peak value. Then it drops back down to zero and goes down to the negative value of its max peak value. The emf generated is equal to π = π¬πππππΆ = π¬π πππ ππ = π¬π πππ (ππ ππ) Figure 1. Definitions The sinusoidal waveform in Fig.1 with its additional notation will now be used as a model in defining a few basic terms. NVSU-FR-ICD-05-00 (081220) Page 3 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Waveform: The path traced by a quantity, such as the voltage in Fig. 2, plotted as a function of some variable, such as time (as above), position, degrees, radians, temperature, and so on. Instantaneous value: The magnitude of a waveform at any instant of time; denoted by lowercase letters (e1, e2). Peak amplitude: The maximum value of a waveform as measured from its average, or mean, value, denoted by uppercase letters [such as Em for sources of voltage and Vm for the voltage drop across a load]. For the waveform in the average value is zero volts, and Em is as defined by the figure. Peak value: The maximum instantaneous value of a function as measured from the zero volt level. For the waveform the peak amplitude and peak value are the same since the average value of the function is zero volts. Peak-to-peak value: Denoted by Epp or Vpp, the full voltage between positive and negative peaks of the waveform, that is, the sum of the magnitude of the positive and negative peaks Periodic waveform: A waveform that continually repeats itself after the same time interval. The waveform in Fig.2 is a periodic waveform. Period (T): The time of a periodic waveform. Cycle: The portion of a waveform contained in one period of time. Figure 3 Frequency ( f ): The number of cycles that occur in 1 s. The frequency of the waveform in Fig. 4a is 1 cycle per second, and for Fig. 4b is 2.5 cycles per second. If a waveform of similar shape had a period of 0.5 s. Figure 4c the frequency would be 2 cycles per second. Figure 4 NVSU-FR-ICD-05-00 (081220) Page 4 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 The unit of measure for frequency is the hertz (Hz), where 1 hertz (Hz) = 1 cycle per second (cps) π»= π π π= π π» Given the period T= 1 s what is the frequency? Solution: π= π π» = π ππ so the frequency is 1 Hz What is the period when the frequency is 25 Hz, 50 Hz, and 100 kHz? Solution: π= 1 π = 1 = 0.04 s 25π»π§ π= 1 π = 1 = 0.02 s 50π»π§ π= 1 π = 1 = 100ππ»π§ 0.00001 s or 10 microseconds 1. Calculate the period if frequency is: a. 455 Hz b. 689 Hz c. 967 Hz 2. Find the frequency if period is : a. .005 s b. 65 ms c. .04 ms Frequency And Electrical Degrees An ac generator having two poles will develop one cycle of voltage variations for each revolution of the rotating machine. If rotation occurs at rhe rate of one revolution per second, the voltage has a frequency of one cycle per second. If each cycle is generated for every pair of pole, then for a multipolar alternator will generate P/2 cps per revolution and the frequency will be (P/2)x (rps) cycles per second. Or f will be π= NVSU-FR-ICD-05-00 (081220) π· π πππ π Page 5 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 or if the rotating speed is in revolution per minute(rpm) π= π· π πππ πππ Sinusoidal Alternating Currents The Sinusoidal Waveform The sinusoidal waveform is the only alternating waveform whose shape is unaffected by the response characteristics of R, L, and C elements. The unit of measurement for the horizontal axis can be time (as appearing in the figures thus far), degrees, or radians. The term radian can be defined as follows: If we mark off a portion of the circumference of a circle by a length equal to the radius of the circle, as shown in Fig. 5 the angle resulting is called 1 radian. The result is 1 rad = 57.296° Figure 5 For comparison purposes, two sinusoidal voltages are plotted in Fig. 6 using degrees and radians as the units of measurement for the horizontal axis. It is of particular interest that the sinusoidal waveform can be derived from the length of the vertical projection of a radius vector rotating in a uniform circular motion about a fixed point. NVSU-FR-ICD-05-00 (081220) Page 6 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Angular velocity( ω) -refers to how quickly the wave moves, it is directly proportional to the frequency. π = ππ π (radians per second.) the angle(θ) that has been covered in a particular amount of time is equal to π½ = ππ Determine the angular velocity of a sine wave having a frequency of 60 Hz. Solution: π = ππ π = 2 ∗ π ∗ 60 π»π§ = 377 πππ/π ππ If the frequency is 350 Hz. Find the angular velocity, then find the angle covered in 2 seconds. Solution: π = 2 ∗ π ∗ 350 π»π§ = 2199.11 ππππ /π ππ π = 2199.11 ππππ /π ππ ∗ 2 = 4398.22 ππππ Given π = 200 rad/s, determine how long it will take the sinusoidal waveform to pass through an angle of 90°. Solution: π = 200 πππ π ππ π 2 π = 90 πππππππ = ( ππ πππππππ ) π πππ π½(ππ πππ ) 2 π= = = π. πππππ−π πππ π πππ πππ /πππ NVSU-FR-ICD-05-00 (081220) Page 7 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Find the angle through which a sinusoidal waveform of 60 Hz will pass in a period of 5 ms. Solution: π‘ = 5 ππ π = 60 π»π§ π = 2ππ = 2 ∗ π ∗ 60 = 377 πππ/π ππ π½ = ππ‘ = 377 πππ 180 πππππππ ∗ 5ππ ππ = 1.885 πππ ∗ = πππ π ππππππ π ππ π πππ Find the angular velocity when the angle is covered in .5 seconds. d. 699 Hz e. 357 Hz f. 865 Hz GENERAL FORMAT FOR THE SINUSOIDAL VOLTAGE OR CURRENT The basic mathematical format for the sinusoidal waveform is π¨πππππΆ where Am is the peak value of the waveform and a is the unit of measure for the horizontal axis, as shown in Fig 6. The equation πΆ = ππ states that the angle a through which the rotating vector in Fig. 6 will pass is determined by the angular velocity of the rotating vector and the length of time the vector rotates. Figure 6. For electrical quantities such as current and voltage, the general format is π = π¬π ππππΆ π = π°π ππππΆ NVSU-FR-ICD-05-00 (081220) Page 8 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 π= Where: πΌπ = π π¬πππππΆ = = π°π ππππΆ πΉ πΉ πΈπ π Given π = 5 π ππ πΌ, determine e at πΌ = 40° and πΌ = 0.8π Solution: a. π = 5 π ππ πΌ = 5 sin(40) = 3.2139 π b. π = 5 π ππ πΌ = 5 sin(0.8π) = 2.93 π Note! if the given value for α is in degrees set your calculator in degree(D) mode and if α is in rad set your calculator in to rad(R) mode Given π = 6 π₯10−3 π ππ 1000π‘, determine i at π‘ = 2 ππ . Solution: Since the ω is in rad/sec in the equation π = πΌππ ππππ‘, set your calculator to rad mode π = 6 π₯10−3 π ππ 1000π‘ = 6 π₯10−3 π ππ 1000(2 π₯10−3 ) = π. ππ ππ¨ Or you can convert the rad unit to degree and set your calculator to D mode π = 6 π₯10−3 π ππ 1000π‘ = 6 π₯10−3 π ππ 1000(2 π₯10−3 )π₯ 180 = π. ππ ππ¨ π πππ PHASE RELATION Thus far, we have considered only sine waves that have maxima at π/2 and 3π/2, with a zero value at 0, π, and 2π. If the waveform is shifted to the right or left of 0°, the expression becomes π¨ππππ(ππ ± π½) where θ is the angle in degrees or radians that the waveform has been shifted. If the waveform passes through the horizontal axis with a positive going (increasing with time) slope before 0°, as shown in Fig.7, the expression is π¨ππππ(ππ + π½) NVSU-FR-ICD-05-00 (081220) Page 9 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Figure 7 If the waveform passes through the horizontal axis with a positive-going slope after 0°, as shown in Fig. 8, the expression is π¨ππππ(ππ − π½) Figure 8 The terms leading and lagging are used to indicate the relationship between two sinusoidal waveforms of the same frequency plotted on the same set of axes. In Fig. 9 the cosine curve is said to lead the sine curve by 90°, and the sine curve is said to lag the cosine curve by 90°. The 90° is referred to as the phase angle between the two waveforms. In language commonly applied, the waveforms are out of phase by 90°. Note that the phase angle between the two waveforms is measured between those two points on the horizontal axis through which each passes with the same slope. If both waveforms cross the axis at the same point with the same slope, they are in phase. Figure 9 NVSU-FR-ICD-05-00 (081220) Page 10 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 The geometric relationship between various forms of the sine and cosine functions can be derived as πππ πΌ = π ππ (πΌ + 90°) π πππΌ = πππ (πΌ − 90°) −π πππΌ = π ππ (πΌ ± 180°) −πππ πΌ = π ππ (πΌ − 90°) An angular measurement that tells the position of a sine wave with respect to something, usually the origin, is referred to as phase. Basically the phase is the angular offset of one wave relative to another. One wave could lead or lag another. 0 o 1 0 0 o Figure 10 Figure 10 shows that wave 1 leads wave 2 by 100o, or you can say that wave 2 lags by 100o 1 0 0 0 o o Figure 11 Figure above shows that wave 1 leads wave 2 by 100o, or wave 2 lags wave 1 by 100o. What is the phase shift between wave Y and Z NVSU-FR-ICD-05-00 (081220) Page 11 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Figure 12 The difference between the two waves is 45 o, and wave Y is before Z. Therefore ,Wave Y leads wave Z by 45o. Or Wave Z lags wave Y by 45 o. What is the phase relationship between the sinusoidal waveforms of each of the following sets? a. π£ = 10 π ππ(ππ‘ + 30°) π = 5 π ππ(ππ‘ + 70°) Solution i leads v by 40β°, or v lags i by 40β°. Figure 13 b. π = 15 π ππ(ππ‘ + 60°) π£ = 10 π ππ(ππ‘ − 20°) Solution: i leads v by 80β°, or v lags i by 80β°. NVSU-FR-ICD-05-00 (081220) Page 12 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Figure 14 c. π = 2 cos(ππ‘ + 10°) π£ = 3 π ππ(ππ‘ − 10°) Solution: Note! Convert the cosine function to sine function π = 2 πππ (ππ‘ + 10°) = 2 π ππ (ππ‘ + 10° + 90°) π = 2 π ππ (π£π‘ + 100°) i leads v by 110β°, or v lags i by 110β°. Figure 15 PHASORS The addition of sinusoidal voltages and currents is frequently required in the analysis of ac circuits. A shorter method uses the rotating radius vector This radius vector, having a constant magnitude (length) with one end fixed at the origin, is called a phasor when applied to electric circuits. The addition (or subtraction) of two sinusoidal voltages of the same frequency and phase angle is simply the sum (or difference) of the peak values of each with the sum (or difference) having the same phase angle. NVSU-FR-ICD-05-00 (081220) Page 13 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 For in phase and out of phase phasors the resultant vector can be computed using the general equation for adding vectors as ππ = ππ + ππ +ππ + β― + ππ Figure 16. The sinusoidal waveform of figure 16 a can be drawn to its phasor diagram in figure 16 b. Getting the resultant of v1 and v2 ππ = ππππππ = π∠π and ππ = ππππππ = π∠π ππ = ππ + ππ = π∠π + π∠π = π∠π NVSU-FR-ICD-05-00 (081220) Page 14 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Figure 17. The resultant phasor for the out of phase phasor voltage v1 and v2 can be computed using the general equation ππ = ππ + ππ +ππ + β― + ππ Since ππ = ππππππ = π∠π π½ And ππ = ππππ(ππ + ππ) = π∠ππ π½ ππ = ππ + ππ = π∠π + π∠ππ = π. ππ∠ππ. ππ π½ NVSU-FR-ICD-05-00 (081220) Page 15 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Find the sum of the following sinusoidal functions π1 = 5 π ππ(ππ‘ + 30°)π΄ π2 = 6 π ππ(ππ‘ + 60°)π΄ Solution: π1 = 5 π ππ(ππ‘ + 30°)π΄ = 5∠30 π΄ π2 = 6 π ππ(ππ‘ + 60°)π΄ = 6∠60 π΄ ππ = ππ + ππ = π∠ππ + π∠ππ = ππ. ππ∠ππ. ππ π¨ Find the current i2 for the network in figure 18 Figure 18 Root- Mean- Square(RMS) or Effective Value Of A Sinusoidal Wave Defined as the square root of the average of the squares of the given quantity taken over a period of time πππ(πππ πππππ) = πππππππ πππππ √π Average Value Of A Sinusoidal Wave πππ = π. πππ π πππ. πππππ Convert the following to peak and rms values: 10 V pp, 40 V pp, and 10 A pp. Solution: NVSU-FR-ICD-05-00 (081220) Page 16 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Peak values: rms values: 10 πππ / 2 = 5 ππ 5 ππ ∗ 0.707 = 3.535 π πππ 40 πππ / 2 = 20 ππ 20 ππ ∗ 0.707 = 14.14 π πππ 10 π΄ππ / 2 = 5 π΄π 5 π΄π ∗ 0.707 = 3.535 π΄ πππ Convert the following to peak and peak to peak values: 20 A rms, 50 A rms, and 100 V rms. Solution: Peak values: Peak to Peak values: 20 π΄ πππ ∗ 1.414 = 28.28 π΄ π 28.28 π΄ π ∗ 2 = 56.56 π΄ ππ 50 π΄ πππ ∗ 1.414 = 70.7 π΄ π 70.7 π΄ π ∗ 2 = 141.4 π΄ ππ 100 π πππ ∗ 1.414 = 141.4 π π 141.4 π π ∗ 2 = 282.8 π ππ If resistor is 100 ohms and AC voltage source of 25 Vrms What is the peak current? Solution: πΌπππ = πΈπππ 25 π = = 0.25 π΄ π 100 β¦ πΌ ππππ(πππ₯) = 1.414 ∗ 0.25 π΄ πππ = 0.3535 π΄ If current is 50 mA(peak to peak) and two resistors 7.5 kilo ohms each are in series, What is the voltage in rms and peak value? NVSU-FR-ICD-05-00 (081220) Page 17 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Solution: π π‘ = 7.5 ππππ πβππ ∗ 2 = 15 ππππ πβππ πππ = 0.05 π΄ ∗ 15000 πβππ = 750 πππ π½ππππ = πππ π½ = πππ π½ π π½πππ = π½π ∗ π. πππ = πππ. πππ π½ A 500 ohm, 300 ohm, and 1,000 ohm resistors are in parallel. A 10 V peak voltage is applied. What is the value of current in rms, peak, and peak to peak? Solution: π π‘ = πΌπ‘ = 1 1 1 1 + + 500β¦ 300β¦ 1000β¦ = 157.895β¦ 10 π = 0.0633 π΄ 157.895β¦ π° ππ = π. ππππ π¨ ∗ π = π. ππππ π¨ ππ π°πππ = π. ππππ π¨ π ∗ π. πππ = π. πππππ π¨ πππ 1. Convert to peak value: a. 40 V pp b. 25 V rms c. 10 V ave 2. Convert to rms value: a. 20 V p b. 15 V pp c. 9 V ave 3. Convert to ave value: a. 10 V p b. 8 V rms c. 12 V pp NVSU-FR-ICD-05-00 (081220) Page 18 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 VII. LEARNING ACTIVITIES Problem Solving. Solved what is ask. Show your complete solution and box your final answer. 1. What is the frequency in MHz of the following: a. 0.345 nanoseconds b. 50 nanoseconds 2. What is the frequency in kHz? a. 45 microsecond b. 33.65 microsecond 3. What is the frequency in Hz? a. 12.43 ms b. 36.71 ms 4. What is the period in milliseconds? a. 245 Hz b. 500 Hz 5. Convert the following voltages to peak value. a. 50 V pp b. 25 V rms 6. Convert to peak value for the following currents: a. 16.3 A rms b. 24.6 A pp 7. A 2, 300 ohm resistor in series and a 9 V rms source (see figure below). Find the current in rms. 8. A 400 and 600 ohm resistors in parallel and a 10 V pp source (see figure below). Find the current in rms. NVSU-FR-ICD-05-00 (081220) Page 19 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 9. A 500 ohm and a 1000 ohm resistor in parallel, a 4 volt peak to peak value. What is the resulting current in root-mean-square value? 10. A 450 ohm, 340 ohm, and a 200 ohm resistor are in parallel with each other, with a 600 ohm resistor in series (see figure below). If a 6 volt rms was applied, what would the resulting current be in peak value? 11. A voltage wave has the equation π = 170π πππΌ. Calculate the instantaneous values of voltage for the following angles: 30β°. 75β°, 105β°, 280β° and 330β°. 12. A current wave has the equation π = 21.2π πππΌ. At what angles be the instantaneous current be equal to 7.25 A, 13.6 A, -13.6 A and -7.25 A. 13. A 25 cycle sinusoidal emf wave has a maximum value of 340 V. Determine the instantaneous values of emf a) 0.007 sec after the wave passes through zero in a positive direction, b) 0.01 sec after the wave passes through zero in a negative direction. 14. The following information is given in connection with an ac source that delivers the current to three loads in parallel. Ia= 20 A; Ib= 12 A and lags behind Ia by 30 electrical degrees; and Ic= 32 A and leads Ia by 60 electrical degrees. Using Ia as the reference phasor, determine the resultant current phasor. 15. The current in a given conductor is represented by the geometric sum of two current whose equations are π1 = 25 π ππ377π‘ and π2 = 15 πππ 377π‘. What is the rms value of the current in the conductor? π 4 16. The equations of the currents in two parallel connected electric devices are π1 = 11.3 sin (377π‘ + ) And π2 = 8.5π ππ(377π‘ − π/4). What is the rms value of the resultant current. REFERENCES B. L.Theraja, A.K. Theraja, Textbook of Electrical Technology Volume I –, S. Chand & Co. Charles k. Alexander & Matthew N. Sadiku. (2000). Fundamentals of Electric Circuits. New York: Mc Graww Hill Companies Inc. NVSU-FR-ICD-05-00 (081220) Page 20 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.” Republic of the Philippines NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya INSTRUCTIONAL MODULE IM No.01: EE 1-2nd Semester-2020-2021 Siskind, C. S. (1982). Electrical Circuits. Johannesburg: Mc- Graw Hill International Book Company. NVSU-FR-ICD-05-00 (081220) Page 21 of 21 “ In accordance wit Section 185. Fair use of copyrighted Work of Republic Act 8293, the copyrighted works included in this material may be reproduced for educational purposes only and not for commercial distribution.”
