Alternating Current Network PDF

1 Chapter Two Alternating Current Network Types of Alternating Waveforms. 1- SINUSOIDAL: Consider the sinusoidal voltage v (t) = Vm si...

1 Chapter Two Alternating Current Network Types of Alternating Waveforms. 1- SINUSOIDAL: Consider the sinusoidal voltage v (t) = Vm sin ωt where Vm = the amplitude of the sinusoid ω = the angular frequency in radians/s ωt = the argument of the sinusoid It is evident that the sinusoid repeats itself every T seconds; thus, T is called the period of the sinusoid. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  2 Figure 1: A sketch of Vm sin ωt: (a) as a function of ωt, (b) as a function of t. 𝛼 α= ωT, ω= 𝑇 2𝜋 2𝜋 We observe that ωT = 2π, 𝑇 = , ω= 𝑤 𝑇 That is, v has the same value at t + T as it does at t and v (t) is said to be periodic. The reciprocal of this quantity is the number of cycles per second, known as the cyclic frequency f of the sinusoid. Thus, 1 𝑓= 𝑎𝑛𝑑 𝑤 = 2𝜋𝑓 𝑇 While ω is in radians per second (rad/s), f is in hertz (Hz). 1 Hz = 1 cycle per second. EXAMPLE: - Determine the angular velocity of a sine wave having a frequency of 60 Hz. Solution: 𝑤 = 2𝜋𝑓 = (2π)(60 Hz) =≈ 377 rad/s ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  3 EXAMPLE: - Determine the frequency and period of the sine wave of Fig. 2𝜋 Solution: Since ω= 𝑇 Let us examine the two sinusoids shown in Fig. 2: v1 (t) = Vm sin ωt and v2 (t) = Vm sin (ωt + φ) Therefore, we say that v2 leads v1 by φ or that v1 lags v2 by φ. If φ ≠ 0, we also say that v1 and v2 are out of phase as shown in Fig. 2. If φ = 0, then v1 and v2 are said to be in phase ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  4 Figure 2 Sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B ∓ sin A sin B With these identities, it is easy to show that: Sin (ωt ± 180◦) = - sin ωt cos (ωt ± 180◦) = - cos ωt sin (ωt ± 90◦) = ± cos ωt Cos (ωt ± 90◦) = ∓ sin ωt 2- Generating AC Voltages: One way to generate an AC voltage is to rotate a coil of wire at constant angular velocity in a fixed magnetic field, Figure 3. (Slip rings and brushes connect the coil to the load.) The magnitude of the resulting voltage is proportional to the rate at which flux lines are cut (Faraday’s law), and its polarity is dependent on the direction the coil sides move through the field. Since the coil rotates continuously, the voltage produced will be a repetitive, periodic waveform as you saw in Figure 2. The time for one revolution of 600 rpm is one tenth of a second, i.e., 100 ms. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  5 FIGURE 2 Cycle scaled in time. At 600 rpm, the cycle length is 100 ms. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  6 FIGURE 3. Generating an ac voltage. The 0° position of the coil is defined as in (a) Where the coil sides move parallel to the flux lines. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  7 FIGURE 4. Coil voltage versus angular position. As Figure 4 shows, the coil voltage changes from instant to instant. The value of voltage at any point on the waveform is referred to as its instantaneous value. 3- Definitions Related to Alternating Waveforms. Definitions:-The sinusoidal waveform of Fig. 5 with its additional notation will now be used as a model in defining a few basic terms. These terms, however, can be applied to any alternating waveform. It is important to remember as you proceed through the various definitions that the vertical scaling is in volts or amperes and the horizontal scaling is always in units of time. Waveform: The path traced by a quantity, such as the voltage in Fig. 5, 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). ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  8 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 of Fig.5, 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 of Fig. 5, 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 Ep-p or Vp-p, 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 of Fig. 5 is a periodic waveform. Period (T): The time interval between successive repetitions of a periodic waveform (the period T1 = T2 = T3 in Fig.5), as long as successive similar points of the periodic waveform are used in determining T. Cycle: The portion of a waveform contained in one period of time. The cycles within T1, T2, and T3 of Fig.5 may appear different in Fig. 6, but they are all bounded by one period of time and therefore satisfy the definition of a cycle. Frequency (f): The number of cycles that occur in 1 s. The frequency of the waveform of Fig.7 (a) is 1 cycle per second, and for Fig. 7 (b), 21⁄2 cycles per second. If a waveform of similar shape had a period of 0.5 s [Fig. 7 (c)], the frequency would be 2 cycles per second. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  9 EXAMPLE: - Find the period of a periodic waveform with a frequency of? a. 60 Hz. b. 1000 Hz. Solutions: 1 1 a. T = 𝑓 = =0.01667 s or 16.67 ms 60 𝐻𝑍 (a recurring value since 60 Hz is so prevalent) 1 1 b. T = 𝑓 = = 10−3 s = 1 ms 1000 𝐻𝑍 4- AC Waveforms and Average Value. Ac quantities are generally described by a group of characteristics, including instantaneous, peak, average, and effective values. To find the average value of a waveform, divide the area under the waveform by the length of its base. Areas above the axis are counted as positive, while areas below the axis are counted as negative Because a sine wave is symmetrical, its area below the horizontal axis is the same as its area above the axis; thus, over a full cycle its net area is zero, independent of frequency and phase angle. Thus, the average of sin wt, sin (wt ± 𝜃), sin 2wt, cos wt, cos (wt ± 𝜃), cos 2wt, and so on are each zero. The average of half a sine wave, however, is not zero. Consider Figure 8. The area under the half-cycle may be found using calculus as Figure 8 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  10 Two cases are important; full-wave average and half-wave average. The full-wave case is illustrated in Figure 9-a. The area from 0 to 2п is 2(2Im) and the base is 2п. Thus, the average is For the half-wave case (Figure 9 -b), (a) (b) Figure (9) The corresponding expressions for voltage are Sometimes ac and dc are used in the same circuit. Figure 10 shows superimposed ac and dc. It does not alternate in polarity since it never changes polarity to become negative. Figure 10 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  11 EXAMPLE: - Determine the average value of the sinusoidal waveform In Fig. EXAMPLE: - Determine the average value of the waveform in Fig. 5-Effective Values. An effective value is an equivalent dc value: it tells you how many volts or amps of dc that a time-varying waveform is equal to in terms of its ability to produce average power Effective values are also called rms values root mean square for reasons discussed shortly. Figure 11 First, consider the dc case. Since current is constant, power is constant, and average power is. 𝑃𝑎𝑣𝑔 = 𝑃 = 𝐼 2 𝑅 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  12 Now consider the ac case. Power to the resistor at any value of time is p (t) =i2R, where i is the instantaneous value of current. Since i = Im sin wt, Note that the average of cos 2wt is zero Equate two eq. The √2 relationship holds only for sinusoidal waveforms. For other waveforms, you need a more general formula. Using calculus, it can be shown that for any waveform 𝑇 1 𝐼𝑒𝑓𝑓 = √ ∫ 𝑖 2 𝑑𝑡 𝑇 0 The integral of i2 represents the area under the i2 waveform. Thus, ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  13 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑖 2 𝑐𝑢𝑟𝑣𝑒 𝐼𝑒𝑓𝑓 =√ 𝑏𝑎𝑠𝑒 To use this equation, we compute the root of the mean square to obtain the effective value. For this reason, effective values are called root mean square or rms values and the terms effective and rms are synonymous. EXAMPLE: - Find the rms values of the sinusoidal waveform in each part of Fig. EXAMPLE: - Find the rms value of the waveform in Fig. 13.60. Solution: v2 (Fig. 13.61). ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  6- Representing AC Voltages and Currents by Complex Numbers. 14 The sinusoidal voltage e(t) =200 sin(wt + 40°) of Figure 12(a) and (b) can be represented by its phase equivalent, 𝐸 = 200𝑉∟40, as in (c). Figure 12 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  15 To convert between forms: 𝐶 = 𝑎 + 𝑗𝑏 (𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚) 𝐶 = 𝑐∠𝜃 (𝑝𝑜𝑙𝑎𝑟 𝑓𝑜𝑟𝑚) 𝑏 𝑎 = 𝑐 𝑐𝑜𝑠𝜃, 𝑏 = 𝑐 𝑠𝑖𝑛𝜃, 𝑐 = √𝑎2 + 𝑏 2 , and 𝜃 = 𝑡𝑎𝑛 −1 𝑎 −1 𝑗 = √−1, 𝑗 2 = −1, 𝑗 3 = −𝑗, 𝑗 4 = 1, and 𝑗= 𝑗 The conjugate of a complex number (denoted by an asterisk *) is a complex number with the same real part but the opposite imaginary part. Thus, the conjugate of 𝐶 = 𝑐∠𝜃 = 𝑎 + 𝑗𝑏 is 𝐶 ∗ = 𝑐∠ − 𝜃 = 𝑎 − 𝑗𝑏. For example, if 𝐶 = 3 + 𝑗4 = 5∠53.13° , then 𝐶 ∗ = 5∠ − 53.13° = 3 − 𝑗4. EXAMPLE:- For Figure 13, 𝑣1 = √2(16)𝑠𝑖𝑛𝑤𝑡 𝑉, 𝑣2 = √2(24) sin(𝑤𝑡 + 90° ) 𝑉, 𝑣3 = √2(15)sin (𝑤𝑡 − 90° ) 𝑉. Determine source voltage e. Figure 13 Solution: Applying KVL: 𝑒 = 𝑉1 + 𝑉2 + 𝑉3 = 16∠0° + 24∠90° + 15∠−90° = 16 + 𝑗0 + 24𝑗 − 15𝑗 = 16 + 9𝑗 (𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚) = √162 + 92 = 18.36 9 θ = tanh−1 = ∠29.36° 16 e= √2(18.36) sin(𝑤𝑡 + 29.36° ) 𝑉 e= 18.36∠29.36° (𝑝𝑜𝑙𝑎𝑟 𝑓𝑜𝑟𝑚) ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  16 7- Ohm’s Law for AC Circuits. -Resistors When a resistor is subjected to a sinusoidal voltage as shown in Figure 15, the resulting current is also sinusoidal and in phase with the voltage. Figure 15 The sinusoidal voltage v=Vmsin (wt+θ) may be written in phasor form as V=V∠𝜃 °.Whereas the sinusoidal expression gives the instantaneous value of voltage for awaveform having an amplitude of Vm (volts peak), the phasor form has a magnitude which is the effective (or rms) value. The voltage and current phasors may be shown on a phasor diagram as In Figure 16. Figure 16 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  17 Example: - The current through a 5 Ω resistor is given. Find the sinusoidal expression for the voltage across the resistor for i =40 sin (377t + 30°). Solution: Vm = Im R = (40 A) (5 Ω) = 200 V (V and i are in phase), resulting in v = 200 sin (377t + 30°) EXAMPLE: - The voltage across a resistor is indicated. Find the sinusoidal expression for the current if the resistor is 10 Ω. Sketch the curves for v and i. a. v = 100 sin 377t b. v = 25 sin (377t + 60°) ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  18 -Inductors When an inductor is subjected to a sinusoidal current, a sinusoidal voltage is induced across the inductor such that the voltage across the inductor leads the current waveform by exactly 90°. If we know the reactance of an inductor, then from Ohm’s law the current in the inductor may be expressed in phases form as: For an inductor, vL leads iL by 90°, or iL lags vL by 90°. iL = Im sin (ωt ±Ɵ ) vL = ωLIm sin(ωt ± Ɵ ± 90°) In vector form, the reactance of the inductor is given as: 𝑋𝐿 = 𝑤𝐿 = 2𝜋𝑓𝐿 ohms, Ω 𝑉𝑚 XL = , Vm = Im XL 𝐼𝑚 EXAMPLE: - The current through a 0.1-H coil is provided. Find the sinusoidal expression for the voltage across the coil. Sketch the v and i curves. a. i =10 sin 377t b. i =7 sin(377t -70°) Solution: a. XL =ωL = (377 rad/s)(0.1 H) = 37.7 Ω Vm = Im XL = (10 A) (37.7 Ω) = 377 V And we know that for a coil v leads i by 90°. v =377 sin (377t + 90°) ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  19 b- XL remains at 37.7Ω Vm =Im XL = (7A)(37.7Ω) = 263.9 V And we know that for a coil V leads I by 90̊. Therefore, V= 263.9 sin(377t - 70̊ +90̊ ) V= 263.9 sin(377t +20̊ ) The curves are sketched in Fig. H.W EXAMPLE:-The voltage across a 0.5-H coil is provided below. What is the sinusoidal expression for the current? v = 100 sin 20t ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  20 - Capacitors: When a capacitor is subjected to a sinusoidal voltage, a sinusoidal current Results. The current through the capacitor leads the voltage by exactly 90°. If We know the reactance of a capacitor, then from Ohm’s law the current in the Capacitor expressed in phasor form is For a capacitor, iC leads vC by 900, or vC lags iC by 90°. VC = Vm sin (ωt ± Ɵ) iC =ωC Vm sin (ωt ± Ɵ + 90°) 𝑉𝑚 𝑉𝑚 1 = = 𝐼𝑚 𝜔𝐶 𝑉𝑚 𝜔𝐶 1 1 𝑋𝐶 = 𝑤𝐶 = 2𝜋𝑓𝐶 Ohms, Ω 𝑉𝑚 Xc = Ohms, Ω 𝐼𝑚 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  21 H.W EXAMPLE: - The current through a 100-mF capacitor is given. Find the sinusoidal expression for the voltage across the capacitor. i =40 sin (500t +60°). ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  22 8. Power in AC Circuits: At any given instant, the power to a load is equal to the product of voltage times current: 𝑃 = 𝑣𝑖 Watts This is illustrated in Fig. 19, where we have multiplied voltage time's current point by point to get the power waveform. Figure 19 Thus, during positive parts of the power cycle, power flows from the source to the load, while during negative parts, it flows out of the load back into the circuit. Thus, if P has a positive value, it represents the power that is really dissipated by the load. For this reason, P is called real power. In modern terminology, real power is also called active power. Thus, active power is the average value of the instantaneous power, and the terms real power, active power, and average power mean the same thing. First consider power to a purely resistive load (Fig. 20). Here, current is in phase with voltage. Assume 𝑖 = 𝐼𝑚 𝑠𝑖𝑛𝑤𝑡, 𝑎𝑛𝑑 𝑣 = 𝑉𝑚 𝑠𝑖𝑛𝑤𝑡, 𝑡ℎ𝑒𝑛 𝐼𝑚 𝑉𝑚 𝑝 = 𝑣𝑖 = (𝐼𝑚 𝑠𝑖𝑛𝑤𝑡)(𝑉𝑚 𝑠𝑖𝑛𝑤𝑡) = 𝐼𝑚 𝑉𝑚 𝑠𝑖𝑛2 𝑤𝑡 = (1 − 𝑐𝑜𝑠2𝑤𝑡) 2 𝐼𝑚 𝑉𝑚 Where is called average power. 2 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  23 Figure 20 Note that p is always positive. This means that power flows only from the source to the load. We therefore conclude that power to a pure resistance consists of active power only. For a purely inductive load as in Fig. 21 (a), current lags voltage by 90°. If we select current as reference 𝑖 = 𝐼𝑚 𝑠𝑖𝑛𝑤𝑡, 𝑎𝑛𝑑 𝑣 = 𝑉𝑚 sin (𝑤𝑡 + 90° ).A sketch of p versus time (obtained by multiplying v times i) then looks as shown in (b). Thus, the average power to an inductance over a full cycle is zero, i.e., there are no power losses associated with a pure inductance. Consequently, P=0 W and the only power flowing in the circuit is reactive power. (a) Figure 21 (b) With𝑖 = 𝐼𝑚 𝑠𝑖𝑛𝑤𝑡, 𝑎𝑛𝑑 𝑣 = 𝑉𝑚 sin (𝑤𝑡 + 90° ), 𝑝 = 𝑣𝑖 becomes 𝑝𝐿 =(𝐼𝑚 𝑠𝑖𝑛𝑤𝑡)(𝑉𝑚 sin (𝑤𝑡 + 90° ) After some trigonometric manipulation, this reduces to 𝑝𝐿 = 𝐼 𝑉 sin2𝑤𝑡 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  24 The product VI is defined as reactive power and is given the symbol QL, its unit is VAR (volt-amps reactive). 2 𝑉2 𝑄𝐿 = 𝐼 𝑋𝐿 = (𝑉𝐴𝑅) 𝑋𝐿 For a purely capacitive load, current leads voltage by 90°. Taking current as reference𝑖 = 𝐼𝑚 𝑠𝑖𝑛𝑤𝑡, 𝑎𝑛𝑑 𝑣 = 𝑉𝑚 sin (𝑤𝑡 − 90° ). Multiplication of v times i yields the power curve of Fig. 22. This means that the average power to a capacitance over a full cycle is zero, i.e., there are no power losses associated with a pure capacitance. Consequently, P=0 W and the only power flowing in the circuit is Reactive power. Figure 22 This reactive power is given by: 𝑝𝐶 = 𝐼𝑚 𝑉𝑚 𝑠𝑖𝑛𝑤𝑡 sin (𝑤𝑡 + 90° ) Which reduces to: 𝑝𝐶 = −𝑉𝐼 𝑠𝑖𝑛2𝑤𝑡 Now define the product VI as QC which can be expressed as: 𝑉2 𝑄𝐶 = 𝐼 2 𝑋𝐶 = (VAR) 𝑋𝐶 When a load has voltage V across it and current I through it, the power that appears to flow to it is VI. However, if the load contains both resistance and reactance, this product represents neither real power nor reactive power. Since it appears to represent power, it is called apparent power. Apparent power is given the symbol S and has units of volt- 𝑉2 amperes(VA).Thus, 𝑆 = 𝑉𝐼 = 𝐼 2 𝑍 = (𝑉𝐴) 𝑍 Note that these represent P, Q, and S respectively as indicated in figure 23. This is called the power triangle. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  25 (d) Capacitive case Figure 23 Q = Reactive power P = Active power S = apparent power. From the geometry of this triangle, you can see that =√𝑝 2 + 𝑄 2 , and 𝑃 = 𝑉𝐼 𝑐𝑜𝑠𝜃 = 𝑆 𝑐𝑜𝑠𝜃 (𝑊) 𝑄 = 𝑉𝐼 𝑠𝑖𝑛𝜃 = 𝑆 𝑠𝑖𝑛𝜃 (𝑉𝐴𝑅) The quantity 𝑐𝑜𝑠𝜃 is defined as power factor and is given the symbol FP. Thus, 𝑃 𝑃 𝐹𝑃 = 𝑐𝑜𝑠𝜃, 𝑐𝑜𝑠𝜃 = , 𝑎𝑛𝑑 𝜃 = 𝑐𝑜𝑠 −1 ( ) 𝑆 𝑆 Angle θ is the angle between voltage and current. Thus, an inductive circuit has a lagging power factor, while a capacitive circuit has a leading power factor. A load with a very poor power factor can draw excessive current. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  26 9. A.C Series Circuit: Impedance is a term used to collectively determine how the resistance, capacitance, and inductance “impede” the current in a circuit. The symbol for impedance is the letter Z and the unit is the ohm (Ω). Because impedance may be made up of any combination of resistances and reactance's, it is written as a vector quantity Z, where 𝒁 = 𝑍∠𝜃 Resistive impedance ZR is a vector having a magnitude of R along the positive real axis. Inductive reactance ZL is a vector having a magnitude of XL along the positive imaginary axis, while the capacitive reactance ZC is a vector having a magnitude of XC along the negative imaginary axis as shown in figure 24. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  27 Figure 24 EX:-For a series ac circuit consisting of n impedances, the total impedance of the circuit is found as the vector sum: 𝑍𝑇 = 𝑍1 + 𝑍2 + ⋯ + 𝑍𝑛 Thus we may determine the total impedance of figure 25: 𝑍𝑇 = 3 + 𝑗0 + 0 + 𝑗4 = (3 + 𝑗4)Ω = 5Ω∠53.13° The above quantities are shown on an impedance Figure 25 Diagram as in Figure26. Figure 26 The rectangular form of an impedance is written as: 𝑍 = 𝑅 ± 𝑗𝑋 If we are given the polar form of the impedance, then we may determine the equivalent rectangular expression from 𝑅 = 𝑍 𝑐𝑜𝑠 𝜃, 𝑎𝑛𝑑 𝑋 = 𝑍 sin 𝜃 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  28 In the rectangular representation for impedance, the resistance term, R, is the total of all resistance looking into the network. The reactance term, X, is the difference between the total capacitive and inductive reactance's. The sign for the imaginary term will be positive if the inductive reactance is greater than the capacitive reactance. In such a case, the impedance vector will appear in the first quadrant of the impedance diagram and is referred to as being an inductive impedance. If the capacitive reactance is larger, then the sign for the imaginary term will be negative. In such a case, the impedance vector will appear in the fourth quadrant of the impedance diagram and the impedance is said to be capacitive. The polar form of any impedance will be written in the form 𝒁 = 𝑍∠𝜃. The value Z is the magnitude (in ohms) of the impedance vector Z and is determined as follows:𝑍 = √𝑅 2 + 𝑋 2 (Ω) The corresponding angle of the impedance vector is determined as: 𝑋 𝜃 = ±𝑡𝑎𝑛−1 ( ) 𝑅 Example: Consider the network of Fig. 27. a. Find ZT. b. Sketch the impedance diagram for the network and indicate whether the total Impedance of the circuit is inductive, capacitive, or resistive. C. Use Ohm’s law to determine I, VR, and VC. Solution: a. 𝑍𝑇 = 25 + 𝑗200 − 𝑗225 = (25 − 𝑗25)Ω = 35.35Ω∠−45° b. The corresponding impedance diagram is shown in Fig. 28. Because the total impedance has a negative. Fig. 27. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  29 Reactance term (j25), ZT is capacitive. ° 10𝑉∠0 ° 𝐶. 𝐼 = ° = 0.2828𝐴∠45 35.35Ω∠−45 ° ° 𝑉𝑅 = (0.2828𝐴∠45 ) (25Ω∠0 ) ° = 7.07𝑉∠45 ° ° 𝑉𝐶 = (0.2828𝐴∠45 ) (225Ω∠−90 ) = 63.63𝑉∠−45° Figure28 In the simple series circuit shown in Fig. 31, we know that only the resistor will dissipate power. Figure 31 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  30 𝑉𝑅2 𝑉 𝑃 = 𝑉𝑅 𝐼 = = 𝐼 2 𝑅, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝐼 = , 𝑡ℎ𝑢𝑠 𝑅 𝑍 𝑉2 𝑉2 𝑅 𝑃 = 2𝑅 = = 𝑍 𝑍 𝑍 𝑅 𝑉2 From figure 32 we have 𝑐𝑜𝑠𝜃 = 𝑡ℎ𝑢𝑠 𝑃 = 𝑉𝐼 𝑐𝑜𝑠𝜃 = 𝐼 2 𝑍 𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠𝜃 𝑍 𝑍 Example: Refer to the circuit of Figure 32. a. Find the impedance Z. b. Calculate the power factor of the circuit. c. Determine I. d. Sketch the phasor diagram for E and I. e. Find the average power delivered to the circuit by the voltage source. f. Calculate the average power dissipated by both the resistor and the capacitor ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  31 Figure 32 Solutions: 𝑎. 𝑍𝑇 = 𝑅 + 𝑋𝐶 = (3 − 𝑗4)Ω = 5Ω∠−53.13° 𝑅 3 b. 𝐹𝑃 = 𝑐𝑜𝑠𝜃 = 𝑍 = 5 = 0.6 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 c. The phasor form of the applied voltage is: √2 × 20 𝐸= ∠0 = 20𝑉∠0 √2 20𝑉∠0 𝐼= ° = 4𝐴∠53.13° 5Ω∠−53.13 d.The phasor diagram is shown in figure 33 𝑒. 𝑃 = 𝑉𝐼 𝑐𝑜𝑠𝜃 = 20 × 4 × cos(53.13) = 48 𝑊 f. 𝑃𝑅 = 42 × 3 × cos(0) = 48 𝑊 𝑃𝐶 = 42 × 4 × cos(90) = 0 𝑊 Figure 34 Example: Consider the circuit of Figure 35. a. Find ZT. b. Determine the voltages VR and VL using the voltage divider rule. ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering  32 c. Verify Kirchhoff’s voltage law around the closed loop. Solution: a. 𝑍𝑇 = 5𝑘Ω + 𝑗12𝑘Ω = 13𝑘Ω∠67.38° b. 5𝑘Ω∠0° 𝑉𝑅 = ° × 26𝑉∠0° = 10𝑉∠ − 67.38° 13𝑘Ω∠67.38 12𝑘Ω∠90° 𝑉𝐿 = ° × 26𝑉∠0° = 24𝑉∠22.62° 13𝑘Ω∠67.38 C. Figure 35 V (t) = √𝑉𝑟 2 + 𝑉𝑙 2 = 26 = √102 + 242 = √100 + 576 = 26 26𝑉∠0° − 10𝑉∠ − 67.38° − 24𝑉∠22.62° = 0 (26 + 𝑗0) − (3.846 − 𝑗9.231) − (22.145 + 𝑗9.231) = 0 (26 − 3.846 − 22.145) + 𝑗(0 − 9.231 + 9.231) = 0 0 + 𝑗0 = 0 ASSISTANT LECTURER Omar AL-Azzawi Department of Computer Technical Engineering