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Contents Introduction 3 Impedance 4 Resistance and inductance in series 6 The practical inductor 6 The voltage triangle 10 The impedance triangle 12 Added series resistance to inductor 16 Resistance and capacitance in series 24 Summary 37 Answers 38 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 1 2 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Introduction This section deals with the calculations needed to solve for voltage, current, impedance and power in series R-L and R-C circuits. At the end of this section you should be able to: apply Pythagoras Theorem to a right angle triangle. define impedance determine the impedance, current and voltages for a series RL and RC circuits given the resistance, inductance and supply voltage. draw and label the an impedance triangle for a series RL and RC circuits draw the equivalent circuit of a practical inductor draw phasor diagrams for a series RL and RC circuits give examples of inductive and capacitive components in power circuits and systems and describe their effect on the phase relationship between voltage and current. state the AS/NZS 3000 requirements for the installation of capacitors. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 3 Impedance Resistance and reactance are special cases of a more general property called impedance. You have learnt that: Ohms Law can be used for resistive ac circuits just as with dc circuits. Relations similar to Ohms Law can also be used for capacitors and inductors. Capacitive reactance XC or inductive reactance XL can be used instead of resistance in Ohm’s Law. Pure (ideal) inductors cause the current to lag the voltage by 90°, and capacitors cause the current to lead the voltage by 90°. Next we will proceed to analyse circuits that contain both resistive and reactive components. Impedance is the property that opposes current in an ac circuit. Impedance is like a ‘generalised resistance’, and obeys and equation similar to Ohm’s Law: Z V I where: Z = impedance in ohm V = rms voltage in volt I = rms current in ampere Note that: 4 In a resistive circuit, the impedance is equal to the resistance. In a pure reactive circuit (pure capacitance or inductance), the impedance is equal to the reactance. In circuits containing resistance and reactance in series, the impedance is obtained by combining these two quantities. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 For circuits that are series connected we can calculate impedance from the resistance and reactance. The 90° phase difference between resistor and reactor voltages allows use of the Pythagorean theorem. Symbols introduced for the right-angled triangle were: Z the hypotenuse X the opposite side to the angle R the side adjacent to the angle . The expression to calculate impedance in a circuit using the right-angled triangle becomes: Z R2 X 2 where: Z = impedance in ohm R = resistance in ohm X = reactance in ohm The right-angle triangle as shown in Figure 1 is known as the impedance triangle. Remember that this is used only for series circuits. Figure 1: Impedance triangle EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 5 Resistance and inductance in series In a pure resistive circuit, the current is in phase with the voltage, and in a pure inductive circuit, the current lags the voltage by 90°. Therefore, in an RL circuit, the current will lag the voltage by some angle φ in the range 0° to 90°. Figure 2 shows the relationship for the resistive/inductive circuit. Figure 2: Voltage and current in resistive inductive circuits The practical inductor As an inductor is made as a coil of wire it has some resistance even though this value may be very low. When dealing with inductors in practice both the resistance and inductance must be considered. We therefore model a real inductor component using an ideal inductor in series with a resistance. The circuit model of a practical inductor is shown in Figure 3. Figure 3: Schematic representation of an inductor To determine the impedance of an inductor in terms of resistance and inductive reactance, we can consider the individual voltage drops across 6 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 each equivalent component. These cannot all be measured but can be calculated Figure 4: Three voltages to be considered for an inductor Since this is a series circuit, we take the current as the reference phasor, and then consider the two voltage drops VR and VXL. These voltages are shown in the diagram in Figure 5. The phase relationship of VR and VXL must be considered: The voltage and current for a resistor are in phase, so VR must be in phase with the circuit current. In a pure inductor, the voltage leads the current by 90 degrees, so VXL leads the circuit current by 90 degrees. This is shown in Figure 5. Note that VXL therefore leads VR by 90°. Figure 5: Voltage drops in an inductor The first step is to measure the dc parameters by applying a DC supply and measuring the current. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 7 Figure 6: Circuit for measuring DC values We can now determine the ac parameters by applying a variable ac supply voltage to give the same current as obtained on dc. Figure 7: Circuit for measuring ac values This value is now used in our phasor diagram. The results for an inductor we have tested are as follows: 10 A dc at 30 V (Figure 6) ac voltage 50 V (Figure 7) gave 10 A Using the tip-to-toe phasor diagram, plot the resistive voltage drop VR = 30 V and the direction of the voltage VXL as in Figure 8. 8 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 8: Plotting the resistive voltage drop The applied voltage VAC (50 V) is now included by drawing an arc from the origin which cuts the line representing the direction of VXL. The VAC phasor is then drawn between the origin and the point where the arc intersects VXL. The complete phasor diagram now shows VAC applied, VR and VXL. Figure 9: Complete phasor diagram for an inductor on ac The phase angle can also be measured. The voltage VXL is the perpendicular phasor and its value, measured from the diagram, is 40 V. Note that this value could also be determined from Pythogoras’ Theorem. The final figures for our example are: VAC = 50 V VR = 30 V VXL = 40 V ϕ = 53.l° with current lagging applied ac voltage EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 9 As briefly introduced earlier, the inductor has impedance, resistance and reactance. From our measurements we can determine these parameters. Z Vac I ac 50 10 5 V R dc I dc 30 10 3 V X L XL I 40 10 4 The voltage triangle In the drawing of the tip-to-toe voltage phasor diagram for an inductor, we developed a right angle triangle which has the applied voltage on the hypotenuse, the voltage drop due to resistance on the adjacent side and the voltage drop due to inductive reactance on the opposite side. Figure 10: Voltage triangle for inductor 10 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 These three voltage values may be expressed as: V = IZ VR= IR VXL= IXL The sides of the voltage triangle may be labelled again using these values. Figure 11: Equivalent voltage triangle The equivalent voltage triangle is used in the development of the impedance triangle for a series R-L circuit. The current can be cancelled from each side, leaving the resistance, reactance and impedance values forming a right angle triangle. Student exercise 1 1 Using the voltage triangle find the values VR and VXL if V = 100 V at a 60° (I lagging) angle. ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 11 The impedance triangle If each side of the equivalent voltage triangle of Figure 11 is divided by I, each side will be labelled as the: hypotenuse Z adjacent R opposite XL. Figure 12: Impedance triangle for series R-L circuit Because this triangle no longer represents the voltages, no arrowheads are used. Each side now represents the electrical properties of: Z = impedance R = resistance XL = inductive reactance This figure is known as the impedance triangle, and it allows us to calculate any one of the sides, knowing the other two. To express the impedance of an inductor in terms of its resistance and reactance, use Pythagoras’s theorem: From the triangle we can see: Z2 = R2 + XL2 Impedance Z can then be calculated. 12 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Z R2 X L2 ohm Note: This expression is only for series combinations and should not be applied to parallel circuit calculations. The circuit phase angle is maintained in the impedance triangle and is the angle enclosed by impedance Z and resistance R. This phase angle may be determined by using either the sine ratio or the cosine ratio. XL Z R cos Z _ X sin 1 L Z _ R cos 1 Z sin The function sin–1(x) means ‘the angle whose sine is x’ and the function cos–1(x) means ‘the angle whose cosine is x’. These functions are known as the ‘inverse sine’ and ‘inverse cosine’ respectively. The phase angle for any inductor must always be less than 90° due to the resistance of the inductor winding. Example An inductor having a resistance of 47 ohm and an inductance of 1.8 henry is connected to a 240 V, 50 Hz supply. Determine the: (a) inductor impedance (b) circuit current (c) circuit phase angle. Draw the phasor diagram. Solution (a) Before we can calculate the impedance, the inductive reactance has to be calculated. X L 2 fL 314 1.8 = 565 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 13 Z R 2 X L2 47 2 5652 567 (b) Circuit current V Z 240 = 567 0.423 A I (c) Phase angle The phase angle has a cosine ratio of resistance over impedance. R Z _ 47 cos 1 567 _ 1 cos 0.08289 85.24 lagging (I lags V) cos _ 1 Note: it is important that the mode of your calculator is set to degrees when you calculate sine and cosine, or inverse sin and cosine. The phasor diagram is shown below: Figure 13: Using a phasor diagram Example An inductor, when tested on a 20 volt dc supply had a 300 mA current, and when tested on a 20 volt ac supply had a current of 35 mA? Find the inductance. 14 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Solution The calculations to determine the answer use the inductive reactance expression. X L 2 fl L XL 2 f To determine XL we can use the impedance expression. Z R 2 X L2 X L Z 2 R2 Finding the values for impedance we have test figures from the ac test. Z VAC I AC 20 0.035 571 The resistance is calculated using the DC test figures. We assume here that the ac resistance is equal to the dc resistance. R VAC I AC 20 0.3 66.7 We can now calculate the value of XL. X L Z 2 R2 5712 67 2 567 The inductance may now be calculated. L XL 2 f 567 2 3.14 50 1.8 H EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 15 Added series resistance to inductor We will now consider a series circuit which has a resistor and inductor. A schematic diagram in Figure 14 shows the circuit properties to be considered. Figure 14: Resistor in series with inductor Circuit problems Given the data in Figure 14, we will be able to solve problems to determine: inductor impedance and phase angle circuit impedance, current and phase angle voltage drops VR andVL to draw the phasor diagram. Inductor Z: X L 2 fL 2 3.14 50 0.5 157 Z L RL 2 X L 2 302 157 2 159.8 16 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Inductor : L cos _ 1 RL ZL 30 159.8 1 cos 0.1877 79.2 cos 1 Circuit Z: ZT RT 2 X L 2 RT R RL 120 30 150 ZT 150 2 157 2 = 217 Circuit I IT V ZT 240 217 1.1 A Circuit T cos _ 1 RT ZT 150 217 _ 1 cos 0.6912 46.3 lag cos _ 1 Note: The inductor phase angle is 79.2°. With the added resistance the total circuit phase angle has been reduced to 46.3°. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 17 Phasor diagram Calculate VR and VL VR IR = 1.1 × 120 = 132 V VL IZ L = 1.1 × 159.8 = 175.78 V The reference phasor is the current phasor and this need not be drawn to scale. The scale for the voltages should be 2 V/mm. Figure 15 Draw reference phasor (see Figure 15). Include VR in phase with I. 132 2 = 66 mm length of phasor VR = Include VL (see Figure 16) 176 2 = 88 mm length of phasor VL = This voltage leads I by L, that is, 79° 18 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 16 Figure 17 When the complete phasor diagram has been drawn the total circuit voltage and phase angle can be measured. Figure 17 shows the length of V to be 120 mm, that is, 240 volt. The phase angle is 46° with current lagging voltage. cos = 0.69 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 19 Example A 100 ohm resistor is connected in series with an inductor. The inductor has an inductance of 0.95 henry and a resistance of 25 ohm. If the circuit is connected to a 150 V, 50 Hz supply, calculate: (a) inductor impedance and phase angle (b) circuit impedance and phase angle (c) circuit current (d) inductor voltage drop and resistor voltage drop. Draw the circuit phasor diagram tip-to-toe. Solution (a) Inductor impedance and phase angle: X L 2 fL 314 0.95 298.3 Z L R 2 X L2 252 2982 299 25 299 _ 1 cos 0.0835 85.2 L cos (b) _ 1 Circuit impedance and phase angle: Z L R 2 X L2 where: RT R RL 100 25 125 Therefore: Z 1252 2982 323 20 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 cos _ cos _ 1 1 RT Z 125 323 _ cos 1 0.387 67.24 (c) Circuit current V Z 150 323 0.464 A I (d) Voltage drops: VR IR 0.464 100 46.4 V VL IZ L 0.464 299 138.7 V Figure 18: Circuit diagram for the above example showing V, VR and VL The circuit phasor diagram (Scale: 1 mm = 2 V) : EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 21 Figure 19 The VR phasor is drawn 23 mm in phase with the reference. Tip-to-toe, draw the voltage phasor for the voltage VL 69 mm at 85° leading the reference. Draw the applied voltage phasor from the origin to the tip of VL. This, when measured, is 75 mm and the circuit phase angle can be measured as 67° with current lagging. 22 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Student exercise 2 From the data given in Figure 20 answer the following questions. Figure 20 Rl = 250 ohm, RL = 50 ohm, L = 1 henry Find the circuit current IT. Draw the phasor diagram to a scale of 2 V/mm. ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ 2 Calculate the impedance of an inductor if its resistance is 5 ohm and reactance 12 ohm. ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 23 Resistance and capacitance in series Figure 21 below shows a series resistive-capacitive circuit. Figure 21: Series resistive-capacitive circuit You should recall that: In a purely resistive circuit, the current is in phase with the voltage. In a purely capacitive circuit the current leads the voltage by 90°. Therefore all ac circuits that have a combination of resistance and capacitance will have the current leading the voltage by some angle between 0° and 90° as shown in Figure 22. The angle depends on the relative values of resistance and capacitive reactance in the circuit. The difference between R-L and R-C circuits is that current leads the voltage in the R-C circuit and lags in the R-L circuit. 24 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 22: Phase relationship of an R-C circuit The selection of the reference phasor depends on whether the voltage or current is common. In series circuits the current is common to all components. Current is therefore the reference phasor for series circuits. A real capacitor component can be considered to be ‘ideal’ because the resistance is so low it may be neglected. The phasor treatment for this series RC circuit is very similar to the series RL circuit: Firstly, the reference phasor will be the current, as this is common to all components in a series circuit. The voltage VR is in phase with the current The voltage VC lags the current by 90°, so it is plotted downwards on the diagram. The supply voltage V is the phasor sum of VR and VC. As these two are displaced by 90°, total voltage may be calculated using Pythagoras’ equation. V VR 2 VC 2 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 25 The phasor diagram for an R-C series circuit is shown in Figure 23. Figure 23: Phasor diagram for an R-C series circuit The angle (is the phase angle between circuit voltage V and current I. I leads V. This angle may be calculated as the angle whose cosine is the ratio of VR to V. cos _ 1 VR V The voltage triangle may now be drawn. This is merely a reproduction of the phasors of the circuit voltages. Figure 24: Series R-C circuit voltage triangles If we now divide each side of the triangle by I, the result is the impedance triangle. Note that the impedance triangle applies to series circuits only. 26 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 25: The impedance triangle Example A series R-C circuit comprises a 60 ohm resistor and an 80 microfarad capacitor. The supply is 240 volt, 50 hertz. Determine: 1 circuit impedance 2 circuit current 3 circuit phase angle 4 voltage drops VR and VC 5 draw the tip-to-toe phasor diagram with scale 2 V/mm. Solution Figure 26: Phasor diagram 1 Circuit impedance: First, we require the capacitive reactance: XC 1 2 fC 1 2 50 80 106 39.82 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 27 Therefore the impedance is: Z R2 X C 2 602 39.82 72 2 Circuit current: V Z 240 72 3.33 A I 3 Phase angle: R Z 60 = cos -1 72 33.6 cos _ 1 4 Voltage drops: VR IR 3.33 60 200 V VC IX C 3.33 39.8 132.5 V 5 Diagram: Phasor diagram, with scale 2 V/mm VR = 200 V 100 mm V = 240 V 120 mm VC = 132.5 V 66 mm 28 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 27: Phasor diagram EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 29 Student exercise 3 Figure 28 For the circuit in Figure 28 determine: 1 circuit Z ____________________________________________________________________ 2 circuit I ____________________________________________________________________ 3 VR and VC ____________________________________________________________________ 4 Draw the phasor diagram with scale 2 V/mm Check your answers with those given at the end of the section. 30 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Check your progress Part A In questions 1 – 23, place the letter matching your answer in the brackets provided. 1 An ac circuit contains resistance and inductance in series. The phase angle will be increased if the: (a) current is increased (b) voltage is increased (c) frequency is increased (d) resistance is increased 2 ( ) The current in a series circuit containing resistance and inductance, with an ac supply, lags the voltage by: (a) 90 (b) between 0 and 90 (c) always less than 70 (d) between 90 and 180 3 ( ) ( ) ( ) In a series ac circuit, the ratio of the applied voltage to the current is equal to: (a) impedance (b) reluctance (c) resistance (d) inductance 4 The phase angle between supply voltage and current, in an ac series circuit, if resistance and reactance are equal is: (a) 0 (b) 45 (c) 60 (d) 90 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 31 5 The three sides of an impedance triangle represent: (a) resistance, reactance and impedance (b) reactance, inductance and impedance (c) resistance, inductance and impedance (d) reactance, inductance and resistance 6 ( ) ( ) ( ) The equivalent circuit of a practical inductor is represented by: (a) resistor and inductor in series (b) inductor and capacitor in series (c) resistor and inductor in parallel (d) resistor, capacitor and inductor in series 7 An ac circuit contains resistance and capacitance in series. The phase angle will increase if the: (a) current is decreased (b) voltage is increased (c) frequency is decreased (d) resistance is increased 8 An ac series circuit contains resistors and a capacitor. You obtain total resistance by: (a) averaging the resistors (b) adding the resistors as phasor quantities (c) finding the sum of the resistor reciprocals (d) simply adding all the resistor values together 9 ( ) ( ) The current in an ac circuit, consisting of resistance and inductance in series, lags the applied emf by an angle (a) of 90 (b) between 0 and 90 (c) between 90 and 180 (d) always less than 60 32 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 10 Current in a series ac circuit with resistance and reactance can be determined from: (a) V Z (b) V R (c) V X (d) V X R ( ) ( ) ( ) 11 Increasing the frequency of the ac supply to a capacitor (a) reduces the capacitative reactance (b) increases the capacitative reactance (c) does not affect the capacitative reactance (d) varies the capacitance 12 In an impedance triangle the hypotenuse always represents: (a) impedance (b) reactance (c) resistance (d) reluctance Part B 1 An inductor when tested is found to have a dc resistance of 5 ohm and an impedance of 13 ohm at 50 hertz. Determine its inductive reactance and phase angle at 50 hertz. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 33 2 An inductor having a resistance of 50 ohm and an inductance of 1.5 henry is to be connected to a 240 volt, 50 hertz supply. What will be the circuit current and phase angle? _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 3 A non-inductive electric furnace is designed to draw a current of 20 ampere from a 240 volt, 50 hertz supply. Determine the inductive reactance of an ideal choke reactor, connected in series, to provide the same current on a 260 volt, 50 hertz supply. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 4 A series R-L circuit has the following properties: the resistor value is 100 ohm the inductor has a resistance of 100 ohm and an inductance of 0.25 henry supply voltage is 240 volt, 50 hertz. Calculate the: (a) total circuit impedance _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (b) total circuit current _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 34 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 (c) Draw the phasor diagram and determine the phase angle for the inductor alone. 5 Calculate the current drawn from a 240 volt, 50 hertz supply by an 80 microfarad capacitor. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 6 A series R-C circuit consists of a 150 ohm resistor and a 40 microfarad capacitor connected to a 240 volt, 50 hertz supply. Determine: (a) circuit impedance _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (b) circuit current _____________________________________________________________________ _____________________________________________________________________ EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 35 7 An air-cored inductor is connected to a 240 volt dc supply and draws a current of 48 ampere. It is then connected to a 240 volt, 50 hertz supply and draws a current of 4.8 ampere. What is its reactance at 50 hertz? _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 8 An ac circuit has a resistance of 30 ohm connected in series with an inductance of 0.127 henry. When connected to a 230 V supply the current taken is 4.6 A. Determine the frequency of the supply. _____________________________________________________________________ _____________________________________________________________________ 36 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Summary The impedance of a series circuit may be calculated using Pythagoras’ equation: Z R2 X 2 where: R resistance in ohms X reactance X C or X L in ohms Current or in a series R-L or R-C circuit may be calculated using an equation similar to Ohm’s law. I V Z where: V voltage in rms volts Z total impedance in ohms I current in rms amperes EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 37 Answers Student exercise 1 1 VR V cos VXL = 100 × 0.5 = 50 V V sin = 100 × 0.866 = 86.6 V Student exercise 2 1 ZL = 318 L = 81 ZT = 434.3 IT = 0.553 A VRI = 138.3 V VL = 175.8 V T = 46.3 cos = 0.6908 2 38 13 ohm EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 Figure 29 Student exercise 3 1 50 Ω 2 2A 3 80 V: 60 V Figure 30 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812 39 Check your progress Part A Part B 1 (c) 1 12 , 67 2 (b) 2 0.506 A, 84 3 (a) 3 5 , 4 (b) 4 (a) 215 Ω 5 (a) 6 (a) 5 6A 7 (c) 6 (a) 170 8 (d) 9 (a) 10 (a) (b) 1.12 A (b) 1.41 A 7 49.7 Ω 8 50 Hz 11 (c) 12 (a) 40 EGG202A: 4 Analyse series ac circuits NSW DET 2017 2006/060/04/2017 LRR 3812