DOCSLIB.ORG

RL Circuit.Pdf

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
RL Circuit.Pdf RL Circuit Equipment: Capstone with 850 interface, 2 voltage sensors, RLC circuit board, 2 male to male banana leads approximately 80 cm in length 1 Introduction The three basic linear circuit elements are the resistor, the capacitor, and the inductor. This lab is concerned with the characteristics of inductors and circuits consisting of a resistor and an inductor in series (RL circuits). The primary focus will be on the response of an RL circuit to a step voltage and a voltage square wave. 2 Theory 2.1 Inductors An inductor is a 2-terminal circuit element that stores energy in its magnetic field. Inductors are usually constructed by winding a coil with wire. Initially, as the current increases in the coil, a magnetic field begins building across the coil. When the current stops flowing, the inductor will oppose the current change (Lenz's law) and try to keep the flow of current. This will be done producing a current in the opposite direction by the magnetic field it had stored. To increase the magnetic field inductors that are used for low frequencies often have the inside of the coil filled with a magnetic material. (At high frequencies such coils can be too loosy.) Inductors are the least perfect of the basic circuit elements due to the resistance of the wire they are made from. Often this resistance is not negligible, which will become apparent when the voltages and currents in an actual circuit are measured. If a current I is flowing through an inductor, the voltage VL across the inductor is proportional dI to the time rate of change of I, or dt . We may write dI V = L ; (1) L dt where L is the inductance in henries (H). The inductance depends on the number of turns of the coil, the configuration of the coil, and the material that fills the coil. A henry is a large unit of inductance. More common units are the mH and the µH. A steady current through a perfect inductor (no resistance) will not produce a voltage across the inductor. The sign of the voltage across an inductor depends on the sign of dI=dt and not on the sign of the current. The figure on page 2 shows the relationship between the current and voltage for a resistor and an inductor. 1 General Physics II Lab: RL Circuit Relationship between current and voltage in a resistor and inductor The arrow indicates the positive direction of the current through the two devices (resistor and inductor). That is, if the current is in the direction of the arrow, then I > 0. [Note that just because the arrows point down doesn't mean that the current is flowing in the downward direction. The current can also flow in the upwards direction, but in that case I < 0.] As mentioned above, the coil of wire making up an inductor has some resistance RL. This resistance acts like a resistor RL that is in series with an ideal inductor. (An ideal inductor is one with no resistance.) In this situation, the voltage across the inductor is dI V = L + R I: (2) L dt L 3 RL Circuits A series RL circuit with a voltage source V(t) connected across it is shown in Fig. 1. Figure 1: The loopy arrow indicates the positive direction of the current. The + and − signs indicate the positive values of the potential differences across the components. The voltage across the resistor and inductor are designated by VR and VL, and the current around the loop by I. The signs are chosen in the conventional way. I is positive if it is in the direction of the arrow. Kirchoff's law, which says that the voltage changes around the loop are zero, may be written VL + VR = V: (3) 2 General Physics II Lab: RL Circuit V0 R Voltage V across the resistor V R 0 V0 L Voltage V across the inductor V L 0 0 time Figure 2: Voltages VR across the resistor R and VL across the inductor L as a function of time after the source voltage is switched from V = 0 to V = V0. dI Assuming RL = 0 and letting VL = L dt and VR = IR Eq.(3) becomes dI L + RI = V: (4) dt When the applied source is shorted. The solution to the equation (V(t)=0=a short circuit) is − t I(t) = I0e L=R , where I0 is the current through the circuit at time t=0. This solution leads − t − t immediately to VR = I0Re L=R and VL = −I0Re L=R . The solution decays exponentially with a time constant of L/R. − t When a constant voltage is appled to an RL circuit e L=R describes the time dependence of the circuit. It is only necessary to use physical intuition and put the appropriate 2 constants into the solution. One guiding principle is that the current and VR do not change the instant the voltage across an RL circuit is changed. VR and I are continuous. The entire voltage change must appear across the inductor. After a constant voltage V is applied, the current in the circuit will exponentially approach V/R if the inductor has no resistance, or V/(R+RL) if the inductor has a resistance RL. Consider an RL circuit where V=0 and there is no current. We assume that RL = 0. If at time t=0 a constant voltage V is put across the circuit, VL and VR are given by, for t≥0, − t − t VL = V e L=R and VR = V (1 − e L=R ): (5) This behavior is illustrated in Fig. 2. The current is not plotted, but remember that the current is proportional to VR. Initially all of V appears across L because the current just before and after the application of V is zero. As the current exponentially builds up the voltage across the resistor increases and the voltage across the inductor decreases. If we wait a time T/2 where T/2 is many ∼ time constants (reminder time constants are L/R, 2L/R, 3L/R, etc) we will have VR = V and ∼ VL = 0. If now V is set equal to 0 (this is equivalent to shorting the circuit) VL and VR will be given by − t − t VL = −V e L=R and VR = V e L=R : (6) The response of an RL circuit to a constant voltage which is at first V and then 0 can be observed by applying a square wave to the circuit where one leg of the square wave has zero voltage and the other has voltage V. QUESTION. How would Fig. 2 be modified if the inductor had some 3 General Physics II Lab: RL Circuit resistance? QUESTION. If T L=R, what is the response of an RL circuit to a symmetric square wave that oscillates between +V and -V (assume RL = 0)? If a high frequency square wave, such that T L=R, is applied to an RL circuit, the changes in current and VR are minimal. There is not enough time for the current through the inductor to change much before the voltage is reversed. If the square wave is not symmetric with respect to ground the average VR will be the average voltage of the square wave, assuming RL = 0. 4 Measuring RL In this section the DC value of RL will be measured. Run the Capstone software. Go to the hardware setup window. Click on the output of the 850 interface and select Output Voltage - Current Senor. In the tools column, select Signal Generator icon and click on the 850 output 1. In waveform, pick DC, program it for 2 volts DC, and click on Auto. Next, setup a digits display and in select measurements pick Output Current. Program the digits display for at least 2 decimal places. Connect the output of the 850 interface directly across the coil (see Fig. 3) and click Record and then Stop. Inductor Resistors Inductor Core Figure 3: Picture of the circuit board From your data calculate RL, the resistance of the coil. Remove the wires from the coil. During your experiments compare the value of RL to the values of the resistances used in the RL circuits. Why do we need to measure RL? 4 General Physics II Lab: RL Circuit 5 Experiments In this section, the response of an RL circuit will be examined experimentally using the signal generator, 2 voltage sensors, and the oscilloscope display. In Capstone, go to the Hardware Setup window. Click on the output of the 850 interface and select Output Voltage - Current Senor. Program the analog inputs of the 850 interface for the two voltage sensors. Next, drag the scope icon to the center of the white screen. Resize the windows by using the orange tack. Add two additional y axis's to the scope by clicking the Add new y-axis to scope display icon twice. On the scope display, click Select Measurement on each axes, and set one as Output Voltage, and the other two as the analog voltages. In the experiment, various resistors will be connected in series with an 8.2 mH inductor. Hook the circuit and voltage sensors up as shown in Fig. 7, paying attention to the polarities of the leads and terminals. The polarities are very important in this experiment. A positive voltage from a voltage sensor will correspond to a positive voltage as given by the convention of Fig. 2. The input square wave to the RL circuit will be in the Output Voltage channel, VL and VR will be on the other two channels.
Load more

Recommended publications
  • Phasor Analysis of Circuits
    Phasor Analysis of Circuits Concepts Frequency-domain analysis of a circuit is useful in understanding how a single-frequency wave undergoes an amplitude change and phase shift upon passage through the circuit. The concept of impedance or reactance is central to frequency-domain analysis. For a resistor, the impedance is Z ω = R , a real quantity independent of frequency. For capacitors and R ( ) inductors, the impedances are Z ω = − i ωC and Z ω = iω L. In the complex plane C ( ) L ( ) these impedances are represented as the phasors shown below. Im ivL R Re -i/vC These phasors are useful because the voltage across each circuit element is related to the current through the equation V = I Z . For a series circuit where the same current flows through each element, the voltages across each element are proportional to the impedance across that element. Phasor Analysis of the RC Circuit R V V in Z in Vout R C V ZC out The behavior of this RC circuit can be analyzed by treating it as the voltage divider shown at right. The output voltage is then V Z −i ωC out = C = . V Z Z i C R in C + R − ω + The amplitude is then V −i 1 1 out = = = , V −i +ω RC 1+ iω ω 2 in c 1+ ω ω ( c ) 1 where we have defined the corner, or 3dB, frequency as 1 ω = . c RC The phasor picture is useful to determine the phase shift and also to verify low and high frequency behavior. The input voltage is across both the resistor and the capacitor, so it is equal to the vector sum of the resistor and capacitor voltages, while the output voltage is only the voltage across capacitor.
    [Show full text]
  • Network Analysis
    LECTURE NOTES ON NETWORK ANALYSIS B. Tech III Semester (IARE-R18) Ms. S Swathi Asistant professor ELECTRICAL AND ELECTRONICS ENGINEERING INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) DUNDIGAL, HYDERABAD - 50043 1 SYLLABUS MODULE-I NETWORK THEOREMS (DC AND AC) Network Theorems: Tellegen‘s, superposition, reciprocity, Thevenin‘s, Norton‘s, maximum power transfer, Milliman‘s and compensation theorems for DC and AC excitations, numerical problems. MODULE-II SOLUTION OF FIRST AND SECOND ORDER NETWORKS Transient response: Initial conditions, transient response of RL, RC and RLC series and parallel circuits with DC and AC excitations, differential equation and Laplace transform approach. MODULE-III LOCUS DIAGRAMS AND NETWORKS FUNCTIONS Locus diagrams: Locus diagrams of RL, RC, RLC circuits. Network Functions: The concept of complex frequency, physical interpretation, transform impedance, series and parallel combination of elements, terminal ports, network functions for one port and two port networks, poles and zeros of network functions, significance of poles and zeros, properties of driving point functions and transfer functions, necessary conditions for driving point functions and transfer functions, time domain response from pole-zero plot. MODULE-IV TWO PORTNETWORK PARAMETERS Two port network parameters: Z, Y, ABCD, hybrid and inverse hybrid parameters, conditions for symmetry and reciprocity, inter relationships of different parameters, interconnection (series, parallel and cascade) of two port networks, image parameters. MODULE-V FILTERS Filters: Classification of filters, filter networks, classification of pass band and stop band, characteristic impedance in the pass and stop bands, constant-k low pass filter, high pass filter, m- derived T-section, band pass filter and band elimination filter. Text Books: 1.
    [Show full text]
  • 33. RLC Parallel Circuit. Resonant Ac Circuits
    University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II -- Lecture Notes PHY 204: Elementary Physics II (2021) 12-4-2020 33. RLC parallel circuit. Resonant ac circuits Gerhard Müller University of Rhode Island, [email protected] Robert Coyne University of Rhode Island, [email protected] Follow this and additional works at: https://digitalcommons.uri.edu/phy204-lecturenotes Recommended Citation Müller, Gerhard and Coyne, Robert, "33. RLC parallel circuit. Resonant ac circuits" (2020). PHY 204: Elementary Physics II -- Lecture Notes. Paper 33. https://digitalcommons.uri.edu/phy204-lecturenotes/33https://digitalcommons.uri.edu/ phy204-lecturenotes/33 This Course Material is brought to you for free and open access by the PHY 204: Elementary Physics II (2021) at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: Elementary Physics II -- Lecture Notes by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected]. PHY204 Lecture 33 [rln33] AC Circuit Application (2) In this RLC circuit, we know the voltage amplitudes VR, VC, VL across each device, the current amplitude Imax = 5A, and the angular frequency ω = 2rad/s. • Find the device properties R, C, L and the voltage amplitude of the ac source. Emax ~ εmax A R C L V V V 50V 25V 25V tsl305 We pick up the thread from the previous lecture with the quantitative anal- ysis of another RLC series circuit. Here our reasoning must be in reverse direction compared to that on the last page of lecture 32. Given the
    [Show full text]
  • ELECTRICAL CIRCUIT ANALYSIS Lecture Notes
    ELECTRICAL CIRCUIT ANALYSIS Lecture Notes (2020-21) Prepared By S.RAKESH Assistant Professor, Department of EEE Department of Electrical & Electronics Engineering Malla Reddy College of Engineering & Technology Maisammaguda, Dhullapally, Secunderabad-500100 B.Tech (EEE) R-18 MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY II Year B.Tech EEE-I Sem L T/P/D C 3 -/-/- 3 (R18A0206) ELECTRICAL CIRCUIT ANALYSIS COURSE OBJECTIVES: This course introduces the analysis of transients in electrical systems, to understand three phase circuits, to evaluate network parameters of given electrical network, to draw the locus diagrams and to know about the networkfunctions To prepare the students to have a basic knowledge in the analysis of ElectricNetworks UNIT-I D.C TRANSIENT ANALYSIS: Transient response of R-L, R-C, R-L-C circuits (Series and parallel combinations) for D.C. excitations, Initial conditions, Solution using differential equation and Laplace transform method. UNIT - II A.C TRANSIENT ANALYSIS: Transient response of R-L, R-C, R-L-C Series circuits for sinusoidal excitations, Initial conditions, Solution using differential equation and Laplace transform method. UNIT - III THREE PHASE CIRCUITS: Phase sequence, Star and delta connection, Relation between line and phase voltages and currents in balanced systems, Analysis of balanced and Unbalanced three phase circuits UNIT – IV LOCUS DIAGRAMS & RESONANCE: Series and Parallel combination of R-L, R-C and R-L-C circuits with variation of various parameters.Resonance for series and parallel circuits, concept of band width and Q factor. UNIT - V NETWORK PARAMETERS:Two port network parameters – Z,Y, ABCD and hybrid parameters.Condition for reciprocity and symmetry.Conversion of one parameter to other, Interconnection of Two port networks in series, parallel and cascaded configuration and image parameters.
    [Show full text]
  • How to Design Analog Filter Circuits.Pdf
    a b FIG. 1-TWO LOWPASS FILTERS. Even though the filters use different components, they perform in a similiar fashion. MANNlE HOROWITZ Because almost every analog circuit contains some filters, understandinghow to work with them is important. Here we'll discuss the basics of both active and passive types. THE MAIN PURPOSE OF AN ANALOG FILTER In addition to bandpass and band- age (because inductors can be expensive circuit is to either pass or reject signals rejection filters, circuits can be designed and hard to find); they are generally easier based on their frequency. There are many to only pass frequencies that are either to tune; they can provide gain (and thus types of frequency-selective filter cir- above or below a certain cutoff frequency. they do not necessarily have any insertion cuits; their action can usually be de- If the circuit passes only frequencies that loss); they have a high input impedance, termined from their names. For example, are below the cutoff, the circuit is called a and have a low output impedance. a band-rejection filter will pass all fre- lo~~passfilter, while a circuit that passes A filter can be in a circuit with active quencies except those in a specific band. those frequencies above the cutoff is a devices and still not be an active filter. Consider what happens if a parallel re- higlzpass filter. For example, if a resonant circuit is con- sonant circuit is connected in series with a All of the different filters fall into one . nected in series with two active devices signal source.
    [Show full text]
  • AC Power • Resonant Circuits • Phasors (2-Dim Vectors, Amplitude and Phase) What Is Reactance ? You Can Think of It As a Frequency-Dependent Resistance
    Physics-272 Lecture 20 • AC Power • Resonant Circuits • Phasors (2-dim vectors, amplitude and phase) What is reactance ? You can think of it as a frequency-dependent resistance. 1 For high ω, χ ~0 X = C C ωC - Capacitor looks like a wire (“short”) For low ω, χC∞ - Capacitor looks like a break For low ω, χL~0 - Inductor looks like a wire (“short”) XL = ω L For high ω, χL∞ - Inductor looks like a break (inductors resist change in current) ("XR "= R ) An RL circuit is driven by an AC generator as shown in the figure. For what driving frequency ω of the generator, will the current through the resistor be largest a) ω large b) ω small c) independent of driving freq. The current amplitude is inversely proportional to the frequency of the ω generator. (X L= L) Alternating Currents: LRC circuit Figure (b) has XL>X C and (c) has XL<X C . Using Phasors, we can construct the phasor diagram for an LRC Circuit. This is similar to 2-D vector addition. We add the phasors of the resistor, the inductor, and the capacitor. The inductor phasor is +90 and the capacitor phasor is -90 relative to the resistor phasor. Adding the three phasors vectorially, yields the voltage sum of the resistor, inductor, and capacitor, which must be the same as the voltage of the AC source. Kirchoff’s voltage law holds for AC circuits. Also V R and I are in phase. Phasors R ε ω Problem : Given Vdrive = m sin( t), C L find VR, VL, VC, IR, IL, IC ε ∼ Strategy : We will use Kirchhoff’s voltage law that the (phasor) sum of the voltages VR, VC, and VL must equal Vdrive .
    [Show full text]
  • Chapter 14 Frequency Response
    CHAPTER 14 FREQUENCYRESPONSE One machine can do the work of fifty ordinary men. No machine can do the work of one extraordinary man. — Elbert G. Hubbard Enhancing Your Career Career in Control Systems Control systems are another area of electrical engineering where circuit analysis is used. A control system is designed to regulate the behavior of one or more variables in some desired manner. Control systems play major roles in our everyday life. Household appliances such as heating and air-conditioning systems, switch-controlled thermostats, washers and dryers, cruise controllers in automobiles, elevators, traffic lights, manu- facturing plants, navigation systems—all utilize control sys- tems. In the aerospace field, precision guidance of space probes, the wide range of operational modes of the space shuttle, and the ability to maneuver space vehicles remotely from earth all require knowledge of control systems. In the manufacturing sector, repetitive production line opera- tions are increasingly performed by robots, which are pro- grammable control systems designed to operate for many hours without fatigue. Control engineering integrates circuit theory and communication theory. It is not limited to any specific engi- neering discipline but may involve environmental, chemical, aeronautical, mechanical, civil, and electrical engineering. For example, a typical task for a control system engineer might be to design a speed regulator for a disk drive head. A thorough understanding of control systems tech- niques is essential to the electrical engineer and is of great value for designing control systems to perform the desired task. A welding robot. (Courtesy of Shela Terry/Science Photo Library.) 583 584 PART 2 AC Circuits 14.1 INTRODUCTION In our sinusoidal circuit analysis, we have learned how to find voltages and currents in a circuit with a constant frequency source.
    [Show full text]
  • LABORATORY MANUAL Network Theory Lab EE-223-F (3Rd Semester)
    Electrical & Electronics Engineering Department BRCM COLLEGE OF ENGINEERING & TECHNOLOGY BAHAL – 127028 ( Distt. Bhiwani ) Haryana, India Laboratory LABORATORY MANUAL Network Theory Lab EE-223-F (3rd Semester) Prepared By: Vivek Kumar(A.P) B. Tech. (EEE), M. Tech. (EEE) Department of Electrical & Electronics Engineering BRCM College of Engineering &Technology Bahal-127028 (Bhiwani). 1 Electrical & Electronics Engineering Department BRCM COLLEGE OF ENGINEERING & TECHNOLOGY BAHAL – 127028 ( Distt. Bhiwani ) Haryana, India NETWORK THEORY LAB Index S. No Name of Experiment Page No 1 To find resonance frequency , Bandwidth , Q - factor of RLC series circuit 2 To study and plot the transient response of RL circuit 3 To study and plot the transient response of RC circuit. 4 To calculate and verify 'Z' parameters of two-port network 5 To calculate and verify 'Y' parameters of two-port network 6 To calculate and verify 'ABCD' parameters of two-port network 7 To determine equivalent parameters of parallel connection of two-port network 8 To plot the frequency response of High pass filter and determine the half-power frequency 9 To plot the frequency response of Low pass filter and determine the half- power frequency To study frequency response of Band pass filter 2 Electrical & Electronics Engineering Department BRCM COLLEGE OF ENGINEERING & TECHNOLOGY BAHAL – 127028 ( Distt. Bhiwani ) Haryana, India NETWORK THEORY LAB EXPERIMENT NO :1 AIM: To find resonance frequency , Bandwidth , Q - factor of RLC series circuit APPARATUS REQUIRED : Power Supply, Function Generator, CRO, Series Resonance kit, Connecting Leads. BRIEF THEORY : The ckt. is said to be in resonance if the current is in phase with the applied Voltage .
    [Show full text]
  • LECTURE 19 RL CIRCUITS and AC CIRCUITS I Lecture 19
    LECTURE 19 RL CIRCUITS AND AC CIRCUITS I Lecture 19 2 ! Reading chapter 28-8, 29-1, and 29-4. ! Concepts: ! RL circuits (DC) ! LC circuits ! RLC circuits ! AC circuit with resistor ! RMS RL circuit 3 ! A circuit containing a resistor and an inductor is called an RL circuit. ! Current in a RL circuit cannot change discontinuously. ! In a dc circuit, an inductor acts as a short circuit long time after the switch is closed/opened. Energizing an RL circuit 4 ! The switch is closed at t = 0. dI E0 − IR − L = 0 From Kirchhoff’s loop rule dt ! The current in the circuit grows and approaches the final value If. E I t 0 1 e−t /( L/ R) I 1 e−t /τ ( ) = ( − ) = f ( − ) R ! The time constant of the circuit is given by L τ = R Current after short time 5 ! The switch was open, then closed. ! Initially the inductor can be treated as an open circuit. Current after long time 6 ! A long time after the switch is closed, the inductor can be treated as a short circuit. Quiz: 19-1 & 19-2 De-energizing an RL circuit 8 ! The switch has been at position e for a long time, then switched to position f at t = 0. ! The current in the inductor starts out at I0 and exponentially decays. dI −IR − L = 0 From Kirchhoff’s loop rule dt E I t = I e−t /τ = 0 e−t /τ ( ) 0 R + R 1 Demo: 1 9 ! RL circuit ! An inductor in series with a resistor, a battery and a switch.
    [Show full text]
  • First Order Circuits
    First Order Circuits EENG223 Circuit Theory I First Order Circuits A first-order circuit can only contain one energy storage element (a capacitor or an inductor). The circuit will also contain resistance. So there are two types of first- order circuits: z RC circuit z RL circuit Source-Free Circuits A source-free circuit is one where all independent sources have been disconnected from the circuit after some switch action. The voltages and currents in the circuit typically will have some transient response due to initial conditions (initial capacitor voltages and initial inductor currents). We will begin by analyzing source-free circuits as they are the simplest type. Later we will analyze circuits that also contain sources after the initial switch action. SOURCE-FREE RC CIRCUITS z Consider the RC circuit shown below. Note that it is source-free because no sources are connected to the circuit for t > 0. Use KCL to find the differential equation: dv 1 t = 0 + += v(t) 0 for t ≥ 0 dt RC _+ VX R C v (t) z and solve the differential _ equation to show that: -t RC v(t) = VXe for t ≥ 0 SOURCE-FREE RC CIRCUITS Checks on the solution z Verify that the initial condition is satisfied. z Show that the energy dissipated over all time by the resistor equals the initial energy stored in the capacitor. First Order Circuits General form of the D.E. and the response for a 1st-order source-free circuit z In general, a first-order D.E. has the form: dx 1 += x(t) 0 for t ≥ 0 dt τ Solving this differential equation (as we did with the RC circuit) yields: -t
    [Show full text]
  • Phasor Diagram
    Electric Circuits II Phasor Diagram Dr. Firas Obeidat 1 Phasor diagram for the Passive Circuit Elements The Resistor Let 풊(풕) = 푰풎풄풐풔⁡(흎풕 + 흓) In polar form But Vm∟θ and Im∟ 흓 merely represent the general voltage and current phasors V and I. Thus The angles θ and 흓 are equal, so that the current and voltage are always in phase. The Inductor 풋(흎풕+흓) Let 풊(풕) = 푰풎풄풐 풔 흎풕 + 흓 = 푰풎풆 2 Dr. Firas Obeidat – Philadelphia University Phasor diagram for the Passive Circuit Elements We obtain the desired phasor relationship Note that the angle of the factor jωL is exactly +90◦ and that I must therefore lag V by 90° in an inductor. The Capacitor 풋(흎풕+흓) Let 풗(풕) = 푽풎풄풐 풔(흎풕 + 흓) = 푽풎풆 풅풗(풕) 풊 풕 = 푪 푰 풆풋휽 = 풋흎푪푽 풆풋흓 풅풕 풎 풎 ퟏ 푽 = I 푰 = 푗휔퐶푽 푗휔퐶 Note that the angle of the factor 1/jωC is exactly -90◦ and that I must therefore lead V by 90° in an Capacitor. 3 Dr. Firas Obeidat – Philadelphia University Phasor diagram for series RL circuit Example: for the circuit shown in figure (a), draw the phasor circuit , impedance diagram and voltages phasor diagram. V=100∟0, so the phasor circuit is shown in figure (b). o ZT=ZR+ZL=3Ω+j4Ω =5∟53.13 . Impedance diagram is shown in figure (c). 표 푉 100∟0 o 퐼 = = o= 20∟−53.13 푍푇 5∟53.13 o o VR=IZR=(20∟-53.13 A)(3∟0Ω)=60∟-53.13 V. o o VL=IZL=(20∟-53.13 A)(4∟90Ω)=80∟36.87 V.
    [Show full text]
  • Electrical Circuits Lab. 0903219 Series RL Circuit Phasor Diagram
    Electrical Circuits Lab. 0903219 Series RL Circuit Phasor Diagram - Simple steps to draw phasor diagram of a series RL circuit without memorizing: * Start with the quantity (voltage or current) that is common for the resistor R and the inductor L, which is here the source current I (because it passes through both R and L without being divided). Figure (1) Series RL circuit * Now we know that I and resistor voltage VR are in phase or have the same phase angle (there zero crossings are the same on the time axis) and VR is greater than I in magnitude. * Since I equal the inductor current IL and we know that IL lags the inductor voltage VL by 90 degrees, we will add VL on the phasor diagram as follows: * Finally, the source voltage VS equals the vector summation of VR and VL: Figure (2) Series RL circuit Phasor Diagram Prepared by: Eng. Wiam Anabousi - Important notes on the phasor diagram of series RL circuit shown in figure (2): A- All the vectors are rotating in the same angular speed ω. B- This circuit acts as an inductive circuit and I lags VS by a phase shift of Ө (which is the current angle if the source voltage is the reference signal). Ө ranges from 0o to 90o (0o < Ө <90o). If Ө=0o then this circuit becomes a resistive circuit and if Ө=90o then the circuit becomes a pure inductive circuit. C- The phase shift between the source voltage and its current Ө is important and you have two ways to find its value: a- b- = - = - D- Using the phasor diagram, you can find all needed quantities in the circuit like all the voltages magnitude and phase and all the currents magnitude and phase.
    [Show full text]