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The phasor voltage Vab in the circuit shown in (Figure 1) is 300 0V (rms) when no external load is connected to the terminals a,b.When a load having an impedance of 200j500 is connected across a,b,the value of Vab is 156+-42V (rms).a) find the impedance that should be connected across a,b for maximum average power transfer. results for this questionHow do you calculate branch currents in phasor?How do you calculate branch currents in phasor?The branch currents are also indicated in phasor form this time.We can easily calculate the three branch impedances Figure 3.8.The same circuit in phasor notation.(3.131) Z 3 = R 3 + j L 3.Once the impedances are found,the circuit equations can be set up.Phasor Notation - an overview ScienceDirect Topics

If V ab = 400 V in a balanced Y-connected three-phase generator,find the phase voltages,assuming the phase sequence is What is the phase sequence of a balanced three-phase circuit for which V an = 160 solvedthe phasor voltage v ab in the circuit shown in f#176;30 V and V cn = 160 solvedthe phasor voltage v ab in the circuit shown in f#176; 90 V? Find V bn .Chapter 12,Solution 2.Since phase c lags phase a by 120 solvedthe phasor voltage v ab in the circuit shown in f#176;,this is an acb sequence.1.For the circuit below Vs = 12 volts,R1 = R2 = R3 = 24K B) What is the voltage at terminals A-B? C) Sketch the Thevenin equivalent circuit and calculate the Thevenin resistance and voltage.Show all work.2.In the circuit below Vi = 3 volts.What is the value of Vo.Show all work.3.For each circuit below determine the output voltage V o if V i = 3 volts.A)

Solution for For the full wave rectifier circuit shown,if Vs=15sin(120nt) (V),R=10o11,assuming constant voltage drop with VDo=0.7 J.1.LDraw clearly andBalanced Three-Phase Circuitsphase internal voltage of the generator is specified as the reference phasor.Construct the aphase equivalent circuit of the system.Calculate the three line currents I aA,I bB,and I cC.a)Calculate the three phase voltages at the load.V AN,V BN,and V CN.b)Calculate the line voltages V AB,V BC,and V

Here the circuit forms a voltage divider.We can use the voltage divider formula with the impedances.V O = V S Z C Z C + Z R = (12640 V) j2 8 j2 = 2:916 35:96 V You should then move the phasor result back to the time-domain v O(t) = 2:91cos(4t 35:96 ) V This can be done in MATLAB using the script shown below.1 clear all%ClearallvariablesintheWorkspaceChapter12 - 518 PART 2 12.7 AC Circuits In a system a phaseTake Z = 12 j 15,Z Y = 4 + j 6,and Z = 2.12.20 A balanced delta-connected source has phase voltage V ab = 416 30 V and a positive phase sequence.If this is connected to a balanced delta-connected load,find the line and phase currents.Take the load impedance per phase as 60 30 and line impedance per phase as 1 + j 1.

ab AB an AN V V VE IV Zaan an / Learning Objectives a.Derive the relationship between line to line voltages and line to neutral voltages in a balanced Y-Y three phase circuit b.For a balanced Y-Y three phase circuit convert the line to line voltage phasor to the line to neutral voltage phasor and vice versaET 332a Ac Motors,Generators and Power SystemsExample 5-1 A balanced delta connected three-phase load draws 200 A per phase with a leading power factor of 0.85 from a 12.47 kV line to line system.Determine the following a) The line current magnitude of the load b) the phase voltage magnitude of the load c) the total apparent power of the load d) the total real power drawn by the load

Example The three-phase balanced load shown in the circuit has impedance per phase of ZY=8+j6 .If the load is connected to 208-V lines,predict the readings of the wattmeters W 1 and W 2.Find P T and Q T.The impedance per phase is The pf angle is 36.87oExperiment 4 (5V supplies) Op-Amp Circuitsomitted from simple circuit schematics to improve clarity.Figure A-4.The balance (or null offset) pins (1 and 8) provide a way to eliminate any offset in the output voltage of the amplifier.The offset voltage (usually denoted by Vos) is an artifact of the integrated circuit.The offset voltage is additive with V O (pin 6 in this case).It

Thus the line voltage lags the corresponding phase voltage by 30 solvedthe phasor voltage v ab in the circuit shown in f#176;.The phasor diagram for this connection is shown below.Van Vbn Vcn Vab V bc 2-5.Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure P2-3.SOLUTION Note that because this load is -connected,the line and phase voltages are Lesson 4 Three Phase Sources and LoadsExample 4-2 For the delta connected load shown,a) find the phasor values of phase and line currents for the circuit.b) draw a phasor diagram of the computed currents and given voltages R = 15 ohms in each phase.V ab = 240 0o V bc = 240 o-120 V ca o= 240 -240 240 0o 240 -240o 240 -120o Example 4-2 Solution (1) 20 Lesson 4_et332b.pptx

v an = 2 V i sin (t),v bn = 2 V i sin (t 2 3),v cn = 2 V i sin (t 4 3),v ab = v an v bn = 6 V i sin (t + 6) The waveforms that will be used for the analysis of this section are the same as the ones shown in Fig.4.10(d)(h) .

Consider two AC voltages,V 1 having a peak voltage of 20 volts,and V 2 having a peak voltage of 30 volts where V 1 leads V 2 by 60 o.The total voltage,V T of the two voltages can be found by firstly drawing a phasor diagram representing the two vectors and then constructing a parallelogram in which two of the sides are the voltages,V 1 and Phasor Domain - an overview ScienceDirect TopicsSolution The peripheral impedance is Z(f)=V(f)/I(f) where V(f) is the phasor representation of the aortic pressure and I(f) is the phasor notation for the blood flow.(Since this example describes frequency ranges in Hz instead of radians,we use f instead of .) One approach is to apply a V(f) having a known value and solving for the current,I(f).The impedance then is just the ratio.

When using phasor notation,first one waveform must be chosen as the reference.In this example,the reference will be waveform A.The reference waveform phasor,E A,is then positioned along the X axis,as shown in Figure 4.36,at the zero-degree rotational reference.This phasor is a vector representing the voltage of an ac generator as its conductors are rotated through a magnetic field.Phasor Voltage - an overview ScienceDirect Topicsthe line voltage is 3 times the phase voltage and leads it by 30 solvedthe phasor voltage v ab in the circuit shown in f#176; (V L = 3V PH 30 solvedthe phasor voltage v ab in the circuit shown in f#176;); the phase current is equal to the phase voltage divided by the phase impedance (I PH =V PH /Z PH); the phase angle is given by = tan 1 (X L /R L) whereX L andR L are the load reactance and resistance,respectively; the power factor is given by cos .

The dashed-line phasor is the same angle as the solid-line phasor reaching from point B to point A,just moved so that its tail rests at the circles center in order to plot its vertical projection on the rectangular graph to the right of the circle.As you can see,the new phasor \(V_{AB}\) is 180\(^{o}\) shifted from the old phasor \(V_{BA}\).Q31.1 A resistor is connected across an ac source assource as shown.For this circuit,what is the relationship between the instantaneous current i through the resistor and the instantaneous voltage v ab across the resistor? Q31.1 A.i is maximum at the same time as v ab B.i is maximum one-quarter cycle before v ab C.i is maximum one-quarter cycle after v ab D.not enough information given to

In an RL series circuit,a pure resistance (R) is connected in series with a coil having the pure inductance (L).To draw the phasor diagram of RL series circuit,the current I (RMS value) is taken as reference vector because it is common to both elements.Voltage drop V R is in phase with current vector,whereas,the voltage drop in inductive reactance V L leads the current vector by 90 o Section 8-2 and 8-3 Average and Complex Power(e) Vrms =1260 V,Irms =230 A.Solution (a) Vrms = 100 2 ej30 V,Irms = 2.5 2 ej60 V.S =VrmsI rms = 100 2 ej30 solvedthe phasor voltage v ab in the circuit shown in f#215; 2.5 2 ej60 =125ej30 (VA) S =|S=125 VA Pav =Re[S]=125cos30 =108.25 W Q =Im[S]=125sin30 =62.5 VAR s =v i =30 +60 =30 (hence pf is lagging) pf =cos30 =0.866.(b) Vrms = 25 2 ej40 V,Irms = 0.2 2 ej10 A,S =VrmsI

Oct 25,2020 solvedthe phasor voltage v ab in the circuit shown in f#0183;So draw the voltage phasor,V R along same axis or direction as that of current phasor i.e V R is in phase with I.Step III.We know that in inductor,voltage leads current by 90 solvedthe phasor voltage v ab in the circuit shown in f#176; so draw V l (voltage drop across inductor) perpendicular to current phasor in leading direction.Solved The Phasor Voltage V_ab In The Circuit Shown In (F The phasor voltage V_ab in the circuit shown in (Figure 1) is 250 angle 0 degree V(rms) when no external load is connected to the terminals a,b.When a load having an impedance of 200 - j500 ohm is connected across a,b,the value of V_ab is 130 - j35 V(rms).

The phasor voltage V_ab in the circuit shown in (Figure 1) is 275 solvedthe phasor voltage v ab in the circuit shown in flt; 0 degree V (rms) when no external load is connected to the terminals a,b.When a load having an impedance of 200 - j500 Ohm is connected across a.b,the value ot V_ab is 143 - j38.5 V (rms).Solved The Phasor Voltage V_ab In The Circuit Shown In Fi The Phasor Voltage V_ab in the circuit shown in Fig.3 is 240 0 degree V (rms) when no external load is connected to the terminals a,b.When a load having an impedance of 90 - j30 Ohm is connected across a,b,the value of V_ab is 115.2 - j 86.4 V (rms).a) Find the impedance that should be connected across a,b for maximum average power transfer.

Topic 13 Phasor Analysis Given the circuit shown in Figure 1,find v,the voltage across the current source.Figure 1.A circuit in steady-state The first step is to convert the actual circuit to its phasor equivalent.The circuit shown in Figure 2 includes these conversions.Figure 2.The phasor equivalent of the Figure 1 circuit

voltage value Vd and power Pd.S5.2 A single-phase diode rectifier A single-phase diode rectifier is shown in the figure below.The rms value of the grid voltage is Vs = 230V.Assume an ideal grid (Ls=0).Assume that the load is represented by a constant dc current,Id = 10A.The grid frequency is equal to 50 Hz.fHz IA LmH V d s s 50 10 5 230 = = = = vs(t)Thevenin equivalent circuits - Iowa State UniversityJun 10,2014 solvedthe phasor voltage v ab in the circuit shown in f#0183;The open-circuit voltage / short-circuit current approach can be used to calculate the Thevenin equivalent for a known circuit.Consider the circuit from slide 4 + V S R 1 R 2 I S 9V 6 mA 1.5 k! 3 k! Open-circuit voltage Use whatever method you prefer.Well use node voltage in this case.+ V S R 1 R 2 I S v a + v oc YRF= YD

Assume a three-phase system with voltages as shown in fig.20-32.The applied voltage is shown in fig.20-32.The voltage vectors are drawn in fig.20-33b in the directions (1),(2) and (3) as they occur spatially and as applied to the stator coils.What is Short Transmission Line? Phasor Diagram Aug 15,2020 solvedthe phasor voltage v ab in the circuit shown in f#0183;This is for lagging power factor load or inductive load.Phasor OI,OA,AB,BC,AC and OC represents load current I.V S is sending end voltage and V R is receiving end voltage.IR is resistive drop in the line.IX is reactive drop in the line.IZ is impedance drop.From phasor diagram,sending end voltage V

A synchronized UJT trigger circuit using a UJT is shown below.Diodes D 1 D 4 rectifies ac to dc.Resistors R 1 lowers V dc to a suitable value for the Zener diode and UJT.Zener diode Z functions to clip the rectified voltage to a standard level V z,which remains constant except near the V dc zero.It prevents erratic firing.

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