Introduction Of Active And Passive Components

Basic Electrical & Electronics lS.P.I.TEXPERIMENT No:-4DATE:-// 2011HALF WAVE & FULL WAVE RECTIFIERAim :To study half wave and full wave rectifier circuit with and without capacitor filter.Apparaturus & Components :- C.R.O.Center tap alf wave Rectifier:- When the switch(S1) is open the circuit acts as a half waverectifier (HWR). During the positive half cycle of input ac voltage ‘A’ becomes positivew.r.t ‘B’which makes the diode (D1) forward biased & hence it conducts current. Duringthe negative half cycle the diode is reversed biased and no current flows in the circuit.Therefore, current flows through diode only during the half cycle. Hence dc current isobtained across load.Full wave Rectifier:- When the switch(S1) is closed the circuit acts as a fullwave rectifier (FWR). During the positive half cycle of the secondary voltage diode(D1) forward biased & diode (D2) become reverse biased. Therefore D1 conducts and D2does not conduct. During the negative half cycle D2 is forward biased & D1 is reversebiased. The load current flows in Both half cycles of ac voltage and in the same directionthrough the load resistance. Hence we get the rectified output across the load.Circuit operation with filter:- In capacitor filter, capacitor is connected inparallel with load R. During positive half cycle, diode D1 is forward biased & D2 isreverse biased. Therefore current flows in D1 charging the capacitor to maximum valuesay V. Since capacitor and load resistance are in parallel, voltage across capacitor willappear as output Voltage. Now the capacitor C starts discharging through load resistanceRL. At the same time another diode gets forward biased & starts conducting. Thecapacitor C again starts charging and quickly gets charged Through.The forward biasedPage23

Basic Electrical & Electronics lS.P.I.Tdiode having small forward resistance. The time required by the capacitor to change topeak value is quite small & diode current again reduces zero.Theoretical calculation:-Procedure :a. Connect the circuit as shown in the figure.b. For HWR keep switch “S1” open while for FWR keep the switch“S1” closed.c. Observe the waveform for both HWR & FWR with and without filter andplot the graphs.Page24

Basic Electrical & Electronics lS.P.I.TObservation table:Primary Input Voltage (Vp) 230 VSecondary Voltage (Vs) Without Filter quencyHzFWRWith Filter :Vr P-P(volts)Vm(volts)Vdc Vm - Vr P-P /2(volts)HWRFWRPage25Vr RMS (Vr P-P / 2 3)r Vr RMS /Vdc

Basic Electrical & Electronics lResultS.P.I.T:Conclusion:Page26

Basic Electrical & Electronics lS.P.I.TSERIES AND PARALLEL RLC RESONANCEPage27

Basic Electrical & Electronics lS.P.I.TCIRCUIT DIAGRAM:RCLmASignalGeneratorFig.4.1 RLC series circuitRLmACSignalGeneratorFig.4.2 RLC parallel circuitPage28

Basic Electrical & Electronics lS.P.I.TEXPERIMENT No:-5DATE:-// 2011RLC series and parallel circuitAim:- To study the Resonance phenomena of Series and parallel RLC circuit.Apparatus :Variable resistorsInductorCapacitorSignal GeneratorMilliammeters.Theory:- Series RLC circuitIn an RLC series circuit impedance is given by Z (R2 (XL - XC))The frequency at which the reactive part in the impedance becomeszero is calledresonant frequency. So at resonanceXL XC or Z R2 л f0 L 1/(2 л f0C)f0 1/(2 л LC)At resonance current is maximum where as impedance ‘Z’ is minimum. Bandwidthof circuit is given by band of frequencies which lies between the two points at which the currentis 1/ 2 times the current at resonant frequency.Parallel RLC circuit:In a parallel RLC circuit when the voltage and total current are in phase at aparticular frequency then parallel circuit is said to be at resonance. The frequency at which theparallel resonance occurs is called resonant frequency. In the circuit , current drawn by inductivebranch is IL V/ZLwhere ZL R jXL and current drawn by capacitive branch IC V/XCwhere XC 1/(2 л f C) IL lags voltage ‘V’ by angle ΦLPage29

Basic Electrical & Electronics lS.P.I.TVC/XC ( V/ZL).(XL/ZL) ( VXL)/ZL2ZL2 XL . XCR2 (2 л f0 L)2 ( 2 л f0 L) . (1/2 л f0 C)(2 л f0 L)2 (L/C) – R2f0 (1/2 л ) ((1/LC)-(R2/L2))Now if R is very small as compared to L and C then (R2/L2) (1/LC) and f0 1/(2 л LC)At parallel resonance current is minimum while total impedance of circuit is maximum.Procedure :1) Connect the Series RLC circuit as shown in figure 1.2) Make sure that the milliammeter is in series with the circuit.3) Keep the values fixed of R, L and C.4) Keep varying frequency of the input signal from zero onwards5) Measure frequency ‘f’ and current ‘I’ in mA.6) Tabulate readings and draw graphs.7) Repeat the above procedure for Parallel RLC circuit.Page30

Basic Electrical & Electronics lS.P.I.TSERIES AND PARALLEL RLC RESONANCEObservation Table:Sr. No.Series ResonanceF (KHz)I age31Parallel ResonanceF (KHz)I (mA)

Basic Electrical & Electronics lS.P.I.TResult :Conclusion :Page32

Basic Electrical & Electronics lS.P.I.TTHREE PHASE POWER MEASUREMENTUSING TWO WATTMETER METHODPage33

Basic Electrical & Electronics lS.P.I.TCircuit Diagram :-Fig. 5.1 Power measurement using 2 wattmeter methodObservation VI(W)W1(W)W2(W)W1 W2(W)Error%

Basic Electrical & Electronics lS.P.I.TEXPERIMENT NO.6DATE:// 20113 phase power measurement using 2 wattmeter methodAim:- To measure power in a 3 phase circuit using two wattmeter method.Apparatus:- Two wattmeter:Ammeters:Lamp loadsVoltmeter(DMM)Theory :If the 3 phase load is balanced or even ubalanced, the power in the load canbe found by two wattmeter method. Let VR, VY & VB be rms value of phase voltages andIR, IY & IB be the rms value of phase currents. Assume the load to be inductive & hencephase currents will lag respective phase voltages by some phase angle as in fig. below.Current through wattmeter W1 IR. Potential Difference (P.D) across voltage coil of W1 VRB VR -VB. Angle between the current through the current coil & the voltage acrossthe pressure (voltage) coil is (30- ф).Reading of W1 IR VRB cos (30 - ф). Similarly, if current through wattmeter W2 IYPotential Difference (P.D) across voltage coil of W2 VYB VY -VBFrom the phasor diagram, phase difference between VYB & IY 30 фReading of W2 IY VYB cos (30 ф). Since load is balanced therefore VYB VRB VYR VL (Line voltage) & IR IY IB IL (Line current). W1 VL IL cos (30 - ф),Page35

Basic Electrical & Electronics lS.P.I.TW2 VL IL cos (30 ф)W1 W2 VL IL cos (30 - ф) VL IL cos (30 ф) VL IL [cos 30 cos ф sin 30 sin ф cos 30 cos ф - sin 30 sin ф] VL IL [2 cos 30 cos ф] 3 VL IL cos фW1 W2 Total power in 3 phase loadNote: For resistive load ф 0, therefore cos ф 1Procedure: 1) Connect the circuit as shown in the circuit diagram.2) Adjust the 3-ф transformer voltage to 200volts.3) Vary the lamp load in 3 phases.4) Measure VR, VY, VB, IR, IY & IB .5) Take 3 to 4 such ------------Conclusion:- e36

Basic Electrical & Electronics lS.P.I.TO.C. & S.C. TEST OF A SINGLE Φ TRANSFORMERPage37

Basic Electrical & Electronics lS.P.I.TCircuit Diagram for Open circuit test :-Circuit Diagram for Short circuit test :-Page38

Basic Electrical & Electronics lS.P.I.TEXPERIMENT No:-07DATE:-// 2011O.C & S.C. test on 1- Φ transformerAim :- To carry out open circuit & short circuit test in a single phase transformer andfind percentage efficiency, percentage regulation and equivalent circuit.Apparaturus & components required :Single Φ transformer:Variable AC source:AC wattmeter:Voltmeters:Theory:A transformer is a device used to change voltages and currents of AC electricpower. In the simplest version it consists of two windings wrapped around a magneticcore; windings are not electrically connected, but they are coupled by the magnetic field,as it shown in Figure. When one winding is connected to the AC electric power, theelectric current is generated. This winding is called the primary winding. The primarycurrent produces the magnetic field and the magnetic flux links the second winding,called the secondary winding. The AC flux through the secondary winding produces anAC voltage, so that if some impedance is connected to the terminals, an AC electriccurrent is supplied.Page39

Basic Electrical & Electronics lS.P.I.TTRANSFORMEROPEN CIRCUIT TESTIt determines the iron loss & No load current (IO). One winding is left open andanotherwinding of transformers connected to voltage supply. Flux is set up in the core dependsupon the applied voltage & iron loss occurs. IO is very small and Cu loss is negligible inprimary and absent in secondary. The wattmeter readings give us the iron losses. In orderto simplify calculation voltage, current and impedance are either primary or secondary IOis no load current. RO and XO are resistance and reactance of circuit.SHORT CIRCUIT TESTThis test determines Cu loss & short circuit current (ISC) which offers the leastimpedance (ZSC) & RSC. In this one winding is short circuited and another windingconnected to very low voltage supply. The wattmeter readings give us the copper losses.The applied voltage is very less so flux set up is very less since iron loss will benegligible and the current flowing through short circuited winding and hence the rated orfull load current through the wattmeter which represents copper losses.Procedure :A: Open circuit test1) Make connections as per the circuit diagram of Figure.2) Adjust the autotransformer to the values specified in the Table and record Io,V1 and Wi for each voltage step.3) Set the autotransformer to zero position. Turn the power off.4) Remove the connections for the circuit.B: Short circuit test1) Make connections as per the circuit diagram of Figure. A short wire can beused for the short circuit connection of the S1 and S2 terminals.2) Make sure that the 1-Ф variable autotransformer (VARIAC) is kept in ZEROPosition initially.Page40

Basic Electrical & Electronics lS.P.I.T3) Turn the power on. Slowly vary the 1-Ф autotransformer (VARIAC) to obtainthe current close enough to the calculated values. Large current variationsshould be avoided.4) Record V1, Isc and Psc for each “% of rated I1current” step calculated.5) Set the autotransformer to zero position. Turn the power off.6) Remove the connections for the circuit.Calculations :For Open-Circuit TestWi V1IOcosØO, since, Wi , V1 & IO can be read from the meters,The no load power factor cosØO Wi / V1IO can be determine,Iµ IOsinØO and IW IOcosØOXOC V1/ Iµ & ROC V1 / IWFor Short-Circuit TestZSC VSC / ISCWSC I2SC. RSC.RSC WSC / I2SCX SC (ZSC )2 (R SC )2Page41

Basic Electrical & Electronics lS.P.I.TEquivalent circuit at no load(open circuit)circuitObservation Table :A: Open circuit testB: Short circuit testPage42Equivalent circuit at short

Basic Electrical & Electronics lResultsS.P.I.T:-ROC XOC R SC X SC ZSC % Regulation:-%ή for load :-%ή max:-Conclusion:-Page43

Basic Electrical & Electronics lS.P.I.TLOAD TEST ON A SINGLE PHASE TRANSFORMERBY DIRECT LOADINGPage44

Basic Electrical & Electronics lS.P.I.TCircuit Diagram:-Observations :V1 Vs 220 Volts ConstantPage45

Basic Electrical & Electronics lS.P.I.TEXPERIMENT No:-08DATE:-// 2011LOAD TEST ON A SINGLE PHASE TRANSFORMERBY DIRECT LOADINGAim: - To determine the transformer efficiency curve by direct loading of thetransformer.Apparaturus & components required:AC ter:DVM:lamp load:Theory:Transformer is a device used predominantly for two purposes:-.a) Change the voltage level in AC circuit.b) Obtain electrical isolation between two circuits.Construction:Physically, a basic transformer is made up of a ferromagnetic core andcopper windings on it. Core is made up of stacked Cold Rolled Grain Oriented (CRGO)steel plates. This material increases the permeability of the flux path, thereby reducingthe leakage flux as well as the hysteresislosses. The stacking of plates helps reducethe eddy current losses. Each of the plate is called a stamping. Each one is coated withvarnish on either side to insulate it from the adjacent stampings. Also, the stampingwidth is varied across the cross section to make it resemble as closely as possibleto a circle. For a given area, circle has the smallest possible perimeter (circumference).Economically, this makes a big impact in larger transformers where a considerable chunkof money is spent in copper windings.Page46

Basic Electrical & Electronics lS.P.I.TEfficiency :Efficiency of a device is calculated as the ratio of the output power to the inputpower. When calculated in percentage, it is known as percentage efficiency. This ratio isobviously less than 1, but for transformers, it is of the order of 0.95 to 0.99 (at the ratedload). In other words, the transformer is a highly efficient device. A way of determiningthe output is to calculate the input and subtract losses from it. The transformer being astatic device, has no moving parts in it. This leaves only two avenues for losses to takeplace, the magnetic circuit and the electric circuit. In the magnetic part( core ), the lossesthat take place are eddy current losses and hysteresis losses. These losses are independentof the load and remain constant. The eddy current losses are reduced by using stampedcores and the hysteresis losses are reduced by using better materials. These lossestogether are known as core loss. In addition to this, whenever a current flows in one orboth the windings, there is an associated I2R loss. This loss is as evident completelydependent on the load known as copper loss.Regulation :The output voltage changes with the load current. This variation is termed asvoltage regulation. This takes place because of the drop across the transformer windingresistances and leakage reactance. The drop across the resistance increases with the loadcurrent. On the other hand, increases in load increases the leakage reactance. The leakagereactance is nothing but a circuit representation of such a flux. Percentage regulation iscalculated as the ratio of change in voltage from no load to the value to the no load value.There is a circular former placed on the stacked core, in order to insulate the corefrom the windings. The copper windings are wrapped over the former. Most of thetransformers have two windings, though there can be either one or more than two. Thetwo windings are called primary and secondary. The primary winding is connected to theAC supply and load is connected to the secondary winding. Although the two windingsare shown on two separate limbs, practically, the windings are not segregated as such.Both of them are wound on each of the limb, in order to reduce leakage flux. The copperwindings have paper insulator all along their length.Page47

Basic Electrical & Electronics lS.P.I.TOperation :The transformer action is based on two mutually coupled coils, with a changingmagnetic flux linking them. Simplistically speaking, primary winding is energized froman AC voltage source. This voltage causes a current to flow in it. This current produces aflux in the ferromagnetic core on which it is wound. This flux is also alternating in naturebecause of the current. This alternating flux also links with the secondary windingbecause they share a common magnetic path. This changing flux induces an emf in thesecondary coil. The magnitude is governed by Farady’s Law and direction by Lenz’sLaw. The voltage across the coil depends on the number of turns in the secondarywinding.Procedure:1) Connect the circuit as shown in the diagram.2) Increase the input voltage to the transformer rated primary voltage.3) Vary the lamp load in step from no load to the full load value.4) Take the readings of the input wattmeter (W), load current (IL) and loadvoltage (V2). (Minimum 10 readings).5) Calculate efficiency and regulation.6) Plot percentage efficiency and regulation v/s load current.Calculations:% Re gulation % Efficiency Page48V2 ( 0 ) V2V2 ( 0 )V2 I 2x100W1x100

Basic Electrical & Electronics lResultsS.P.I.T:-Conclusion :Page49

Basic Electrical & Electronics lS.P.I.TMAXIMUM POWER TRANSFER THEOREMPage50

Basic Electrical & Electronics lS.P.I.TCircuit Diagram:-Observation Table:V1 (volts) ; V2 (volts)For maximum power transfer,RL (observed) ( Ω) (from the graph)Page51

Basic Electrical & Electronics lS.P.I.TEXPERIMENT NO:-09DATE:-// 2011MAXIMUM POWER TRANSFER THEOREMAim: To experimentally verify the maximum power transfer theorem.Apparatus:DC Voltage source:Ammeter:DVM:Resistor:Breadboard.Theory:The power in load resistance is given by I2R, where I is the currenttroughtheresistance R. Also any complex circuit can be reduced to its equivalent Thevenin’scircuit. The maximum power transfer theorem states that the power transferred to the loadis maximum when the value of the load resistance is equal to the internal resistance of thesource.In other words, the power dissipated in any resistor in a given resistive networkwill be maximum when the rest of the network is replaced by its Thevenin’s equivalentand the value of the load resistance is equal to the Thevenin’s resistance.The general Expression for the power in the load resistor is(VTH ) 2PL I RL * RL( RTH RL ) 22LTo find the value of RL for which PL is maximum dPL 2 RL12 VTH 0 32 dRL(R R)(R R) TH LTHLdPL2 2 RL RTH RL VTH 03dRL(R R) THLRL RTHPMAX(VTH ) 2(VTH ) 2 * RTH ( RTH RTH ) 24 RTHPage52

Basic Electrical & Electronics lS.P.I.THence the maximum power is transferred to the load when the loadresistance value is equal to the internal resistance of the Thevenizedcircuit.Procedure:1) Connect the voltage sources and the milliammeter as shown in the circuit.2) Keep the voltages such that the maximum current does not go beyond therange of the milliammeter ( i.e. resistance minimum).3) Vary the load resistance from minimum to maximum.4) Note the corresponding load current (IL) and voltage across the load VL

Apparaturus & components required :- i. Digital Multimeter ii. Backlite sheet mounted with different active and passive components. Theory: Write theory related with following questions. 1. Define active and passive components. 2. Draw a simple circuit containing 1K resistor and power supply and explain how