4. MEASUREMENT OF ELECTRICAL CURRENT, VOLTAGE AND RESISTANCE

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1 Measurement of Electrical Current, oltage and Resistance / 1 4. MEASUREMEN OF ELECRICAL CURREN, OLAGE AND RESISANCE 4.1. PRINCIPLES OF MEASUREMENS Electrical voltage and current are two important quantities in an electrical network. he voltage is the effort variable without which no current is available. It is measured across an electrical circuit element or branch of a circuit. he device that measures the voltage is the voltmeter. he current is the flow variable that represents net motion of the charged particles (electrons in solids, ions in a liquid) in a given direction. he product of the two yields the instantaneous electrical power. he ratio of the voltage to the current is the impedance. he current is measured by an ammeter (also called an ampermeter). Ammeters are connected in series with the load to measure the current in the load. Eventually, the ammeters R I L A R L Figure 4.1. Connections for an ammeter and a voltmeter. L require breaking the current loop to place it into the circuit. he voltmeter connection is rather easy since it is connected without disturbing the circuit layout. herefore, most electrical measurements require determination of the voltage rather than the current due the ease of measurement. Connections of ammeters and voltmeters are illustrated in Figure 4.1. he current generates a magnetic field around the current carrying conductor. It is also possible to check out the size of the current by sensing the magnetic field strength. his is carried out by clampon type ammeters. he electrical resistance of a circuit component is measured using an ohmmeter that applies a voltage across and determines the current passing through the component. oltmeters and ammeters display the results as deflections of dials on calibrated screens or numerical values on alphanumeric displays as illustrated in Figure 4.. Both types are connected to the circuit via sensing leads and indicate the voltage. However, their Figure 4.. Analog and digital voltmeters

2 Measurement of Electrical Current, oltage and Resistance / internal operations and user interfaces are different. he first type forms the analog meters that will be discussed in this chapter. he second category will be discussed in the next chapter under the title of digital voltmeters. 4.. MC BASED MEASURING INSRUMENS MC in Analog Electrical Measuring Instruments he standard MC instrument indicates positive DC currents (I MC ) as deflection on the scale. he I FSD R MC MC Figure 4.3. Model of MC galvanometer displays both positive and negative currents. he moving coil is usually made up of a very thin wire. he maximum current that gives full scale deflection I FSD is in the order of.1 to 1 ma and coil resistance R MC 1 to 1 Ω. he maximum deflection angle is about 1. he current through the moving coil I MC is limited by the I FSD. A voltage drop MC = I MC R MC occurs across the coil. he moving coil can represented by the full scale deflection current I FSD and coil resistance R MC as shown in Figure Ammeters Basic DC Ammeter (Ampermeter) he current capacity of the meter can be expended by adding a resistor in I R MC I FSD MC R SH (I I FSD ) Figure 4.4. DC Ammeter. parallel with the meter coil as shown in Figure 4.4. he input current is shared between the coil resistance R MC and the parallel resistance that is called the shunt R SH. As the maximum input current I flows in, the coil takes I FSD and remaining (I I FSD ) is taken by the shunt resistor. oltage developed across the meter is MC = I FSDRMC = ( I FSD ) R SH R M MC R M = MC // I I...(4.1) he meter resistance R M seen between the input terminals is = R R... (4.) SH Example 1 Calculate the multiplying power of a shunt of Ω resistance used with a galvanometer of 1 Ω resistance. Determine the value of shunt resistance to give a multiplying factor of 5. I fsd x1 = (I I fsd )x yielding I = 6xI fsd. For I =5xI fsd, 1xI fsd =(51)xI fsd xr sh yielding R sh =1/49 =.41 Ω

3 Measurement of Electrical Current, oltage and Resistance / 3 Multi Range Ammeter he parallel resistance (shunt) can be changed to suit Switch poles different full scale current requirements as indicated in the previous example. he function can be accommodated by Rotary switch using a set of resistors and selecting them one by one. he arm switch however must be of make before break type Figure 4.5. Make before break type (Figure 4.5) that makes the contact with the new position switch. before it breaks the old connection. his eliminates the chance of forcing the full input current through the moving coil during changing the position of the switch. Example Design a multi range DC ammeter using the basic movement with an internal resistance R MC = 5 Ω and full scale deflection current I MC = I FSD = 1 ma. he ranges required 1 ma, 5 ma, 1 ma and 5 ma as illustrated in Figure 4.6. MC = I MC xr MC = 5 m For range1 (1 ma) R SH1 = 5/9 =5.56 Ω I I FSD R MC Rotary selector switch R SH1 R SH R SH3 R SH4 Multirange ammeter circuit 5 ma 1 ma 5 ma 1 ma Multirange ammeter scale Figure 4.6. A multirange ammeter circuit and scale for example. 5 ma 1 ma 5 ma 1 ma For range (5 ma) R SH = 5/49 =1. Ω For range3 (1 ma) R SH3 = 5/99 =.55 Ω For range4 (5 ma) R SH4 = 5/499 =.1 Ω Example 3 Design a multi range DC ammeter using the basic movement with an internal resistance R MC = 5 Ω and full scale deflection current I MC = I FSD = 1 ma. he ranges required.1 A, 1 A, 1 A and 1 A.

4 Measurement of Electrical Current, oltage and Resistance / 4 MC = I MC xr MC = 5 m For range1 (.1 A) R SH1 = 5/9 = 5.56 Ω For range (1 A) R SH =.5/.99 =.55 Ω For range3 (1 A) R SH3 =.5/9.99 =.5 Ω For range4 (1 A) R SH4 =.5/99.99 =.5 Ω oltmeters R I FSD S R MC S R M MC M Figure 4.7. Basic DC voltmeter. Basic DC oltmeter he moving coil can be used as a voltmeter by adding a series resistance R S as illustrated in Figure 4.7. he input voltage is divided between the coil resistance R MC and R S. Current passing through both resistors is I MC which is limited by the full scale deflection current I FSD of the coil. he full scale input voltage M = I FSD (R S R MC )... (4.3) he input impedance seen is R M = R S R MC... (4.4) However, with R S >>R MC, R M is approximately equal to R S and M I FSD R S. Example 4 he coil of a moving coil voltmeter is 4 cm long and 3 cm wide and has 1 turns on it. he control spring exerts a torque of.4x1 4 Nm when the deflection is 1 divisions on the full scale. If the flux density of the magnetic filed in the airgap is.1 Wb/m, estimate the resistance that must be put in series with the coil to give one volt per division. he resistance of the voltmeter coil may be neglected. EM = SP.4x1 4 = 1x.1x1x1 4 xi FSD I FSD = ma. herefore, current per division is. ma. Assuming that R MC is negligibly small compared to R S : R S = 5 kω Example 5 A moving coil instrument gives fullscale deflection of 1 ma when the potential difference across its terminals is 1 m. Calculate: a. he shunt resistance for a full scale corresponding to 1 ma; R SH = 1 / 9 = 1.11 Ω b. he resistance for full scale reading with 1 ; R MC = 1 /1 = 1 Ω; R S R MC = (1 / 1) kω = 1 kω yielding R S = 1 kω (R MC is negligible)

5 Measurement of Electrical Current, oltage and Resistance / 5 c. he power dissipated by the coil and by the external resistance in each case. Power dissipated by the coil, P C = I M xr MC = 1 mw; P SH = M /R SH = 9 mw P S = M /R S = 1 W. Multi Range oltmeter he series resistance can be changed to suit different full scale voltage requirements as shown in Figure 4.8. Resistors are organized either in parallel fashion (conventional connection) as in the case oltage to be measured R S R S1 R MC I FSD R MC Rotary selector switch R S4 R S3 R S R S Multirange voltmeter circuit Parallel connection R S3 R S Multirange voltmeter circuit Series connection M of ammeter and selecting them one by one or all connected in series like a voltage divider (modified Rotary switch arm Resistance seen by the input terminals of the device connection). he switch however must be of break beforemake type (Figure 4.9) that breaks the contact with the old position before it makes it with the new position. his eliminates the chance of forcing a current larger than the full scale current through the moving coil during changing the position of the switch. he resistors are also called the multiplier resistors. R M = M /I FSD... (4.5) and written on the face of the scale as Ω/. he contribution of the coil resistance R MC can be ignored if it is too small compared to R M. Following examples illustrate the selection of multiplier resistors. Figure 4.8. Parallel and series resistance connections for a multirange voltmeter. Switch poles Figure 4.9. Break before make type switch

6 Measurement of Electrical Current, oltage and Resistance / 6 Example 6 A multi range DC voltmeter is designed using a moving coil with full scale deflection current 1 ma and coil resistance 5 Ω. Ranges available: 1, 5, 1, 1. Determine the multiplier resistors and input resistance of the meter using: a. Conventional connection b. Modified connection In conventional connection, resistors are selected one by one to satisfy M = I FSD (R MC R S ) = MC I FSD R S where M is the full scale voltage of the selected range. MC = (1 ma)(5ω) =.5. Hence, R S = ( M.5)/1 kω. Meter resistance seen between the input terminals is R M = R MC R S Range 1 ( 1): R S1 = 9.5/1 =.95 kω = 95 Ω; R M1 = 95 Ω 5 Ω = 1 Ω Range ( 5): R S = 49.5/1 =4.95 kω; R M = 4.95 kω.5 kω = 5 kω Range 3 ( 1): R S3 = 99.5/1 =9.95 kω; R M3 = 9.95 kω.5 kω = 1 kω Range 4 ( 1): R S4 = 999.5/1 =99.95 kω; R M4 = kω.5 kω = 1 kω For the alternative modified arrangement, the resistor for the lowest range is determined and others calculated as added to the total of the previous value. he total resistance seen from the input in all ranges will be the same as those in the previous case. Resistors between stages can be computed as R Sn = R Mn R M(n 1) Range 1 ( 1): R M1 = 1 Ω; R S1 = 1 Ω 5 Ω = 95 Ω Range ( 5): R M = 5 kω; R S = 5 kω 1 kω = 4 kω; Range 3 ( 1): R M3 = 1 kω; R S3 == 1 kω 5 kω = 5 kω; Range 4 ( 1): R M4 = 1 kω; R S4 = 1 kω 1 kω = 9 kω; Example 7 A basic D Arsonval meter movement with an internal resistance R MC = 1 Ω, full scale current I FSD = 1 ma, is to be converted into a multirange DC voltmeter with ranges 1, 5, 5 and 5. Find the values of multiplier resistors using the potential divider arrangement. Four resistors R S1 R S4 are added in series with R MC. In the first range (1 ) only R S1 is used and the maximum voltage drop on R S1 is 1.1=9.9. hus, R S1 = 9.9/1mA = 9.9 kω In the nd range (5 ) R S1 R S is used and the maximum voltage drop on R S is 5 1= 4. hus, R S = 4/1mA = 4 kω

7 Measurement of Electrical Current, oltage and Resistance / 7 In the 3 rd range (5 ) R S1 R S R S3 is used and the maximum voltage drop on R S3 is 55=. hus, R S3 = /1mA = kω In the 4 th range (5 ) R S1 R S R S3 R S4 is used and the maximum voltage drop on R S4 is 55= 5. hus, R S4 = 5/1mA = 5 kω Ohm and OM Meters Analog Ohmmeter Analog ohmmeter can be designed simply by adding a battery and a variable resistor in series with the moving coil instrument as shown in Figure 4.1. he unknown resistance is connected to the Zero adjust Internal battery MC meter R MC 1 1 Basic series ohmmeter circuit Series ohmmeter scale Figure 4.1. Circuit and scale of a basic ohmmeter. terminals of the device to complete the electrical circuit. he output terminals are shorted together with the leads (wires) used in connecting the external resistor. he variable resistance is adjusted until the full scale deflection current passes through the coil. his is marked as the resistance. When the leads are separated from each other, no current flows indicating an open circuit which means infinite resistance. Hence, the scale is non linear with resistance increases on the right side (opposite to ammeter). Multi range ohmmeters can be obtained by combining the circuits of a series ohmmeter and a multi range ammeter. OM Meter he functions of ammeter, voltmeter and ohmmeter can be combined in a multipurpose meter called a OM (volt ohm milliampere) meter, or shortly the OM. It has several multiple scales, usually color coded in some way to make it easier to identify and read. Generally, it has a single multipurpose switch to select the function and the range. Example 8 A moving coil has 1 turns, 5 cm coil area, and airgap magnetic flux density of.1 esla (Wb/m ). he control spring exerts a torque of 5x1 6 Nm at the fullscale deflection of 9. he potential difference across the coil terminals at the fullscale deflection is 1 m. Using the above movement, design a multirange DC ammeter with ranges 5 ma, 1 A and multirange DC voltmeter with ranges 1 and.

8 Measurement of Electrical Current, oltage and Resistance / 8 I FSD = SP /NBA = 1 ma, therefore R MC = MC / I FSD =1 Ω For ammeter ranges: R SH1 = 1 m/ (51) ma =.4 Ω and R SH = 1/999 =.1 Ω For voltmeter ranges: R S1 = (1.1)/1mA = 9.9 kω and R S = kω 4.3. LOADING ERRORS Instrument Loading All measuring instruments draw energy from the source of measurement. his is called the loading effect of the instrument. Hence, all measurements include errors due to instrument loading. If the energy taken by the instrument is negligibly small compared to the energy exists in the source (of course of type measured), then the measurement is assumed to be close to perfect, and the loading error is ignored. R M I M Ideal A MC I M M R M I= M R M Practical ammeter Practical voltmeter Figure Representations of practical ammeters and voltmeters. Ideal ammeter has zero internal resistance and no voltage across it. Ideal voltmeter has infinite internal (meter) resistance and draws no current from the circuit. he practical ammeter can be represented by an ideal ammeter with added series resistance that represent the meter resistance. Similarly, the practical voltmeter can be represented by an ideal voltmeter in parallel with the meter resistance. hese two models are illustrated in Figure Loading Errors in Ammeters R R M I L Any electrical circuit can be modeled by a voltage source A and a series resistance R. he circuit is completed when the load resistance R L is connected across the output terminals R L Figure 4.1. Ammeter loading. and a load current R L flows through the load. An ammeter can be placed in series with the load to measure this current as shown in Figure 4.1. Current in the circuit can be calculated as

9 Measurement of Electrical Current, oltage and Resistance / 9 I L =... (4.6) R R L R M In ideal condition, R M = and the true value of the current is I L R R =... (4.7) L he error is the difference between the measured value and the true value, and generally expressed as the percentile error which is: measured value true value % loading error = x1... (4.8) true value Hence, the loading error due to the ammeter can be found as: % loading error for ammeter = R RL RM R RL 1R x R RL R R R L M 1 =...(4.9) M Loading error can be ignored if R M <<(R R L ) which is satisfied in most applications Loading Errors in oltmeters R R Leff R L R M In voltage measurement, the meter is connected in parallel with load resistor as shown in Figure he true value of the voltage across the resistor is (without the meter) R L L =...(4.1) R RL Figure oltmeter loading. and effective load resistance becomes As the meter is connected, R M becomes in parallel with R L R Leff R R L... (4.11) R R = M L M R Leff R L if R M >>R L. he voltage measured by the meter is L RLRM RL RM = Lind =... (4.1) R RLR M R R L M

10 Measurement of Electrical Current, oltage and Resistance / 1 Lind L % loading error = x1... (4.13) L Examples 9 A 15 DC voltage source is coupled to a 5 kω load resistor through a 1 kω source resistance. wo voltmeters (A) and (B) are available for the measurement. oltmeter A has a sensitivity 1 Ω/, while voltmeter B has a sensitivity Ω/. Both meters have 5 range. a. Calculate reading of each voltmeter. b. Calculate error in each reading expressed in a percentage of the true value. L 15 = x5 = 5 ( 1 5) Input resistance of voltmeter A = sensitivity x range = (1 Ω/)x(5 ) = 5 kω and the effective value of the load resistance is 5//5 = 5 kω 15x5 oltage indicated by voltmeter A; LA = = % age loading error = x 1 = 4% 5 Input resistance of voltmeter B = ( Ω/)x(5 ) = 1 kω and the effective value of the load resistance is 5//1 = 48 kω 15x48 oltage indicated by voltmeter B; LB = = % age loading error = x 1 = 3% 5 Example 1 A voltmeter has a resistance of kω/ is used to measure kω kω the voltage on the circuit shown on a 1 range. Find the percentage loading error. 1 RUE = 1x/4 = 5. With R M = kω, the effective load resistance R Leff = (4/) = kω. herefore, Figure for example 1. MEAS = 1x18.18/38.18 = % loading error can be found as: %error = 1x(4.76 5)/5 = 4.8%

11 Measurement of Electrical Current, oltage and Resistance / 11 Example 11 A generator produces 1 volts DC and has an internal resistance of 1 kω as shown in the figure. he output voltage is measured using several voltage indicating devices. Calculate the output voltage and the percentage loading error for each of the following cases: c. An ideal voltmeter (Ri ) o = 1, Error = % d. A digital voltmeter with Ri = 1 MΩ; o = 1x1/1.1 = 99 volts, % error = 1% e. An oscilloscope (Ri = 1 MΩ); o = 1x1/1.1 = 9.9 volts, % error = 9.1% 1 kω 1 Figure for example 11. f. A moving coil type analog voltmeter with 1 kω/ in 1 volt range Meter resistance is 1x1 kω = 1 kω, yielding o =5 volts, % error = 5 % Example 1 A D Arsonval movement gives full scale deflection of 1 ma when a voltage of 5 m is applied across its terminals. Calculate the resistance that should be added in series with this movement to convert it into a 1 voltmeter. he above 1 voltmeter is used to measure the voltage across the 1 kω resistor in the circuit shown. Determine the percentage loading error. 1 kω 9 1 kω Figure for example 1. Meter coil resistance R M = 5 m / 1 ma = 5 Ω and it s effect can be ignored in finding the series resistance of the voltmeter. hen, R S = 1 / 1 ma = 1 kω. rue value of the voltage on the 1 kω resistance (without voltmeter loading) true = (1/11)x9 = 81.8 With the voltmeter connected, 1 kω resistance will experience a 1 kω meter resistance in parallel yielding 9.9 kω at the output. he measured output voltage becomes: M = 9x (9.9/1.9) = he % error = 1x( )/81.8 =.9 %

12 Measurement of Electrical Current, oltage and Resistance / AC OLMEERS Measurement of AC oltage he voltmeter based on the permanent magnet moving coil (PMMC or D Arsonval) cannot be directly used to measure the alternating voltages. he instruments that are used for measuring AC voltages can be classified as: 1. Rectifier D Arsonval meter. Iron ane (Moving Iron) type meter 3. Electrodynamometer 4. hermocouple meter 5. Electrostatic voltmeter Rectifier type (Rectifier D Arsonval) meter is the extension of the DC voltmeter and it will be explained below. i(t)=i sinωt m ime Figure Alternating current (AC) waveform. Average and alues he moving coil instrument reads the average of an AC waveform. he average of the current waveform i(t) shown in Figure 4.14 is: I A 1 = I m sinωtdt =...(4.14) where is the period and ω = π/ = radial frequency (rad/sec). However, if this current is applied to a resistor R, the instantaneous power on the resistor p(t) = i (t)r... (4.15) he average power over the period becomes: R R PA = Im sin tdt = ω Im... (4.16) Hence, the average power is equivalent to the power that would be generated by a DC current called the effective current that is 1 I = Im...(4.17) m Ieff I = i ( t) dt = =. 77 Due to squaring, averaging (mean) and square rooting operations, this is called the. value of the current and I is the true value of the current that we want to measure.

13 Measurement of Electrical Current, oltage and Resistance / 13 If the voltage is applied to the v i (t)= m sinωt ime A v o (t) Figure AC to DC conversion. current in the resistor becomes: m ime resistor through a diode as shown in Figure 4.15, the negative half cycle is chopped off since the diode can conduct current only in positive direction. his is called the half wave rectifier. he average value of the A 1 m = m sin tdt = π ω = (4.18) m A alternate alternate D4 Ii D1 Full Wave Rectifier he half wave rectifier is used in some voltmeters, but the mostly adapted one uses the full wave rectifier shown in Figure Here, a bridge type fullwave rectifier is shown. For the half cycle the current follows the root ABDC. For the half cycle root CBDA is used. he current through the meter resistor R m is the absolute value of the input current as shown in the inset. he voltage waveform on the meter resistance R m has the same shape as the current. he average value of the voltage becomes: B Rm Im D Input D D3 Figure Bridge type fullwave rectifier. C A m = m sin tdt = =.636 m ω...(4.19) π A is the DC component of the voltage and it is the value read by the moving coil instruments. Hence, the inherently measured value (IM) is the average value, while the

14 Measurement of Electrical Current, oltage and Resistance / 14 true value is the value. he voltage reading will contain reading error (unless it is corrected) as indicated Average true % error = ( ) 1% = ( ) 1% = 1%...(4.) true and the indicated voltage will be 1% less than the true value Form Factor and Waveform Errors For Sinusoidal Waveforms he ratio of the true value to the measured value is called the form factor or safe factor (SF). For sinusoidal signals the form factor is SF = ( / A )... (4.1) In AC voltmeters, the reading is corrected by a scale factor = safe factor (SF) = his can be done either at the calculation of the series resistance or setting the divisions of the scale. Eventually, the error is eliminated as: 1.11 indicated Average true % error = ( ) 1% = ( ) 1% = %...(4.) true his is of course true for sinusoidal signals. For other waveforms, the error may be nonzero indicating erroneous readings. v(t) Figure A triangular waveform signal can be expressed as hus For riangular Waveform A triangular voltage waveform v(t) with amplitude m and period is shown in Figure he negative portion is converted to positive after the full wave rectification. Due to the symmetry of the signal, interval from to /4 can be used for integration in finding the average (DC) and values. In this interval, the v(t) = 4 m /... (4.3) = 4 4 4m m A dt = =.5...(4.4) m his is the inherently measured (IM) value. A meter corrected for sinusoidal waveforms will indicate

15 Measurement of Electrical Current, oltage and Resistance / 15 ind = 1.11x.5 m =.555 m... (4.5) he value can be computed as: 4 m m 416 dt = =. 577 = m...(4.6) 3 Hence, the form factor for the triangular waveform is and 1.11 average.he percentile measurement error: 1.11 indicated Average true % error = ( ) 1% = ( ) 1% = 1% = 3.81%.577 true Correction Factor A correction factor (CF) is used to multiply the reading indicated by the meter to correct the measured value. he correction factor must be determined for every specific waveform individually as: CF ( ) ( SF) waveform waveform IM = =...(4.7) ( SF) sin usoidal ( ) sin usoidal IM he voltage indicated for the triangular waveform using a meter adjusted for a sinusoidal waveform can be written as: = SFx = x( )...(4.8) ind ( IM ) waveform ( ) sin usoidal A A waveform Eventually, ( )( CF) = ( SF) ( ) = ( ) =...(4.9) ind wave IM wave wave true he error without the correction: 1 CF % error = 1%... (4.3) CF For the triangular wave shown in the above example CF =.5 = = yielding the percentile error of 3.81%, same as the one found before.

16 Measurement of Electrical Current, oltage and Resistance / 16 Figure 4.18 shows a pictorial presentation of the scale calibrated for sinusoidal voltage waveforms, model of the AC voltmeter based on the basic D Arsonval meter with samples of input and output waveforms DC readings 1 AC readings 11.1 v(t)= m sinωt v(t) IM ime ime AC oltage Fullwave Rectifier Unidirectional oltage D Arsonval meter (SF = 1.11) Figure Illustration of an AC voltmeter corrected for sinusoidal signals. 1 kω 1 kω s = 8 Circuit for example 13. percentile loading error. m Example 13 A D Arsonval (moving coil) movement based AC voltmeter is calibrated to read correctly the value of applied sinusoidal voltages. he meter resistance is 1 kω/ and it is used in 1 range. a. Find m measured by the meter and the rue value of the voltage true = 8x1/13 = 7.38 ; R m = 1 kω leading to R L = 1x1/ = 54.5 kω. herefore m = 8x54.5/64.5 = Percentile loading error = 8.4%. b. A different periodic waveform is applied and the waveform m (t) shown appears across the meter. i. Calculate for this waveform,

17 Measurement of Electrical Current, oltage and Resistance / m (t) 5 = [ 1 5 = 3 t dt dt ; = 5.7, t Waveform for example 13. measurement. % waveform error = 1x( )/5.7 = 5.3%. ii. How much is the voltage indicated by the meter ( indicated )? 1 1 A [ 1tdt 5dt = 5 ( ) 3 1 = herefore, ind = 1.11x5 = 5.55 iii. Find the waveform error in this (t) 3 1 t Example 14 he voltage waveform shown has a magnitude ± 5 and it is applied to an AC voltmeter composed of a full wave rectifier and a moving coil 5 (t) = 1 (t) 1 (t) Fullwave Rectifier (t) = 1 (t) D Arsonval meter (SF = 1.11) t 5 Waveforms for example 14. Model for example 14. (D Arsonval) meter. It is calibrated to measure voltages with sinusoidal waveforms correctly. c. Find the average and values of 1 (t) = 1( A ) 1 ( t) dt = 5tdt = [1 1] = ( ) = 5 = [1 1] = = 8.87 t dt d. Sketch the waveform for (t) e. Find the average and values of (t). Ans. he value of (t) is the same as that of 1 (t) which is 8.87 volts. he average value can be calculated from the area of the triangle easily as 5/ = 5 volts. f. Find the voltage indicated by the meter. Ans. 5x1.11= 7.75 volts g. Calculate the error due to the waveform and find the correction factor. he % waveform error = 1x[ ]/8.87 = 3.88%

18 Measurement of Electrical Current, oltage and Resistance / 18 Correction factor (CF) = (SF) wave /(SF) sine = (8.87/5)/1.11 = 1.4 m v(t) Signal for example 15. Example 15 A generator with 5 Ω internal resistance has a saw tooth output voltage as shown. he value of this output is to be measured by a moving coil instrument whose internal resistance is 1 kω. he instrument has a full wave rectifier and is calibrated for sinusoidal waveforms. Calculate the error due to the waveform and also the loading error..5 kω (t) R in min Circuit for example 15. t he schematic diagram illustrates the measurement problem. For an ideal voltmeter, the meter resistance R in must be very large (R in ). herefore, the true value of the output voltage v true (t) = v(t). he internal resistance is given as R in = 1 kω yielding v in (t) = (1/1.5)v(t). Hence, 1 1 vin vtrue % ( loading) error = x1 = 1.5 x1 = 4.8% v 1 true he voltage measured using this meter is the average of v in (t) which is: 1 1 m 1 m A = x tdt = x. he reading indicated by the meter is compensated for the sinusoidal waveform and it becomes: 5 m ind = 1.11x =. 59 m 1.5 he true value that must be measured by the meter is the value which is: 1 1 m 1 m = t dt = x = Hence, the waveform error is 1x(.59.55)/.55 = 3.8% If the meter would be ideal (R in ), then = m true = 577m 3 =. Having.59 m indicated by the meter, the total measurement error (loading waveform) becomes 1x( )/.577 = 8.3% m

19 Measurement of Electrical Current, oltage and Resistance / CLAMPON MEERS Clamp on meters are used for measuring AC circuit currents in a non invasive manner which Figure Clampon meter avoids having to break the circuit being measured. he meter clamps on to a currentcarrying conductor and the output reading is obtained by transformer action. he principle of operation is illustrated in Figure 4.19, where it can be seen that the clamp on jaws of the instrument act as a transformer core and the current carrying conductor acts as a primary winding. Current induced in the secondary winding is rectified and applied to a moving coil meter. Although it is a very convenient instrument to use, the clamp on meter has low, sensitivity and the minimum current measurable is usually about 1 amp SOLED QUESIONS 1. A moving coil has 1 turns, 3 cm coil area, and airgap magnetic flux density of.1 esla (Wb/m ). he control spring exerts a torque of 3x1 7 Nm at the fullscale deflection of 1. he potential difference across the coil terminals at the fullscale deflection is 5 m. Using the above movement: a. Find the full scale deflection current and coil resistance; I fsd = SP /NBA =.1 ma, therefore R m = m / I fsd =5 Ω b. Design a DC ammeter with a range 5 ma; R sh1 = 5 m/ (5.1) ma =.1 Ω c. Design a multirange DC voltmeter with ranges 1 and. For voltmeter ranges, R m is negligible: R s1 = 1/.1mA = 1 kω and R s = MΩ d. What would be the deflection angle for an input voltage of 7 in 1 range? Since 1 causes 1, 7 will cause 7 of deflection

20 Measurement of Electrical Current, oltage and Resistance /. A moving coil has 8 turns, 4 cm coil area, and airgap magnetic flux density of.1 esla (Wb/m ). he control spring exerts a torque of 4x1 7 Nm at the fullscale 1 m (t) deflection of 9. he potential difference across the coil terminals at the fullscale deflection is 1 m. Using the 1 3 above movement: 5 a. Find the full scale deflection current and coil resistance; I fsd = SP /NBA =.15 ma = 15 μa, therefore R m = m / I fsd =8 Ω b. Design a DC ammeter with a range 1 ma; R sh1 = 1 m/ (1.15) ma.1 Ω c. Design a multirange DC voltmeter with ranges 1 and. For voltmeter ranges, R m is negligible: R s1 =1/.15mA=8 kω and R s = 1.6 MΩ d. What would be the deflection angle for an input voltage of 65 in 1 range? Since 1 causes 9, 65 will cause 5 kω kω 65x9/1 = 58.5 of deflection 3. A D Arsonval (moving coil) movement based AC s voltmeter is calibrated to read correctly the value of applied sinusoidal voltages. he meter resistance is 4Ω/ and it is used in 5 range. a. Find s if it is sinusoidal and m = 36 () he meter resistance is 4(Ω/)x5 = kω is parallel with kω yielding R L = kω. I = m /R L = 1.98 ma. s = 5x = 45.9 (rms) b. he periodic waveform v m (t) shown is applied to the meter. i. Calculate for this waveform, = [ 1xdt xdt] ; =7.71, 6 m t ii. How much is the voltage indicated by the meter ( indicated )? Average value of the rectified signal = indicated = 1.11x A = 74 iii. Find the waveform error in this measurement. % error = 1x( )/7.71 = 4.65%

21 Measurement of Electrical Current, oltage and Resistance / 1 4. An AC voltmeter calibrated for sinusoidal voltages is used to measure both the input ( 1 ) and output ( ) voltages. It has a scale with 1 divisions and measurement ranges: ( 5) m; ( 1) m; ( 5) m; ( 1) ; ( ) ; ( 5) and ( 1) a. Determine the range that would yield the most accurate reading for 1, the value indicated by the meter for 1 and percentage reading uncertainty (assume that the reading uncertainty is ±.5 division). he meter would indicate 1.11 A =1.11x.636x peak = 8.7 m. Hence, range ( 5) m is the most accurate with uncertainty ±.5 m.88% b. Repeat (a) for. ind = 1.11x.636x1.5 = 1.6 volt; range ( ), uncertainty ±.1.94% 5. An average reading full wave rectifier moving coil AC voltmeter is calibrated to read correctly the value of 5 v(t) applied sinusoidal voltages. he periodic waveform v(t) 1 3 shown is applied to the meter. Calculate for this 5 waveform, indicated and the waveform error in it. he full wave rectifier will convert the input waveform into a saw tooth voltage waveform of question 3 with amplitude 5 volts and period = 1 second. Using the equations in answer 3, A =.5 ; =.577 m =.89. he value indicated by the meter ind = 1.11x A =.775. herefore, %(waveform) error = 1x( )/.89 = 4% 6. Draw the circuit diagram and explain the operation of the full wave rectifier bridge circuit used to convert D Arsonval movement into an AC voltmeter. Please refer to the lecture notes for the operation of the full wave rectifier. a. What is the for a zero averaged square waveform of peak to peak value = 1? What is the value indicated for it by the AC voltmeter calibrated to read applied sinusoidal voltages correctly? What is the percentage waveform error in that value? t 5 5 (t) t (t) Fullwave Rectifier r(t) 5 r(t) t he zero averaged square wave has a magnitude ± 5. he magnitude becomes 5 after the full wave rectification for all times. = A = 5. he meter calibrated for 1 (t) t (t) Fullwave Rectifier 1 r(t) r(t) t

22 Measurement of Electrical Current, oltage and Resistance / sinusoidal voltages will read ind = 1.11x5 = Hence, the % error = 1x(5.55 5)/5 = 11 % b. Repeat (a) if the square wave accepts amplitude values between and 1 volts. he output of the full wave rectifier will be the same as it s input as shown in the figure. A = 5 and ind = he voltage is different as: 1 7 = 1 dt = 7.. Yielding, the % error = 1x( )/7.7 = 1.5% 7. Explain the operation of one circuit through which the D Arsonval movement can be used as a meter for measuring periodic signals. What is the scale factor for calibrating such a meter? he meter based on D Arsonval movement inherently measures (IM) the average value of the input applied. herefore, a zero averaged AC input voltage would cause IM = as the displayed value. he full wave rectifier converts the AC input voltage into a waveform that is equal to the absolute value of the input. Hence, the negative halfcycle also produces a positive voltage at the output. Eventually the average of the output becomes m /π, where m is the peak value of the voltage yielding IM = m /π =.636 m he actual value that we want to measure is the value which is =.77 m If the reading is not corrected, there will be 1% error in it. he scale factor SF = 1.11 = / A is used to correct the reading and eliminate the reading error. 8. What is the for the waveform shown? v(t) 1 What is the value indicated by an AC voltmeter 5 7 calibrated for sinusoidal waveforms? What is the 1 3 percentage waveform error in that value? 1 Due to symmetry, can be calculated from [ 1 1 ] yielding = to 4 seconds as: = x t dt dt = he average value is computed in a similar manner as: A = [x 1tdt 1dt] = = 7.5 he voltage reading indicated by the meter is: ind = 1.11x A =8.35. %error = 1x( )/8.16 =.% t

23 Measurement of Electrical Current, oltage and Resistance / 3 4. MEASUREMEN OF ELECRICAL CURREN, OLAGE AND RESISANCE PRINCIPLES OF MEASUREMENS MC BASED MEASURING INSRUMENS MC in Analog Electrical Measuring Instruments Ammeters oltmeters Ohm and OM Meters LOADING ERRORS Instrument Loading Loading Errors in Ammeters Loading Errors in oltmeters AC OLMEERS Measurement of AC oltage Form Factor and Waveform Errors CLAMP ON MEERS SOLED QUESIONS... 19