UNIT-III�A.C Instruments

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True RMS measurement is more reliable and accurate than RMS measurement. We express the AC voltage and current in its effective value or RMS value. RMS Value of alternating voltage or current is equal to the square root of the average of the square of AC current or voltage value over a period of time.

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What is True RMS Measurement?

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• The RMS meter measures the peak value.��Then, the meter calculates the RMS value. The formula of the RMS value of sinusoidal voltage is as given below.

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What is the difference between RMS and True RMS?

The meter which uses the averaging technique to calculate the RMS value of voltage or current is called RMS measurement. The RMS value of the AC current or voltage has the same heating effect that is equal to the heating effect of DC current or voltage of the same magnitude.

The ratio of RMS to the average value for a perfectly sinusoidal waveform is 1.11. The RMS value is equal to 1.11times of the average value.

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• The true RMS meter calculates the heating effect of the current/voltage waveform by sampling the waveform. The True RMS meter has a high sampling rate. The several samples for a waveform give the exact heating effect. ��For a perfect sinusoidal voltage waveform, the peak value is 1.414 times the RMS value or we can say the RMS value of the perfect sinusoidal waveform is 0.707 times the peak value. Thus, the average responding meter gives the RMS measurement correctly for a pure sinusoidal waveform.

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• True RMS and RMS meters read correctly the perfectly sinusoidal waveform����The average responding meter can measure the voltage and current. ��The average responding meters give accurate readings for linear loads.��The linear loads are heaters, induction motors, and incandescent lamps.��The linear load draws current in phase with the applied voltage.��Therefore, the current drawn by non-linear loads increases in proportion to the voltage.

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• True RMS reads correctly the non-sinusoidal waveform, the RMS meter reads with an error ����The average responding meter reads low. ��We use a true RMS meter to read the current of non-linear loads. ��The non-linear loads are DC Drive, variable frequency drive, and electronic equipment. The semiconductor devices like a diode, SCR, and IGBT has non-linear characteristic. ��The current drawn by these devices is not linear to the applied voltage, and because of this, the current waveform is distorted.

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• Specifications of True RMS Multimeter �4½ Digit True RMS Digital Multimeter �DC Votage range: 200mV/2V/200V/1000V �DC Voltage accuracy: +(1.0%+5) �AC Voltage Range :2V/20V/200V/750V �Frequency Range: 20kHz/200kHz �Frequency Accuracy: +(1.5%+25) �AC Frequency response:40Hz to 400Hz �True RMS �AC Frequency response:40Hz to 400Hz �AC Current Range: 200mA/20A �Resistance:200ohm/2kohm/20kohm/200kohm/2Mohm/20Mohm �

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Scaling Voltage & Current to Engineering Units

It is very often necessary to convert a voltage, millivot or current reading into a more useful value such as PSI, GPM, LBS, etc. For example, if measuring force using a record the data in LBS (pounds) instead of millivolts, which is what the load cell typically produces.

Y=MX+B

Where Y is the output or ENGINEERING UNITS Where M is the slope or the SCALE FACTOR Where X is the INPUT (millivolts, volts, etc) and Where B is the OFFSET

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CURRENT PROBE

Since the properties of electricity are invisible, it’s impossible to immediately determine the nature of problems when they occur. In some cases, it becomes necessary to measure current.

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Current can be measured with a variety of tools, including digital multimeters, clamp meters, and current probes.

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Basics of current measurement

Current measurement is one of the most fundamental techniquesfor measuring electronic devices, and it’s used in many situations.

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A variety of devices are used to measure current. Principal instruments that can measure current include the following

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• Digital multimeters
• Clamp meters
• Current probes (viewing current probe voltage output voltage with an oscilloscope)

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Current probes: Measuring current in the field � without turning off the power

They're capable of more accurate measurement than digital multimeters while minimizing effects on the circuit.

That said, you may wonder why current probes can measure current simply by being affixed to a wire.

They utilize various methods to detect the magnetic field that occurs around the current being measured.

There are also high-sensitivity models that can be connected to an oscilloscope in order to observe current waveforms.

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How to use a current probe

Current probes are used by connecting them to an oscilloscope. This process can be quite simple, as it is when connecting current probes to HiokiMemory, HiCorders, which have one-touch BNC connectors.

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Some models have functionality that makes measurement more efficient,

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Precautions when using a current probe

Some current probes are designed to measure circuits carrying large currents, while others can measure extremely small currents with a high degree of precision. It’s common for the waveforms of excessively small currents to be obscured by noise.

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Using current probes to measure current after � carefully reviewing precautions

Instruments such as digital multimeters, clamp meters, and current probes are used to measure current. Current probes, which can be used with oscilloscopes to provide optional measurement functionality, can measure DC and AC current.

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• ��Applications:���Current Consumption Measurement for Low Energy Devices��Evaluating High-Speed Switching Devices��Automotive Electrical Equipment and ECU Current Measurement��Comprehensive Evaluation of Inverter Motors�

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• Transformers used in conjunction with measuring instruments for the measurement purposes are called Instrument Transformers.
• The transformer used for measurement of current is called Current Transformer or C.T. While the transformers used for the voltage measurement are called Potential Transformers or P.T.

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Instrument Transformers

• Use of Instrument Transformers
• The extension of instrument range , so that current , voltage , power & energy can be measured with instruments or meters of moderate size is of very great importance in commercial metering.
• In power systems the currents and voltages are very large and therefore direct measurements are not possible (as these currents and voltages are far too large for any meter of reasonable size and cost. And so the solution lies in stepping down these currents and voltages with the help of instrument transformers so that they could be metered with instrument of moderate sizes.

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• Fig. below shows C.T.

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• Here the primary winding is so connected such that the currents which is being measured passes through it and the secondary winding is connected to ammeter.
• The C.T steps down the current to the level/range of ammeter.

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• Fig. below shows the voltage measurement with P.T.
• Here the primary winding is connected to the voltage (which is being measured) and the secondary winding to a voltmeter.
• The P.T steps down the voltage to the level of voltmeter.

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• Shunts and multipliers:
• Extension of range could be easily done by the use of shunts for currents and multipliers for the voltage measurement.
• Note: This method is suitable only for small values of current and voltage.
• Disadvantages of shunt:
• 1.The method of using shunts is limited to capacities of a few hundred ampere(at most), since the power consumed by the shunts at large currents would be considerably large.

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• 2. The problem of insulation of instrument and shunt is quiet difficult if measurements are done at high voltages of several hundreds or thousands of volt above ground.
• 3. The measuring circuit is not electrically isolated from the power ckt.
• Disadvantages of multipliers.
• 1. The power consumed by multipliers becomes large as the voltage increases.

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• 2. Care has to be taken to keep leakage currents in high voltage multipliers down to negligible values. But however insulation of multipliers which is required to prevent leakage currents becomes very difficult above a few thousand volt. Special type of construction are needed to prevent the above effects. Hence the construction of multipliers for use at high voltages is very costly and complicated.
• 3. The measuring ckt. is not electrically isolated from the power circuit.

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• Advantages of Instrument Transformers.
• (i) When instruments are used in conjunction with instrument transformers, their readings do not depend upon their constants(R,L,C).
• The instrument transformers produce practically the same instrument readings regardless of the constant of the instrument or infact the no. of instruments connected in the circuit.
• (ii) Current transformers have been standardized at 5A secondary winding current and the voltage transformers at 100 to 120V secondary winding voltage.These are very moderate ratings and hence the instruments for measurements are rated near these.

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• A 5A ammeter may be used to measure 1000A with the help of 1000/5A ratio C.T or
• 110V voltmeter may be used to measure a voltage of 66KV with the help of a 66000/110V P.T.
• Therefore we can say that very cheap moderate instruments may be used to measure large currents and high voltages.
• (iii) The measuring circuit is isolated from the power ckt.The fact the meters in the secondary ckt. of an instrument transformer are isolated electrically from the primary side is of very great importance in high voltage systems. And hence because of this insulation is not a problem and the safety is assured for the operators.

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• Ratio of Instrument Transformers
• Transformation Ratio(R):
• Primary winding current/secondary winding current for a C.T
• Primary Winding voltage/secondary winding voltage (for a P.T)
• Nominal Ratio(Kn): For C.T
• Rated Primary winding current/Rated secondary winding current

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• For P.T
• Rated primary winding voltage/Rated secondary winding voltage.
• Turns ratio: (for C.T) n=
• number of turns of secondary winding/no. of turns of primary winding
• For P.T n= number of turns of primary winding/number of turns of secondary winding.

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• Ratio Correction Factor: The ratio correction factor of a transformer is the transformation ratio/nominal ratio.
• Note: The ratio marked on the transformers is their nominal ratio.
• Current Transformers: Fig. of C.T shown in next slide.The primary winding consists of few turns and therefore there is no appreciable voltage drop across it.
• The ammeter, and the wattmeter current coil are connected directly across the secondary winding terminals. Thus we can say that a current transformer will operates with its secondary winding nearly under the short ckt. conditions.

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Current transformers

The C.T is used with its primary winding connected in series with line carrying the current which is to be measured and therefore the primary current is dependent upon the load connected to the system and is not determined by the load(burden) connected to the secondary winding of the current transformer.

The primary winding consists of very few turns and therefore there is no appreciable voltage drop across it.

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• The secondary winding of a C.T has larger number of turns , the exact number being determined by the turns ratio.
• The ammeter, or wattmeter current coil are connected directly across the secondary winding terminals . Thus a C.T operates with its secondary winding nearly under short ckt. conditions.
• One of the terminals of the secondary winding is earthed so as to protect equipment and personnel in the event of an insulation breakdown in the C.T. Fig on next slide shows a ckt. for measurement of current and power with a C.T

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Use of C.T for current & power� measurement

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• Relationship in a C.T
• Fig. shows the equivalent ckt and

phasor

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• Let

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• Now a cut section is considered

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• From the cut section we get eq (1) as

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• Phase angle: The angle by which the secondary current phasor when reversed differs in phase from the primary current is known as the phase angle of the transformer
• This angle is taken to be +ve if the secondary current reversed leads the primary current. The angle is taken as –ve if secondary current reversed lags behind the primary current.
• The angle between Is reversed and Ip is θ. Therefore the phase angle is θ.

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• From the phasor diagram we have

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• Errors in C.T
• It is clear from the eq.

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• The value of transformation ratio(actual ratio) is not equal to the turns ratio(n).
• Also the value is not constant but depends upon the magnetising and loss components of the exciting current , the secondary winding load current and its power factor. This means that the secondary winding current is not a constant fraction of the primary winding current but depends upon the factors listed above. And this introduces considerable errors into current measurements.

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• In power measurements it is ncessary that the phase of the secondary winding current shall be displaced by exactly 180° from that of the primary winding current. It is seen that the phase difference is different from 180° by an angle θ.
• Thus in power measurements owing to the use of C.T two types of errors are introduced;
• One due to actual transformation ratio(R) being different from turns ratio and the other due to secondary winding current(reversed) not being 180° out of phase with the primary winding current.

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• Approximate value for errors:
• The usual instrument burden is largely resistive with some inductance and therefore δ is positive and is generally small.
• Hence sinδ≈0 and cosδ=1.
• Therefore we can write the eqs. of ratio and phase angle errors as

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Construction of a C.T

• The C.T may be classified as:
• (i) Wound Type: A C.T having a primary winding of more than one full turn wound on core.

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• (ii) Bar Type: A C.T in which the primary winding consists of a bar of suitable size and material forming an integral part of transformer.

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• Characterstics of a C.T
• 1. Effect of change of primary winding current: If the primary winding current changes then the secondary winding current would change proportionally.(from relation Ip=nIs).
• At low values of current Ip(or Is) the Im and the loss component Ie are in the greater proportion of Ip and therefore the errors are greater.
• As the current Ip increases , there is a increase in Is (from relation Ip=nIs) and so there is a decrease in the ratio error and phase angle error.(From R=n(1+Ie/Ip)
• And θ=(180/π)(Im/Ip)degree.

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• 2. Effect of change in the secondary winding ckt. burden.
• An increase in the secondary winding ckt. burden impedence means an increase in the VA rating . And This necissitates an increase in the secondary winding induced voltage (so as to overcome an increase in the secondary winding ckt. burden impedence) and which can be generated by an increased flux (Φ) and hence flux density. [As E α ɸ]
• And therefore because of this both magnetising component (Im) and energy loss component (Ie) are increased. Thus because of this the errors will increase in the secondary winding burden.

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• 3. Effect of change of frequency: The effect of increase in frequency will result in proportionate decrease in flux density. [As f is inversly α to flux(ɸ) [From E= 4.44 ɸ fT volts]
• Causes of errors in C.T.
• In an ideal transformer the actual transformation ratio (R) is equal to the turns ratio(n) and the phase angle(θ) for this case would be zero.
• However practically there are departures from this ideal case and so the errors caused.
• The reasons for this are as follows from next slide

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• 1.There is some exciting mmf required by the primary winding so as to produce the flux and therefore because of this fact the transformer draws a magnetising current Im.
• 2. The transformer input must have a component which supplies the core losses(hysteresis & eddy current losses) and copper losses(I²R losses) of the transformer windings due to flow of current Iₒ.
• Therefore we can say loss component Ie is required to feed the losses which are associated with the flux(Φ) and also associated with the copper loss(I²R loss) in the winding (due to the flow of current Iₒ.)(As current flows Cu loss generates)

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• 3. The flux density in the core is not a linear function of the magnetizing force i.e(the transformer core becomes saturated because of this)
• 4. There is always a magnetic leakage and because of this the primary flux linkages are not equal to secondary flux linkages.

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Means to reduce the errors in C.T

• It is clear from eq’s R=n(1+Ie/Ip)
• And θ=(180/π)(Im/Ip)degree that for usual burdens the difference between actual transformation ratio(R) and the turns ratio(n) depends largely on the loss component (Ie)
• And the transformer phase angle (θ) depends largely on magnetising current(Im).
• It is obvious from the above relation if we want (R≈n) And the phase angle(θ) to be small, then Ie and Im must be small as compared to Ip.

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• There are some design features which help us to minimize the errors and they are;
• Design features of C.Ts.
• 1. Core: In order to minimize the errors the magnetizing current Im and the loss component (Ie) must be kept to a low value.
• This means that the core must have a low reluctance(opposes flux) and a low core loss.
• The reduction of reluctance of flux path can be brought by using materials of high permeability, short magnetic paths.

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• The number of joints in building up cores should be minimum as far as possible because joints produce air gaps (which offers paths of high reluctance for the flux).
• The magneto motive force(mmf)(AT’s) consumed by the joints can be reduced by properly lapping the joints and tightly binding the core.
• The core loss is reduced by choosing materials having low hysteresis and low eddy current losses and by working the core at low flux densities.

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• 2. Primary winding current rating: Whatever equipment a C.T is feeding it is desirable that the ratio of exciting current (Iₒ) to primary current(Ip) should be small(As for a good C.T Iₒ is about 1% of Ip.)
• And this means that the ratio of exciting mmf(AT’s) to the primary winding mmf should be low.
• 3. Leakage reactance: It tends to increase the ratio error. Therefore to avoid this the two windings primary and secondary should be close together .

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• 4. Turns compensation: Actual ratio R=n+Ie/Is.
• Considering a 1000/5A C.T with having a loss component (Ie) equal to 0.6% of the primary winding current(Ip)
• Then its nominal ratio(Kn)=1000/5=200.
• So loss component(Ie)=(0.6/100)x1000 = 6A
• Let the number of primary winding turns Np=1.
• Now if the turns ratio(n) is equal to nominal ratio(Kn) then we have n=200.
• Therefore scondary winding turns Ns=nNp(As n=Ns/Np for a C.T) so Ns=200x1=200.

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• So Actual ratio R=n+Ie/Is.
• =200+6/5
• =201.2
• This is the result when we are using 200 turns for the secondary winding.
• Now suppose we do not use 200 turns for the secondary winding but instead of that we use 199 turns.
• That is here we are compensating the turns.
• So the Actual transformation ratio with turns compensation
• R=n+Ie/Is.
• = 199+6/5
• =200.2

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• Thus we find that by reducing the secondary winding turns slightly(here we have reduced by 1 turn), the actual transformation ratio(R)(200.2) is made nearly equal to the nominal ratio(Kn)(200 in this case).
• Usually the best number of secondary winding turns is one or two less than the number which would making (R=Kn) of the transformer.
• Note: The phase angle error is effected very little by a change of one or two turns in the secondary winding.
• Note:The correction through the reduction of the secondary winding turns technique is exact only for a particular value of current and burden impedence. And the C.T in this case is called compensated C.T.

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• Effect on secondary open ckt:
• Never open the secondary winding ckt of a C.T while its primary winding is energised.
• Note: Failure to observe this precaution may lead to serious consequences(both to the operating personal to the transformer.)
• Under normal operating conditions both the primary and secondary windings produce mmf’s (which as per phasor act against each other).

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• The secondary mmf is slightly less than the primary mmf and consequently the resultant emf is small.(as they are in opposition so thus resultant exist due to difference in primary and secondary mmf)
• And this resultant emf being the magnetising mmf (which is required for the maintainance of flux in the core and to supply the iron losses.
• Note: This resultant mmf is responsible for the production of flux in the core and as this mmf is small(as already told), and so the flux density is quiet low under the normal operating conditions and hence because of this a small voltage is induced in the secondary winding.

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• Now if the secondary is open ckted(when the primary winding is carrying current), then in this case the primary winding mmf remains the same while the opposing secondary winding mmf reduces to zero.(as secondary is open ckted).
• Therefore in this case as the opposing secondary mmf is zero so because of this the resultant mmf is equals to the primary winding mmf(IpNp).(as secondary mmf is zero) and which is very large.

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• And this large mmf will produces a large flux in the core (till it saturates).And this large flux when linking with the turns of the secondary winding, would induce a high voltage in the secondary winding and which could be dangerous to the transformer insulation(although now a days modern CT’s are designed to withstand this voltage) and to the person who has opened the ckt.
• Also under this case the eddy current and the hysteresis losses would be very high under these conditions and due to this the transformer may be overheated and completely damaged.
• And if this does not happens then the core may get permanently magnetised and this will gives appreciable ratio and phase angle errors.

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• Permanent Magnetization and Demagnetization of cores of C.T’s.
• The permanent magnetization of cores of C.T’s may result from any of the following effects;
• 1.When the secondary winding is open ckted. With the primary winding energized . This causes a large magnetization force(mmf) which produces a high value of flux density in the core.
• And when this force is taken off , then it leaves a large residual magnetism.

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• 2. A switching transient may leave behind residual magnetism.
• The presence of permanent magnetization in the core of a C.T may reduce its permeability at the flux densities at which it is normally operated and therefore this results in increase of both its ratio and phase angle errors.
• Thus for the proper operation of the C.T , this permanent magnetization should be removed so that the transformer can be restored to its normal conditions.

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• Demagnetization of core:
• The following method used for the demagnetzation of the core:
• 1. The primary winding is excited with full current and a very high variable resistance of several hundred ohm is connected across the secondary winding.
• And this condition is equals to open ckting the transformer practically.
• Now this resistance is then gradually reduced to zero as uniformly as possible.
• And by this means the magnetization of the transformer core is reduced from a very high value to its normal value.

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Numericals on C.T

• Numerical: A 1000/5A, 50Hz C.T has a seconondary burden comprising a non-inductive impedence of 1.6Ω. The primary winding has one turn . Calculate the flux in the core and ratio error at full load. Neglect leakage reactance and assume the iron loss in the core to be 1.5W at full load. The magnetising mmf is 100A.
• Solution:
• Nominal ratio Kn=1000/5=200.
• The turn ratio is assumed equal to the nominal ratio in the absence of any other data.

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• Numerical 2 : A C.T with a bar primary has 300 turns in its secondary winding. The resistance and reactance of the secondary ckt. are 1.5Ω and 1.0Ω respectively including the transformer winding. With 5A flowing in the secondary winding, the magnetising mmf is 100A and the iron loss is 1.2W. Determine the ratio and phase angle errors.

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• Potential Transformers:
• The design of a potential transformer is quiet similar to that of a power transformer but the loading of a potential transformer is small.
• The secondary winding is designed so that a voltage of 100 to 120V is delivered to the instrument load.
• The normal secondary voltage rating is 110V.
• Equivalent ckt. and phasor is shown in next slide.

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• Relationship in a P.T
• The theory of P.T is essentially the same as that of a power transformer. The main point of difference is that the power loading of a P.T is very small and consequently the exciting current is of the same order as the secondary current while in a P.T the exciting current is a very small fraction of secondary winding load current.
• The equivalent ckt and phasor is shown on next slide

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Equivalent Ckt and phasor

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� Enlarged phasor of P.T�

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Actual Transformation ratio

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Phase Angle

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• Difference between C.T and P.T.
• (1)The potential transformer may be considered as ‘parallel’ transformer with its secondary winding operating nearly under open ckt conditions
• Whereas the current transformer may be considered as a series transformer under short ckt. conditions.
• Thus the secondary winding of a P.T can be open ckted without any damage being caused either to the operator or to the transformer.

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• (2) The primary winding current in a C.T is independent of the secondary winding ckt. conditions.
• While the primary winding current in a P.T depends upon the secondary ckt. burden.
• (3) In a P.T , full line voltage is impressed upon its terminals,
• Whereas a C.T is connected in series with one line and a small voltage exists across its terminals.
• However the C.T carries full line current.

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• Errors in P.T
• 1. Ratio(voltage) Error: The actual ratio of transformation varies with operating conditions and the error in secondary voltage may be defined as
• % Ratio error= (Kn-R)/R X100
• Where
• R=Vp/Vs=
• [n+ [nIs(Rs cos Δ + Xs sinΔ) +Ie.rp + Im.xр]/Vs

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• 2. phase angle error: In an ideal voltage transformer there should not be any phase difference between primary winding voltage and the secondary winding voltage reversed. But however in an actual transformer there exists a phase difference between Vp and Vs reversed.

And θ=[Is/n(Xp cosΔ- Rp sinΔ)+ Ie.xp- Im.rp]/nVs

• Note: The phase angle is taken as +ve when the secondary winding voltage(reversed) leads the primary winding voltage while the angle is –ve when the secondary winding voltage(reversed) lags the primary winding voltage (Refer phasor of P.T)

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• Reduction of errors in P.T:
• Reduction of Magnetising and loss components:
• R-n=[nIs(Rs cos Δ + Xs sinΔ) +Ie.rp + Im.xр]/Vs it is clear that the difference between the actual ratio(R) and the turns ratio(n) is made up of 2 parts.
• One is dependent upon the secondary winding current(Is) and other upon the two components of no load current(Ie and Im)(refer expression). These components should be kept low.
• And such a reduction requires short magnetic paths, good quality core material and low flux density in the core.

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• (2) Reduction of resistance and leakage reactance

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• Winding resistance can be minimized by using thick conductors.
• The leakage reactance of the windings depends upon the magnitudes of the primary and secondary leakage fluxes and therefore we should keep the two windings as close as possible.
• A small no. of turns will results in smaller leakage reactance of the windings.

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• (3) Turns compensation: At no load the actual ratio (R) will exceeds the turns ratio(n) by an amount [Ie.rp + Im.xр]/Vs
• (Refer eq. of R)
• Now with having an inductive or resistive load there is a further increase in the ratio(R) because of the addition of voltage drops in resistance and leakage reactance of the windings.(w.r.t from noload)
• For making R=n(for one particular value and type of burden) thenThe solution lies in making turns ratio(n) less than the nominal ratio(Kn)[as was done in the case of C.T].
• This can be done by reducing the no. of primary winding turns or increasing the no. of secondary winding turns
• Note: The angle (θ) is practically unaffected by a small change in the turns ratio.

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• Construction of P.T’s
• The design and construction of P.T are basically the same as that of power transformers with having following differences;
• (i) power transformers are designed keeping in view, efficiency , regulation and cost.
• The cost being reduced by small core and conductor sizes.
• While in designing a P.T economy in the material used is not the big consideration and the transformers are designed to give the desired performance.
• Compared to power transformers a P.T has larger core and conductor sizes as economic designs may lead to large ratio and phase angle errors and which are undesirable features.

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• (ii) The output of a P.T is always small while its is size is quiet large.
• Therefore the temp. rise is small(because of its large size) and hence there are no thermal problems which are caused by overloads(as in the case of Power transformer)
• The loading of a P.T is limited by the accuracy considerations while in a power transformer the load limitation is on heating basis.
• The P.T are able to carry loads on the thermal basis many times their rated loads. These loads ranges from 2-3 times for a low voltage P.T and upto 30 or more times for the high voltage P.T’s.

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• Core: The core may be of shell type or core type in construction.
• Shell type of construction is normally used for low voltage transformers.
• Special precautions should be taken while assembling the transformer so that air gaps at the joints may be minimized.
• Windings: The primary and secondary windings are coaxial(on the same axis) so as to reduce leakage reactance to minimum.
• In order to simplify the insulation problem the low voltage l.v winding(secondary) is put next to the core.
• (contd.)

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• The primary winding may be a single coil in a low voltage transformers but for the high voltage transformers it must be subdivided into a no. of short coils in order to reduce the insulation needed in between the coil layers.
• Insulation: Cotton tape and varnished cambric are used as insulation for coil construction . Hard fibre separators are used between the coils.
• At low voltages the transformers are usually filled without compound but P.T’s for the use at voltages above 7KV are oil immersed(oil act as a insulator)

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• Bushings: Oil filled bushings are used for oil filled P.T’s as this minimizes the overall size of the transformer.(as bushings also carries oil )
• A 2 winding single phase P.T is shown in next slide:

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Single phase P.T.

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• Protection of P.T
• Pt’s are contineously operated at 1.2 times the rated voltage.
• A short ckt. on the secondary side of a P.T can lead to complete damage of the transformer.
• In order to protect the power system against short ckts in the potential transformers fuses are used on the high voltage (primary) side.
• Fuses are used in the secondary side against faulty switching and defective earthing.

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• Capacitive P.T’s.
• At voltages above 100KV (phase) the conventional electro-magnetic type of P.T becomes expensive owing to the insulation requirements.
• A less expensive alternate is capacitive voltage transformer.
• This consists of a capacitance potential divider used in conjunction with a conventional auxillary transformer.(Fig. shown in next slide)

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Capacitive P.T

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• The capacitance potential divider steps down the voltage (which is to be measured)(say to 10KV).
• And the capacitance divider output voltage is further stepped down by the auxillary transformer to the desired secondary voltage(say110V).
• The auxillary transformer consists of an inductance L which may consists (wholly or partly) of leakage inductance of the windings of auxillary transfomer.
• The value of this inductance L may be adjusted to the value equals to 1/ω²(c₁+c₂)[from ω=1/√LC] so that the voltage drop due to the current which is drained from the divider is largely compensated.

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• Characterstics of P.T.
• 1. Effect of secondary current or VA.
• If we increase the secondary burden, the secondary current is increased and therefore the primary current increases.
• Both primary and secondary voltage drops increase and thus for a given value of Vp the value of Vs decreases and hence the actual ratio(R) increases as the burden increases.
• Because of increase in R the ratio error increases
• [(Kn-R)/R] X100 and becomes more negative with the increase in burden.

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• With regard to phase angle , the Vp is more advanced in phase because of the increased voltage drops with the increase in the secondary burden.
• The phasor Vs(reversed) is retarded in phase because of increase in the secondary winding voltage drops.
• Thus with the increase in burden the phase angle between Vp and Vs reversed increases, (becoming more and more –ve).

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• 2. Effect of p.f on secondary burden: If the p.f(cosΔ)[angle between Vs and Is) of the secondary ckt burden is reduced then angle Δ is increased . This makes current Ip to shift towards the current Iₒ(Refer phasor).
• Because of this the voltages Vp and Vs will comes in phase with Ep and Es respectively.(since the voltage drops (Isrs and Isxs) are almost becomes constant in this case).
• The result of this is increase in Vp(phasorly) w.r.t Ep.
• But as Vp remains constant, so Ep reduces relative to Vp.
• And similarly the voltage Vs reduces relative to Es.
• Therefore the transformation ratio(R) increases as the Δ increases(or if the secondary burden reduces)

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• (3) Effect of frequency. For a constant voltage flux is inversly proportional to frequency.
• Increase in freq. will reduces the flux and because of this reduction in flux Im and Ie are decreased and therefore because of this the voltage ratio (R) decreases.
• Note: This decrease is not so much as with the increase in frequency(as we are increasing the freq.) the leakage reactance(Xʟ=2πfL) and therefore due to this the leakage drops are increased and thus giving an increase to ratio(R).

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• (4) Effect of primary voltage: There is no wide variation of supply voltage to which the primary winding of the P.T is connected.
• Therefore the study of variation of ratio and phase angle errors with supply voltage are of no importance.

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Numericals on P.T

• Numerical: A P.T , ratio 1000/100 volt has the following constants:

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REFERENCE

• A.K Sawhney Electrical & Electronic Measurement & Instrumentation.

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