Chapter Five. instrument transformers. instrument transformers are a special type of transformer used for the measurement of voltage and current. As the name suggests, these transformers are used in conjunction with the relevant instruments such as amateurs volt meters, watt meters and energy meters as well as protective relays. instrument transformers are of two types, there are current transformers and potential Transformers or sometimes called voltage transformers. current transformers are used when the magnitude of the AC current exceeds the safe value of current measuring instruments potential transformers are used where the voltage of the AC circuit exceeds 750 volts as it is not possible to provide adequate insulation on measuring instruments for voltages more than at this level.

These are examples of current transformers that are found in the bushings of a power transformer or circuit breaker. They are known as donut CTS, for obvious reasons they look like a doughnut. The picture on the right is not a CT but typical transformer could be a breaker bushing which would be mounted through the center of a doughnut CT. The secondary winding is wrapped concentrically around a toroid which is usually made up of laminated iron or steel. The primary is a single conductor, usually a bushing mounted through the center of the toroid. The doughnut fits over the primary conductor which constitutes one Primary turn.

The secondary is wild around the toroid, which is usually made up of laminated iron. That concentrate concentrates the magnetic flux and forces it through the secondary turns. Some donut CTS come with a primary conductor incorporated in the CT structure, and connections are made by bolting the primary leads to it. If the toroid is well with 240 secondary turns, then the ratio of the CT is 240 to one or 1200 to five amp. The five amp designates the continuous rating of the secondary winding and is normally five amps at least in North America, and one amp or point five amps in many of the other parts of the world. This type of donut CT is most commonly used in circuit breakers and Transformers this year fits into the bushing turret and the bushing fits through the center of the doughnut.

Up to about four CTS of this type can be installed around each bushing in an oil circuit breaker or a transformer bushing. instrument transformers are used for the following reasons one to insulate the high voltage circuit from the measuring circuit in order to protect the measuring instruments from damage. Also to make it possible to measure high voltages with low range volt meters and high current with low range amateurs. Also, instrument transformers are used in combinations of summing and measuring voltages and currents and you'll see more examples of that in subsequent slides. Let's take a closer look at instrument transformer ratios in an ideal transfer Former with a simple load on the secondary and an AC voltage on the primary, the secondary voltage is determined by the turns ratio such that V s over VP is equal to ns over MP vs being the secondary voltage VP being the primary voltage and as the number of turns of the secondary of the transformer in NP, the number of turns are the primary of the transformer.

Looking at the current the Magnum motive force is given by IP NP or is ns, which is the same for both sides of the transformer which turns out to give us this equation which is IP n p is equal to is ns, where IP and i s are the primary and secondary currents respectively and NP and S are the number of terms of the primary answer Secondary respectively or we can rewrite the equation bringing the secondary current to the left hand side and everything else to the right and that gives us is equal to IP times NP Orion s and NP are the number of terms of the primary is one such as excessive is in a bushing. Type CT, then I asked is given by IP divided by the terms of the secondary only, which is the number of terms on the toroid of a doughnut CT. A current transformer differs from a voltage transformer in that its secondary current is determined entirely by the load on the primary system and not on its own secondary load.

In other words, the secondary current is determined by the current in the primary conductor. The voltage across a secondary load which is usually called the burden will rise and fall depending on The current in the primary. So if a an impedance or a burden or resistance is placed on the secondary of a CT, the higher that resistance The higher the voltage drop will be across it because it's trying to maintain the same current through it, which again worth repeating is dependent on the primary. Both the primary and the secondary of a CP have relatively few turns of heavy wire and low impedance subsequently, a current is readily induced into the secondary proportional to the load or the primary current. There is a unique problem encountered with Transformers in general and current transformers in this case, in order to magnetize the core of a transformer. a certain amount of excitation current is required.

Part of the current induced into the secondary from the primary is used to accomplish this. Since the induced current represents the current flowing in the load circuit, the metering or the meters current coils will be influenced influenced proportionally by the load current minus the excite ad exciting current or excitation current. This represents a very small loss of load current and therefore, affects the accuracy of the recorded power consumption. current transformer loss is called errors vary for different types of transformers and the burden or load on the secondary measurements Canada has established acceptable error limits that a transformer must fall within. Engineers must take this into consideration when designing relays and metering setups. This is very small Indeed, but it can make a large difference in some cases and you must be aware of it.

The transformer ratio of a CT is defined by the rated primary current over the rated secondary current. For a given CT, where the primary current is 100 amps and the secondary current is five amps. The transformer ratio is 100 to five or 20 to one. For the same CT what would happen if the primary was looped back and then fed through the CT one more time the CT would see 100 times two which is equal to 200 amps. The secondary would be 10 amps. The transformer ratio here then would be 100 to 10 or 10 to one, this is sometimes used if the primary current is too low for a meter reading designating the polarity is very important and done by marking the CT primary and secondary keys.

This is especially important for power flow direction and when CTS are used for directional reading. There are standards but they all pretty much mean the same thing. The primary and secondary are mark to indicate what the which direction the current will flow during each house cycle a current a primary terminal is marked to associated With a secondary terminal, and we say that the terminal is Spock with respect to the Associated secondary terminal Mark spot and you'll see how that works in a few minutes. In the case of a donut CT, where the primary is the conductor running through it, one side of the CTS marked with a spot the I triple E Institute of Electrical and Electronic Engineers instead of spots they use markings h one and x one and the IEC which is the International Electrotechnical Commission. Use p one and s one. Suffice it to say that the one side or the primary side, one side terminal is marked with respect to the other and again you'll see how that works.

What this indicates is that when the primary current is positive during the sinusoidal half cycle, in other words current into the spot the secondary current will be out of the secondary spot. Just to repeat that again, the primary current is positive during the sinusoidal half cycle or current into a spot on the primary results in the secondary will have current out of the spot on the secondary. Conversely, what this also indicates that when the primary current is negative during the sinusoidal half cycle Where the current is out of the spot and the primary, the secondary current will be out of the non spot of the secondary. with potential transformers. In general a secondary voltage is proportional to the primary voltage equivalent to the turns ratio such that V s divided by VP is equal to ns divided by NP. This relationship is not completely exact for the following reasons, the exciting current that is necessary to magnetize the iron core causes an impedance drop in the primary winding and the load current that is drawn by the burden causes the impedance drop in both the primary and secondary windings.

Both of these produce an overall voltage drop in the transformer and introduce errors in both ratios and phasing. The net result is that the secondary voltage is slightly less than the ratio of the turns would indicate and there is a slight shift in the phase relationship. These two errors are called ratio errors and phasing layers and maybe represented by this equivalent circuit. These errors are usually very small and for the most part can be neglected. However, we must remain aware of them as they do not become or they do become significant from time to time and may not be neglected. As with CPS, there are polarity standards but they pretty much mean the same thing.

The primary and secondaries are marked indicate which terminals are in phase A primary terminal is marked to associated with a secondary terminal. We say that the primary terminal is spot with respect to The Associated secondary terminal Mark spot. In other words in face. This is done on the PT itself and or on the nameplate. Again this is especially important when PPS are used for relays with direction and revenue meters. As in the case of the CT polarity, the primary and secondary terminals are marked in such a way that when the primary current is positive during the positive half cycle that is current into a polarity mark.

The secondary current will be out of the polarity or spot mark or in terms of voltage seen as we're dealing with voltage transformers here. A voltage rise on a trailer terminal gives a voltage rise on the x one terminal. single phase metering using instrument transformers. Let's put a two element polyphase meter to work using instrument transformers. In this case CTS, the load is being fed from a distribution transformer just as in a residential supply. To meter this load we are using a polyphase two element meter and current transformers on a single phase three wire 122 40 volt supply to the load using measurement Canada standard 1311.

In order to understand how this metering circuit works, let's go back to our earlier analysis of a metered residential load where we consider two independent power supplies. Were the watt meters were arranged like this. And the watt measurement formulas were as shown here. The power to each of these loads can be measured by two light meters. Notice the spot or polarity markings on the light meters. Now, if we reconnect the current coil of wire meter to, which would reverse I two, then the watt meter two would read minus v two times minus i two, which of course is just v two times I two.

With this arrangement, the two meters will read w one equal to v one times I one and W two equal to v two times I two. Now, let's suppose that the current to the load is too high for me. So we introduced CTS and the circuit would look like this, the red color of the I one and I two currents are to let us know that they are in phase with the primary current, but their magnitudes are subject to the CT ratios, you will notice that w one and W two formulas are unchanged. So for simplicity sake, I'm going to leave the CTS out of the next slide and keeping the red color to remind us that we have to apply a multiplier due to the CT ratio in the final analysis. Connecting the circuits together at the neutral course makes no difference or electrical difference to the load.

This arrangement is the same as connecting the loads to the center tap residential distribution transformer where v1 is the line to neutral voltage and v2 is the neutral to line voltage on the secondary side of the transformer. Notice that The load is not included in the equation and makes no difference to the formulas calculating the metering here. Let's trace out the current and voltage starting with the current, the load current i one flows through the CT, the secondary which flows through one element of the meter. The load current two flows through the CT and the secondary which flows through the other element of the meter. Now, if we trace out the voltages v one is the line to neutral voltage connected to the load and V Is the neutral to line voltage on the secondary side of the transformer. The total average power measured to the load is now v1 in RMS times I want in RMS times the cosine of the angle between them plus v two in RMS, times I two in RMS times the cosine of the angle between them.

I want an eye to must be multiplied by the CT ratio. Let's work a problem that has CTE connections worked into the circuit. In this figure we've placed a load of 10 ohms across line one and line two and two line to neutral loads 150 and the other 30 ohms. We are connecting the meter through CTS with turns ratios of 50 to five or 10 to one. We want to calculate what the amount of power used by the load. And we want to find out what is the amount of power measured by each element.

And, of course the total load measured by the meter itself. And as you can see, we're using measurement Canada's standard connection 1311 in this case, which does show the connections for two CTS connected to the meter with a direct connection of the potential coils using ohms law as we did before the current through the load r1 is going to be 240 volts divided by 10 ohms which is 24 amps. The power used is 240 volts times 24 amps which is 5760 watts current through R two is 120 divided by 50 which is 2.4 amps and the power through that that particular load will be 120 volts times 2.4 amps gives us 288 watts current through the r three is 120 volts divided by 30 ohms gives us four amps and the power is 120 volts times the four amps which is 480 watts giving us a total reading for the meter of 6500 Hundred and 28 watt.

The total watt meter reading is going to be made up of the sum of the two elements, the one times I one plus p two times i to the meter will then read for the load r1 120 volts times 2.4 amps for the top element, times 10 for the CT ratio gives us 2880 watts, plus 120 volts times 2.4 amps for the bottom element times 10 for the CT ratio, which is 2880 watts for a total power measurement of 5760 watts for the load r1. The meter for our two will read 120 volts times point two Four amps for the top element, times 10 for the CT ratio gives us 288 watts. For our three, the meter will read 120 volts times point four amps for the bottom element times 10 because of the T CT ratio giving us 480 watts totaling all of the elements together for all of the loads, we get 6528 watts made up of the sum of 5760 plus 288 plus 480.

Let's analyze another solution to metering a three wire single phase load that is being fed from a distribution transformer similar to what is used in residential supply. This time we'll use a single element meter and a three wires CT. A three wire CT is a self contained current transformer with two single bar primaries through the central core such that the polarities of each primary is opposite to the other. This is old school metering, but there are still some in existence and it is still a standard. Notice that one of the primary goes through the CT in the reverse direction, which means the secondary current will be I one minus i two, of course, divided by the CT ratio of the three wire CT in order to understand how this metering circuit works, so let's go back to our earlier analysis of a metering of a meter residential load, where we consider two independent power supplies where the watt meter is were arranged like this.

And the watt meter a watt measurement formulas were as shown here. The power to each of these loads can be measured by two watt meters. And again take note of the spot or polarity markings of the watt meter. As you can remember, we derived the formula for the total watt consumption, which is given by the equation wt is equal to the first element v one times one minus the other element the two times I two, which can be written rewritten because of v one is equal to v two is equal to V T or the total voltage divided by two, which can also be rewritten if we just move and rearrange figures mathematically that the total watts is given by the total voltage times the quantity one over two minus i two all over to the meter thus Reeves vt times the quantity I one divided by two minus i two divided by two.

The fact that I want an eye to R divided by two is handled in the three wire CT ratio. Once again we have the meter reading p average or the average power is the RMS voltage times the rms current times a cosine of the angle between them. multiplied by the CT ratio. Keep in mind that the divide by two in the formula is inherent with a three wire CP. The same situation could be managed With the use of two separate CTS, this is a single phase meter using instrument Transformers to sum the secondary currents to measure the load on a single phase three wire 120 to 40 volt supply load. We can duplicate the configuration of a three wire CPE using two IRC T's the primary of CP one is connected to line one.

The secondary is connected so that the current flows positively through the element of the meter. The primary of CT two is connected to line two. A lead is taken from the unmarked terminal of CP two and is connected to the mark polarity terminal of CP one another lead is taken from the mark polarity terminal of CT two and is connected to the unmark terminal of CP one. The result is the current from CT, two flows backwards through the current element of the meter, giving us the negative for I two which satisfies the equation. This type of wiring is often return referred to as cross connecting. The result is the current element of the meter reads Plus, the current in line one minus the current in line to which gives us the equation eyeline one minus, eyeline two The potential element is connected across line one and line two, which which measures essentially, the T or the total voltage satisfying the equation, the T times the quantity I one minus i two, but the calculated equation calls for a divide by two for the total voltage.

If we look at the standard, the standard says that we must apply if we're using this type of a connection, a meter multiplier of one half, which then takes into account our calculated formula. Total watts is equal to vt over two times the quantity I one minus i to the three wire CP solution uses the standard connection of 30 Keno three, which provides the multiplier because it uses AI one divided by two and AI two divided by two built into its turns ratio. Whereas the two CT solution use used by standard 1309 doesn't have a built in solution and therefore must have a multiplier of one half of course plus any CT ratio. Here's a practical example of a single element meter connected through a three wire CT to measure the power to a load that draws 100 amps when it is connected line to line and we are using measurement Canada standard connection 1303 our task will be to find out what is the amount of power used by the load What is the amount of power measured by the watt meter the load draws 100 amps therefore using the formula for power where you just multiply the voltage times the current, the power would equal 240 volts times 100 amps which would equal 24,000 watts or 24 kilowatts.

At first glance you might deduce that the secondary current out of out to the meter is five plus five amps equaling 10 amps However, this is not the case. The three wire CT windings are in parallel each secondary winding contributes 2.5 amps under 100 amp load. In our illustration, 100 amps of primary current will produce 2.5 amps Plus 2.5 amps equaling equaling the desired five amps secondary current out to the meter. looking closely at the nameplate CT ratio is states at 100 to 100 to five, which means 100 amps in both primaries will produce five amps in the secondary. Therefore the meter sees 240 volts for the potential coil times 2.5 amps plus 2.5 amps in the current coil times 20, which is the CT ratio, which gives 24,000 watts or 24 kilowatts. This ends chapter five