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How much power is dissipated by a circuit with 15000 ohms of resistance and 10mA of current?
Wiki User
∙ 12y ago
15 KOhms times 10 mA = 150 volts.
150 volts times 10 mA = 1.5 watts.
Wiki User
∙ 12y ago
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How much power in a 15000 kva transformer?
The power in a 15 MVA (15000 KVA) transformer depends on the power factor. You did not specify the power factor, so I will assume a power factor of 0.92. Simply multiply MVA by PF and you get 13.8 MW.
How many amps does a 20HP motor draw at 415volt three phase?
Assuming a power factor of approx. 0.8 which depends on motor but this will error on the side of caution IL =S/1.73*VL S=P/pf 20hp= about 15000 W. S=15000/.8 = 18750 VA. IL = 18750/1.73*415 = 26A The motor would draw about 26 A. depending on Load and efficiency
What wire size for 15000 Volts and 550 amp service?
There are two distinct questions here. To determine the wire size to carry 15000 volts the circuits load amperage must be stated. The wire size for a 550 amp service is, an 800 MCM copper conductor with an insulation factor of 90 degrees C. This conductor is rated at 555 amps. Parallel 250 MCM will give you a total capacity of 580 amps. A triple run of 2/0 will give you a total capacity of 585 amps.
Is there a standard and safe range for insulation resistance of motors or it's dependent on type of motorsif 30 Mohm is safe for a motor with 1000 Mohm of insulation resistance in normal conditions?
"INSULATION resistance is not about standard nd safe range because resistance value depend upon the application for which we are providing it in what range the current flows." The answer above is incorrect. Application and current rating & range have nothing to do with insulation resistance. There are standards available for insulation resistance testing both in terms of the test procedure and the values. IEEE Standard 43 provides a basis for insulation resistance testing. Both NETA ATS (Acceptance Testing Specification) and NFPA 70B gives criteria for insulation resistance testing. Insulation Resistance (IR) testing, also known by the slang terms meggering or megging, is a procedure where the quality of the electrical insulation is evaluated to determine if it is acceptable for service. It is also used to compare against previous measured values to determine if there has been any degradation to the equipment being tested. In this case we are talking about motor insulation. The question deals with two comparative readings and wants to know if the value of 30 megohms is acceptable. First we should clarify the test methods and results. NETA ATS 2007. Section 7.15.1 covers AC induction motors and generators. The testing involved is broken down into those motors 200 hp (150 kw) and less, and those > 200 hp (150 kw). The test voltage value is based upon the voltage rating of the motor's winding and is found in table 100.1: Rating = 250V; Test Voltage = 500 VDC; Minimum Resistance = 25 megohms Rating = 600V; Test Voltage = 1000 VDC; Minimum Resistance = 100 megohms Rating = 1000V; Test Voltage = 1000 VDC; Minimum Resistance = 100 megohms Rating = 2500V; Test Voltage = 1000 VDC; Minimum Resistance = 500 megohms Rating = 5000V; Test Voltage = 2500 VDC; Minimum Resistance = 1000 megohms Rating = 8000V; Test Voltage = 2500 VDC; Minimum Resistance = 2000 megohms Rating = 15000V; Test Voltage = 2500 VDC; Minimum Resistance = 5000 megohms Rating = 25000V; Test Voltage = 5000 VDC; Minimum Resistance = 20,000 megohms Rating = 34500V; Test Voltage = 15000 VDC; Minimum Resistance = 100,000 megohms It is important to note that the values given are based on a standard temperature of 40C (or sometimes 20C depending on the engineer's specification). You must correct your readings to a standard temperature as the value on the insulation's resistance is going to vary inversely with temperature. That is as temperature increases the resistance will decrease. The rule of thumb is that the measured value will halve for every 15C above standard and will double for every 15C below standard. As an example let us say that we have a 25 hp induction motor rated 480 VAC. The ambient temperature is 15C. Using our table we would set the tester to 1000 VDC and take a reading for one minute. At the end of the minute we get a reading of 450 megohms. Per NETA ATS Table 100.14 the correction factor is 0.31 so IR = 450 megohms x .031 = 139.5 megohms. The minimum acceptable value is 100 so this motor is acceptable. On the flip side if the motor is in a very warm process area, say 50C, then temperature correction factor is 1.59 thus IR = 450 megohms x 1.59 = 716 megohms. As you can see the temperature makes a very large difference in the results! This discussion up to this point has been about a spot-reading check. However for a true check we want to know the Dielectric Absorption value. There are two different standard tests for Dielectric Absorption. The first is the Dielectric Absorption Ratio (DAR) and the second is the Polarization Index. DAR = Reading @ 60 sec / Reading @ 30 sec. Let us say that the 30 sec reading = 325 megohms and the 60 sec = 450 megohms. Thus: DAR = 450 megohms / 325 megohms = 1.38 The minimum DAR per NETA is 1.4 so this particular motor is borderline at best and should be investigated further. PI = Reading @ 10 min / Reading @ 1 min. Let us say that the 1 min reading = 450 megohms and the 10 min = 1100 megohms. Thus: PI = 1100 megohms / 450 megohms = 2.44 The minimum PI per NETA is 2.0 so in this case the motors is acceptable. One final factor that should be taken into consideration is Relative Humidity (RH). The amount of moisture present in the air also affects the measured test values. The more moisture then the lower the reading. There is no published standard correction factor for RH however when NETA Techs perform these tests then they always record the RH for baseline comparison. The bottom line is that the readings will vary based upon temperature and humidity. A reading on warm humid day may be acceptable whereas the same reading on a cold dry day may not. So the question asked here is unanswerable as there is not enough information given. What were the temperatures at the time of the readings? Was it dry during one and raining during the other? What does the person asking mean by "normal conditions"? I highly recommend that anyone interested in this subject get the free book "A Stitch in Time" by Biddle at: http://www.biddlemegger.com/biddle/Stitch-new.pdf Please note that I also changed the category from Health/pregnancy to Electrical Engineering.
Can a 100 breaker box handle a 15 kw furnace?
The equation for amperage is I = W/E, Amps = Watts/Volts. The amperage for the furnace would be, Amps = 15000/240 = 62.5 amps. So to answer the question, yes a 100 amp breaker with a #4 copper conductor will be sufficient to operate a 15 kW furnace. A 70 amp breaker will work but it is cutting the edge a bit too fine as you don't need the furnace to trip off during the winter months (northern hemisphere) if you are away from your home for an extended period of time.
How many amps in 15KVA circuit at 440 volts?
it is 15000/440 based on basic power definition i.e., power =voltage *current
What is the maximum current demand for a 15000 watt range?
The maximum current will depend on the voltage of the power supply. For a supply at 240 volts, the max current will be 15000/240 = 62.5 amps. For a supply at 110 volts, the max current will be 15000/110 = 136.36amps to two significant figures.
What is Oregon's current population?
Oregon's current population is 15000 people
How many jaguars left?
The current population is over 15000 to 30000.
What fraction is 15000?
15000 = 15000/1
What is 19 percent of 15000?
19% of 15000= 19% * 15000= 0.19 * 15000= 2850
What is 110 percent of 15000?
110% of 15000= 110% * 15000= 1.1 * 15000= 16,500
How much is 0.50 percent of 15000?
750.5% of 15000= 0.5% * 15000= 0.005 * 15000= 75
What is 30 percent 0f 15000?
30% of 15000 = 30% * 15000 = 0.3 * 15000 = 4500
What is 17 percent profit of 15000?
17% of 15000 = 17% * 15000 = 0.17 * 15000 = 2550
What is 1 percent of 15000?
1% of 15000 = 1% * 15000 = 0.01 * 15000 = 150
What is 15000 as a fraction?
15000/1
