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  • J2019L's Avatar
    1 Day Ago
    To solve this question, need to have math and electric concept. In the complex number, R represents real part and Xc, XL represent complex parts. So, R is the number on the real axis and Xc and XL are the number on the complex axis. Xc and XL are the number on complex axis, but XL is positive (to up side) Xc is negative (to down side) on the complex...
    6 replies | 2663 view(s)
  • Kalbi_Rob's Avatar
    1 Day Ago
    FYI, NETA 43-2013 states: 12.2.2 Applicability of polarization index when IR1 is greater than 5000 MΩ When the insulation resistance reading obtained after the voltage has been applied for 1 min (IR1) is higher than 5000 MΩ, based on the magnitude of applied direct voltage, the total measured current (IT) can be in the submicroampere range (see Figure...
    21 replies | 12463 view(s)
  • Kalbi_Rob's Avatar
    1 Day Ago
    Aww, that's the equation I forgot Z=1/(sqrt((1/R)^2+((1/XL)-(1/XC))^2))
    6 replies | 2663 view(s)
  • J2019L's Avatar
    2 Days Ago
    1. Convert Xc from delta connection to Y connection. In Y connection, Xc'= 60*60/(60+60+60)=20 2. Calculate balance 3-phase circuit for two loads, parallel Xc' & XL. Total load 1/X= (1-2)j/40 --> X=40j (reactance) 3. 3-Phase balance current I = 277/40 = 6.925 --Ans
    6 replies | 2663 view(s)
  • Kalbi_Rob's Avatar
    4 Days Ago
    I can't help you, I only get 20.781 A. Now the current flowing through the Wye-connected inductors appears to be 6.93A. I have to ask if you might have misunderstood the question.
    6 replies | 2663 view(s)
  • richardsa's Avatar
    5 Days Ago
    In the old days the utility used synchronous motors to control reactive power. They would not drive a load. By control of the excitation current they could off-set and bring the power factor close to unity.
    36 replies | 19261 view(s)
  • clayton's Avatar
    1 Week Ago
    Ahh, well wouldn't that make a whole lot of sense. I guess I assumed people could read my mind A wye-connected alternator that has phase voltages of 277 V is feeding two loads. The first load is wye-connected inductors that have 40 ohms of XL. The second load is a set of delta-connected capacitors that have XC of 60 ohms. How much is the alternator...
    6 replies | 2663 view(s)
  • Kalbi_Rob's Avatar
    1 Week Ago
    Can you assist us by stating the question you need answering?
    6 replies | 2663 view(s)
  • clayton's Avatar
    1 Week Ago
    I have been stuck on this question and just can't seem to figure out how to get the correct answer of 6.93. Would you be able to show the work needed for this?
    6 replies | 2663 view(s)
  • young_dad's Avatar
    1 Week Ago
    The way I taught my guys that seemed to stick with them was to visualize all transformer ratios as single phase devices. First off, it's important in this example to note that it's a WYE primary, which is not common and is an easy way to make a mistake. With a line voltage 7.2kV, we divide by 1.732 to find a phase voltage ~4.16kV. In the DELTA...
    16 replies | 13321 view(s)
  • Jrmcritical's Avatar
    2 Weeks Ago
    Hey Alex, Just passed my Level I exam last s=Saturday. I found that the majority of the questions were safety related, with most of them pertaining to the principles of operation. For example, they'll ask about how different extinguishing agents actually fight the fire, vs asking about classes of extinguishers. If you took your OSHA 30 within the past...
    3 replies | 3614 view(s)
  • ironshovel77's Avatar
    3 Weeks Ago
    another way to calculate find percent of coils 200v x 60%= 0.6 x 100=60v
    2 replies | 1684 view(s)
  • Eiltsman's Avatar
    3 Weeks Ago
    GPerske and BigJohn are correct. The entire length of the coil is one continuous wire. Therefore regardless of where you stab into the taps with your test voltage, voltage will exist on the entire coil and will be boosted beyond the test points, all the way to each end of the coil. 1500/1200x400=500.
    7 replies | 5956 view(s)
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