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  1. Q = C*V C = Q/V = (1.2*10^-6/10) = 1.2*10^-7 Capacitors in parallel add, so if we need n capacitors C = 1.2*10^-7 = n*0.25*10^-12 n = 1.2*10^-7/(0.25*10^-12) = 4.8*10^5 = 480,000 <<<

    1 Answers · Science & Mathematics · 11/02/2014

  2. Depending on what device and how it was shorted. For pure resistance load shorts, PF = 1 Other partial short on transformer, it is no more equal to one.

    2 Answers · Science & Mathematics · 30/01/2014

  3. ...factor = cos 17.17' = .956 (C) Power dissipated = (VA)*(pf) = (230V)*(7.45A)*(.955) = 1636.4 Watts Power ...

    1 Answers · Science & Mathematics · 20/01/2014

  4. 1. D both B & C ; since 3300 pF = 3300E-12 F = 3.3E-9 F = 3.3nF = 0.0033 ...

    2 Answers · Science & Mathematics · 14/01/2014

  5. ...for series R and X: |Z| = V * I with a phase angle PF (Power factor) = true_watts / apparent_VA = cos(θ) θ = ...

    2 Answers · Science & Mathematics · 18/01/2014

  6. More than anything else the efficiency is dependent on load. Most transformers are designed to reach maximum efficiency at 75 to 80% full load, dropping slightly toward full load and ever more steeply as load is reduced. Here I understand...

    2 Answers · Science & Mathematics · 13/01/2014

  7. ... [kΩ] 3K R2 [kΩ] 300K C1 [pF] 680 pF FREQ.MIN [kHz] 877 VTUN = 1.1...

    2 Answers · Science & Mathematics · 16/01/2014

  8. ...the same for both the kWh and kVARh, then the power factor would be: PF = Cos(ATan(1500/5000)) = 0.96 or 96% You have to recognize that...

    2 Answers · Science & Mathematics · 17/01/2014

  9. It's a ceramic capacitor. Value appears to be 47 nF or 47000 pF. It is probably used to suppress spikes from the motor.

    1 Answers · Science & Mathematics · 11/01/2014

  10. ... Also: (V^2)/(Z) = VA But (VA)*(pf) = power pf = cos arc tan XL/R = cos [arc tan (28.27/60)] = .905 Z...

    5 Answers · Science & Mathematics · 05/01/2014

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