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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

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 · 31/01/2014

...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

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

...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

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

... [kΩ] 3K R2 [kΩ] 300K C1 [

**pF**] 680**pF**FREQ.MIN [kHz] 877 VTUN = 1.1...2 Answers · Science & Mathematics · 16/01/2014

...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

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

... 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