An electrical supply company charges its consumers a penalty if their reactive energy consumption (Kvar-hour) exceeds half their true energy consumption (Kilowatt-hour).
Calculate the value of the parallel capacitor required to be installed at the factory to avoid the penalty.
I know that:
Apparent power S = 32594.05
Active power P = 26535.32
Reactive power Q = 18927.46
Using the formula:
C = 1/w [1/Zt * Sin Q - 1/Zt * Cos Q * Tan Q]
I get C = 0.000181
How do I get P or S from the following:
The supply is 240 Volts ac at 50Hz.
The total load impedance is Zt = 3.04 at 35.5 degrees.
The power factor is 0.81.
I = V/Z
P = sqrt(3) * V * I * Cos(theta)
= sqrt(3) * ...
This solution calculates the value of a capacitor required to be installed at a factory and the variables required.
BJT: capacitor size, voltage gain and impedance
Please see the attachment for circuit diagram, where
R1 =91k ohms
R2 = 20k ohms
Rc = 5K ohms
Re = 1k ohms.
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