# The Value of the Capacitor Size

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.

https://brainmass.com/engineering/electrical-engineering/173260

#### Solution Preview

I = V/Z

P = sqrt(3) * V * I * Cos(theta)

= sqrt(3) * ...

#### Solution Summary

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.

Now add a capacitor from emitter to ground, and size it to give a 3 db attenuation in gain to the collector at 1 kHz, relative to gain at a higher frequency where the gain does not change with frequency. What size is the capacitor? Does the input capacitor size have to change, and if so, to what value? What are the new voltage gain and input and output impedances? Assume that the circuit is driven from a 50 Ohm source resistance generator. What is the (rms) voltage noise at the base from 1 kHz to 11 kHz frequency? What is this noise referred to the output?

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