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Blizzard [7]
3 years ago
8

A balanced bank of delta-connected capacitors is connected in parallel with the load which complex power associated with each ph

ase of a balanced load is 146 191j kVA. The effect is to place a capacitor in parallel with the load in each phase. The line voltage at the terminals of the load thus remains at 2460 V . The circuit is operating at a frequency of 65 Hz . The capacitors are adjusted so that the magnitude of the line current feeding the parallel combination of the load and capacitor bank is at its minimum.
What is the size of each capacitor?
Engineering
1 answer:
Sergio [31]3 years ago
8 0

Answer:

77.2805 μF

Explanation:

Given data :

V = 2460 V

Q = 191  Kva

<u>Calculate  the size of Each capacitor </u>

first step : calculate for the value of Xc  

  Q = V^2/ Xc

  Xc ( capacitive reactance ) = V^2 / Q = 2460^2 / ( 191 * 10^3 ) = 31.683 Ω

Given that  1 / 2πFc = 31.683

∴ C ( size of each capacitor ) = \frac{1}{2\pi *65 *31.683}  =  77.2805 μF

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Answer:

At steady state output will be 2

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We have given transfer function G(S)=\frac{6}{12S+3}

Input is unit step so X(S)=\frac{1}{S}

We know that G(S)=\frac{Y(S)}{X(S)}, here Y(S), is output

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Y(S)=\frac{1}{2S(S+\frac{1}{4})}

Now using partial fraction

\frac{1}{2S(S+\frac{1}{4})}=\frac{A}{2S}+\frac{B}{(S+\frac{1}{4})}

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The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the c
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Answer:

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Explanation:

Given;

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resonance frequency, f₀ = 100 kHz

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Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R

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X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)}  \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH

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