# A three-phase cable is supplying $800\;kW$ and $600\;kVAr$ to an inductive load. It is intended to supply an additional resistive load of $100\;k 0 votes 60 views in General closed A three-phase cable is supplying \(800\;kW$ and $600\;kVAr$ to an inductive load. It is intended to supply an additional resistive load of $100\;kW$ through the same cable without increasing the heat dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the load. Given the line voltage is $3.3\;kV,\;50\;Hz$, the capacitance per phase of the bank, expressed in microfarads, is ________.

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$KV{A_1} = \sqrt {{{800}^2} + {{600}^2}} = 1000\;KVA$

Without excessive heat dissipation means current should be constant (i.e.) KVA

rating must be constant.

In second case Active power, $P = 800 + 100 = 900\;KW$

Reactive power in second case ${Q_2} = \sqrt {{{1000}^2} - {{900}^2}} = 435.889\;KVAR$

Reactive power supplied by the three phase bank $= 600 - 435.889 = 164.11\;KVAR$

$\begin{array}{l} \frac{{{Q_b}}}{{ph}} = \frac{{164.11}}{3} = 54.7\;KVAR\\ \frac{V}{{ph}} = \frac{{3.3}}{{\sqrt 3 }} = 1.9052\;KV\\ \frac{{{Q_c}}}{{ph}} = \frac{{{{\left( {\frac{V}{{ph}}} \right)}^2}}}{{{X_c}}}\\ {X_c} = \frac{{{{\left( {1.9052 \times {{10}^3}} \right)}^2}}}{{54.7 \times {{10}^3}}} = 66.36{\rm{\Omega }}\\ C = \frac{1}{{2\pi f{X_C}}} = \frac{1}{{2\pi \times 50 \times 66.36}} = 47.96\;\mu F \end{array}$