Question

The load connected between each line and the neutral of a 415 volts, 50hertz 3 phase Circuit consist of a capacitor of 31.8 micro ferad in series with a resistor of 100 ohms. CALCULATE the line current and the Total Power.

Answer 1

Resistance, $R=100\Omega$

Capacitance, $C=31.8\mu F$

Line Voltage, $V_L=415V$

Frequency of AC supply, $f=50Hz$

Impedance, $Z=\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=\sqrt{(100)^2+(100.10)^2}$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=141.50\space\Omega$

$V_{Ph}=\dfrac{V_L}{\sqrt{3}}=\dfrac{415}{\sqrt{3}}=239.60\space V$

Power factor, $cos\phi=\dfrac{R}{Z}=0.706$

$I_{Ph}=\dfrac{V_{Ph}}{Z_{Ph}}=1.69\space A$

Line Current, $I_{L}=I_{Ph}=1.69\space A$

Total Power absorbed , $P_{total}=3\times I_{Ph}^2\times R_{Ph}=3\times 286.72=860.16\space W$