A voltage source š£(š”) = 240 sin(314.16š” ā 20Ā°)š is connected to a load having an
impedance of Z. The resulting current through the load is š(š”) = 15 sin(314.16š” + 22.5Ā°)A.
Determine the circuit power factor and identify the elements that constitute the load, given
that Z comprises of only two elements connected in series.
i leads the v (voltage). This is possible for the capacitive loads.
Therefore given load is resistor and capacitor load.
"R=|Z|cos \\phi = 16 cos -42.5=-10.81 \\Omega"
"R=|Z|sin \\phi = 16 sin 42.5=11.81 \\Omega"
"Z=R+jX=(11.81-j10.81) \\Omega"
The power factor is "cos \\phi =cos 42.5 =0.74" leading
"X_c= \\frac{1}{wc} \\implies c= \\frac{1}{X_cw}=\\frac{1}{10.81*314.6}=0.00029404"
"C= 294.04*10^{-6} F"
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