A three-phase, star-connected, 10 kVA , 230 V alternator has an armature resistance of 0.5 \Omega per phase and a synchronous reactance of 1.2 \Omega per phase. Calculate the percent voltage regulation at full load at power factors of (a) 0.8 lagging, (b) 0.8 leading, (c) Determine the power factor such that the voltage regulation becomes zero on full load.
\begin{array}{l}\text { Considering } V \text { as reference phasor, } \quad \bar{V}=V \angle 0^{\circ}=132.8 \angle 0^{\circ}=132.8+j 0\\\text { Synchronous impedance, } \bar{Z}_{s}=R+j X_{S}=0.5+j 1.2=1.3 \angle 67.38^{\prime \prime} \Omega\end{array}
\text { (a) When power factor is } 0.8 \text { lagging }
\begin{aligned}\bar{I} &=I \angle-\cos ^{-1} 0.8=25.1 \angle-36.87^{\prime \prime} A \\E_{0} &=\bar{V}+\bar{I} \bar{Z}_{s} \\&=(132.8+j 0)+\left(25.1 \angle-36.87^{\prime \prime}\right)\left(1.3 \angle 67.38^{\prime\prime}\right) \\&=132.8+32.63 \angle 30.51^{\prime \prime}=132.8+28.1+j 1.6 .56 \\&=160.9+j 16.56=161.75 \angle 5.87^{\prime \prime}\end{aligned}
\begin{aligned}\text { Voltage regulation } &=\frac{E_{0}-V}{V} \times 100=\frac{161.75-132.8}{132.8} \times 100 \\&=21.8 \%\end{aligned}
\text { (b) When power factor is } 0.8 \text { leading }
\begin{aligned}\bar{I} &=I \angle+\cos ^{-1} 0.8=25.1 \angle 36.87^{\prime \prime} A \\E_{0} &=\bar{V}+\bar{I} \overline{Z_{s}} \\&=132.8+\left(25.1 \angle 36.87^{\prime \prime}\right)\left(1.3 \angle 67.38^{\prime \prime}\right) \\&=132.8+32.63 \angle 104.25^{\prime \prime} \\&=132.8-8+j 31.62=124.8+j 31.62 \\&=128.74 \angle 14.2^{\prime \prime} V\end{aligned}
\begin{aligned}\text { Voltage regulation } &=\frac{E_{0}-V}{V} \times 100=\frac{128.74-132.8}{132.8} \times 100 \\&=-3.06 \%\end{aligned}
\text { (c) For zero regulation, let } \phi \text { be the required power-factor angle. }
\begin{aligned}\therefore \quad \bar{I} &=I \angle \phi=25.1 \angle \phi A \\E_{0} &=\bar{V}+\bar{I} \overline{Z_{s}} \\&=132.8+(25.1 \angle \phi)\left(1.3 \angle 67.38^{\circ}\right) \\&=132.8+32.63 \angle\left(\phi+67.38^{\circ}\right) \\&=132.8+32.63 \cos \left(\phi+67.38^{\circ}\right)+j 32.63 \sin \left(\phi+67.38^{\circ}\right)\end{aligned}
\begin{aligned}E_{0}^{2} &=[132.8+32.63 \cos (\phi+67.38)]^{2}+[32.63 \sin (\phi+67.38)]^{2} \\\text { Voltage regulation } &=\frac{E_{0}-V}{V} pu\end{aligned}
\text { For zero voltage regulation } E_{0}=V=132.8 V
\therefore \quad(132.8)^{2}=\left[132.8+32.63 \cos \left(\phi+67.38^{\circ}\right)\right]^{2}+[32.63 \sin (\phi+67.38)]^{2}
\begin{array}{l}\text { or }\\(132.8)^{2}=(132.8)^{2}+2 \times 132.8 \times 32.63 \cos \left(\phi+67.38^{\circ}\right)\\\begin{array}{r}+(32.63)^{2} \cos ^{2}\left(\phi+67.38^{\circ}\right)+(32.63)^{2} \sin^{2}\left(\phi+67.38^{\circ}\right) \\(132.8)^{2}=(132.8)^{2}+2 \times 132.8 \times 32.63 \cos\left(\phi+67.38^{\circ}\right)+(32.63)^{2}\end{array}\end{array}
\begin{array}{l}\text { or}\\\cos \left(\phi+67.38^{\circ}\right)=\frac{-32.63}{2 \times 132.8}=-0.12285=\cos 97^{\prime \prime}\\\therefore \phi =97^{\circ}-67.38^{\circ}=+29.62\\ \text { and } \quad \cos \phi=0.8692 \text { leading }\end{array}