A 1-phase 60 kVA , 220 V , 50 Hz, alternator has an effective armature leakage reactance of 0.07 ohm and negligible armature resistance. Calculate the voltage induced in the armature when the alternator is delivering rated current at a load power factor of 0.7 lagging.
Question 6.16: A 1-phase 60 kVA , 220 V , 50 Hz, alternator has an effectiv...
Here, Rated power =60 kVA =60 \times 10^{3} VA
Terminal voltage, V=220 V
Leakage reactance, X_{L}=0.07 \Omega
\begin{array}{l}\text { Load power factor, } \cos \phi=0 \cdot 7 \operatorname{lag} ; \sin \phi=\sin \cos ^{-1} 0 \cdot 7=0 \cdot 7141\\\text { Full load current, } I=\frac{60 \times 10^{3}}{220}=272 \cdot 72 A\end{array}
\text { Voltage induced in the armature, } E_{0}=\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+I X_{L}\right)^{2}}
=\sqrt{(V \cos \phi)^{2}+\left(V \sin \phi+I X_{L}\right)^{2}}
\text { (since value of } R \text { is not given })
\begin{array}{l}=\sqrt{(220 \times 0 \cdot 7)^{2}+(220 \times 0 \cdot 7141+272 \cdot 72 \times 0 \cdot 07)^{2}} \\=\sqrt{(154)^{2}+(176 \cdot 19)^{2}}=234 V \end{array}
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