Answer should be 589 Volt, any thoughts?
Thanks!!
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Voltage Drop (L-N) = I[RCos(theta) + X sin(theta)]
= 230 (.019x0.75 + .02x 0.661) [ sin (Theta)=0.661]
= 6.31 volts
Voltage Drop (L-L) = 1.732x6.31=10.93 volts
So Terminal voltage = 600-10.93=589.07 volts
(for formula ref - ieee 141)
Thanks!!
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Voltage Drop (L-N) = I[RCos(theta) + X sin(theta)]
= 230 (.019x0.75 + .02x 0.661) [ sin (Theta)=0.661]
= 6.31 volts
Voltage Drop (L-L) = 1.732x6.31=10.93 volts
So Terminal voltage = 600-10.93=589.07 volts
(for formula ref - ieee 141)