Independent power producers (IPP’s)

• So VIPP @ 1.0099

• Then IR = 10(VIPP - 1) @ 10(1.0099 - 1) = 0.099 pu

• So that VD x IR = 1.0 x 0.099 = 0.099 pu = PR

• \ 99% of the IPP’s power makes it to the load.

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