Electrical Engineering

Engineering
Tutor: None Selected Time limit: 1 Day

Apr 13th, 2015

Given:

S1=7KW+3KVAR =7000+3000j VA = 7+3j (KVA)

S2=4KW at 0.85 power factor =4000[cos(theta)+jsin(theta)]

=4000*[0.85+jsin(cos-1(0.85))]

=3400 + j2107 VA =3.4+2.107j  (KVA)

Since loads are in parallel, so total complex power would be:

S=S1+S2

=(7+3j) +( 3.4 + 2.107 j)  (KVA)

=10.4 + 5.107 j (KVA)

=10400+5107 j   (VA)

=11586 (with angle of /_ 26.15 degree )     (VA)   Ans

Please message me if you need more help regarding solution. Thanks

Apr 13th, 2015

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