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Three impedances zL=3+j4 ohms, zC=4-j4 ohms and zR=0+j3 ohms are connected in parallel. solve for the power factor of the combination.
Answers
Answered by
Steve
These two articles should get you started, if you haven't yet read your course materials ...
https://www.electronics-tutorials.ws/accircuits/parallel-circuit.html
https://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/calculating-power-factor/
https://www.electronics-tutorials.ws/accircuits/parallel-circuit.html
https://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/calculating-power-factor/
Answered by
Henry
Given:
zL = 3+j4 Ohms.
zC = 4-j4 Ohms.
zR = 3+j0? Ohms.
(3+j4)(4-j4)/(3+j4+4-j4) = (12-j12+j16+16)/7 = (28+j4)/7=4+j4/7 =
zL in parallel with zC.
Z = 3(4+j4/7)/(3+4+j4/7) = (12+j12/7/(7+j4/7) = 12+j12/(7+j4) = 17[45o]/8.06[29.7 = 2.1Ohms[15.3o]
Power Factor = Cos15.3 = 0.96.
zL = 3+j4 Ohms.
zC = 4-j4 Ohms.
zR = 3+j0? Ohms.
(3+j4)(4-j4)/(3+j4+4-j4) = (12-j12+j16+16)/7 = (28+j4)/7=4+j4/7 =
zL in parallel with zC.
Z = 3(4+j4/7)/(3+4+j4/7) = (12+j12/7/(7+j4/7) = 12+j12/(7+j4) = 17[45o]/8.06[29.7 = 2.1Ohms[15.3o]
Power Factor = Cos15.3 = 0.96.
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