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Question

# lf A=Z1+Z2+Z3, B=Z1+Z2ω+Z3ω2,C=Z1+Z2ω2+Z3ω then Z1,Z2,Z3 in terms of A, B, C are

A
Z1=A+B+C3, Z2=A+Bω2+Cω3,
Z3=A+Bω+Cω23
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B
Z1=A+B+Cω23,Z2=A+Bω+Cω23,
Z3=A+Bω2+Cω3
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C
Z1=Aω+B+Cω23, Z2=A+B+C3,
Z3=Aω2+Bω+Cω3
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D
Z1=A+Bω+Cω23,Z2=Aω2+Bω+C3,
Z3=Aω2+Bω2+Cω3
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Solution

## The correct option is A Z1=A+B+C3, Z2=A+Bω2+Cω3,Z3=A+Bω+Cω23A=Z1+Z2+Z3B=Z1+Z2ω+Z3ω2C=Z1+Z2ω2+Z3ωClearly, A+B+C=3Z1+Z2(1+ω+ω2)+Z3(1+ω+ω2)⇒Z1=A+B+C3∵(1+ω+ω2=0)Now, Bω=Z1ω+Z2ω2+Z3ω3=Z1ω+Z2ω2+Z3Cω2=Z1ω2+Z2ω4+Z3ω3=Z1ω2+Z2ω+Z3So,A+Bω+Cω2=Z1(1+ω+ω2)+Z2(1+ω+ω2)+3Z3⇒Z3=A+Bω+Cω23Similarly, Bω2=Z1ω2+Z2ω3+Z3ω4=Z1ω2+Z2+Z3ωCω=Z1ω+Z2ω3+Z3ω2=Z1ω+Z2+Z3ω2So, A+Bω2+Cω=Z1(1+ω+ω2)+3Z2+Z3(1+ω+ω2)⇒Z2=A+Bω2+Cω3

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