Question

lf $A=Z_1+Z_2+Z_3,B=Z_1+Z_2\omega +Z_3{\omega }^2,C=Z_1+Z_2{\omega }^2+Z_3\omega$ then $Z_1,Z_2,Z_3$ in terms of $A,B,C$ are

• A. $Z_1=\frac{A+B+C}{3},Z_2=\frac{A+B{\omega }^2+C\omega }{3}$,
  $Z_3=\frac{A+B\omega +C{\omega }^2}{3}$
• B. $Z_1=\frac{A+B+C{\omega }^2}{3},Z_2=\frac{A+B\omega +C{\omega }^2}{3}$,
  $Z_3=\frac{A+B{\omega }^2+C\omega }{3}$
• C. $Z_1=\frac{A\omega +B+C{\omega }^2}{3},Z_2=\frac{A+B+C}{3}$,
  $Z_3=\frac{A{\omega }^2+B\omega +C\omega }{3}$
• D. $Z_1=\frac{A+B\omega +C{\omega }^2}{3},Z_2=\frac{A{\omega }^2+B\omega +C}{3}$,
  $Z_3=\frac{A{\omega }^2+B{\omega }^2+C\omega }{3}$

Answer 1

The correct option is A $Z_1=\frac{A+B+C}{3},Z_2=\frac{A+B{\omega }^2+C\omega }{3}$,

$Z_3=\frac{A+B\omega +C{\omega }^2}{3}$
$A=Z_1+Z_2+Z_3$
$B=Z_1+Z_2\omega +Z_3{\omega }^2$
$C=Z_1+Z_2{\omega }^2+Z_3\omega$
Clearly, $A+B+C=3Z_1+Z_2(1+\omega +{\omega }^2)+Z_3(1+\omega +{\omega }^2)$
$\Rightarrow Z_1=\frac{A+B+C}{3}\because (1+\omega +{\omega }^2=0)$
Now,
$B\omega =Z_1\omega +Z_2{\omega }^2+Z_3{\omega }^3=Z_1\omega +Z_2{\omega }^2+Z_3$
$C{\omega }^2=Z_1{\omega }^2+Z_2{\omega }^4+Z_3{\omega }^3=Z_1{\omega }^2+Z_2\omega +Z_3$
$So,A+B\omega +C{\omega }^2=Z_1(1+\omega +{\omega }^2)+Z_2(1+\omega +{\omega }^2)+3Z_3$
$\Rightarrow Z_3=\frac{A+B\omega +C{\omega }^2}{3}$
Similarly,
$B{\omega }^2=Z_1{\omega }^2+Z_2{\omega }^3+Z_3{\omega }^4=Z_1{\omega }^2+Z_2+Z_3\omega$
$C\omega =Z_1\omega +Z_2{\omega }^3+Z_3{\omega }^2=Z_1\omega +Z_2+Z_3{\omega }^2$
So, $A+B{\omega }^2+C\omega =Z_1(1+\omega +{\omega }_2)+3Z_2+Z_3(1+\omega +{\omega }^2)$
$\Rightarrow Z_2=\frac{A+B{\omega }^2+C\omega }{3}$