Question

The value of ${}^{\prime }{Z}^{\prime }$ such that no current flows in branch $Y$ will be

• A. $(3.46+j2)\Omega$
• B. $(13.85+j8)\Omega$
• C. $(6.93+j4)\Omega$
• D. $(27.7+j16)\Omega$

Answer 1

The correct option is C $(6.93+j4)\Omega$



Let, $V_{RY}=V\angle 0^{\circ }V$
$V_{YB}=V\angle -{120}^{\circ }V$
$V_{BR}=V\angle {120}^{\circ }$
Let current through capacitor be $I_C$
$I_C=\frac{V\angle -{120}^{\circ }}{X_c}=\frac{V\angle -{120}^{\circ }}{-j8}$
$\therefore I_C=\frac{V}{8}\angle -{120}^{\circ }+{90}^{\circ }$
$I_C=\frac{V}{8}\angle -{30}^{\circ }A$
Let current though phase $Y$ be $I_Y$
$\therefore I_Y=I_C-I_L$
as $I_Y=0$
$I_C=I_L$
$\begin{array}{cc}\therefore \frac{V}{8}\angle -{30}^{\circ }& =\frac{V\angle 0^{\circ }}{R+jX}\\ \therefore R+jX=8\angle {30}^{\circ }& =8\angle {30}^{\circ }\Omega \\ Z& =(6.93+j4)\Omega \end{array}$