In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced threephase supply of 100√3 V with phase sequence ABC.

The star connected load has Z_A = 10Ω and Z_B =20∠60°Ω. The value of Z_C in Ω , for which the voltage difference across the nodes n and n' is zero, is

The correct answer is 20∠-60°.

To find the value of $Z_C$ for which the voltage difference between the neutral node of the supply ($n$) and the neutral node of the load ($n^{\prime }$) is zero, we can apply Millman's Theorem to the three-phase star-star circuit shown below:

From the diagram, the balanced three-phase supply has phase voltages $E_A$, $E_B$, and $E_C$ connected at supply neutral node $n$. The star-connected load has impedances $Z_A$, $Z_B$, and $Z_C$ meeting at load neutral node $n^{\prime }$.

Since the supply is balanced with a phase sequence of ABC, the phase voltages (taking phase A as the reference) are represented as:
$E_A=E_p\angle 0^{\circ }$
$E_B=E_p\angle -{120}^{\circ }$
$E_C=E_p\angle {120}^{\circ }$

The voltage difference across the nodes $n$ and $n^{\prime }$, denoted as $V_{n^{\prime }n}$, is given by Millman's Theorem:
$V_{n^{\prime }n}=\frac{E_A{Y}_A+E_B{Y}_B+E_C{Y}_C}{Y_A+Y_B+Y_C}$
where the branch admittances are:
$Y_A=\frac{1}{Z_A}$, $Y_B=\frac{1}{Z_B}$, and $Y_C=\frac{1}{Z_C}$.

For the voltage difference to be zero ($V_{n^{\prime }n}=0$), the numerator of the expression must equal zero:
$E_A{Y}_A+E_B{Y}_B+E_C{Y}_C=0$

Given values:
$Z_A=10\angle 0^{\circ }\Omega \Rightarrow Y_A=\frac{1}{10}\angle 0^{\circ }=0.1\angle 0^{\circ }\text{S}$
$Z_B=20\angle {60}^{\circ }\Omega \Rightarrow Y_B=\frac{1}{20}\angle -{60}^{\circ }=0.05\angle -{60}^{\circ }\text{S}$

Substituting the phase voltages and admittances into the balance equation:
$(E_p\angle 0^{\circ })(0.1\angle 0^{\circ })+(E_p\angle -{120}^{\circ })(0.05\angle -{60}^{\circ })+(E_p\angle {120}^{\circ })Y_C=0$

Dividing the entire equation by the common factor $E_p$:
$0.1\angle 0^{\circ }+0.05\angle -{180}^{\circ }+(1\angle {120}^{\circ })Y_C=0$

Converting the terms to rectangular form:
$0.1\angle 0^{\circ }=0.1$
$0.05\angle -{180}^{\circ }=-0.05$
So, the sum of the first two terms is:
$0.1-0.05=0.05$

Substituting this back into the equation:
$0.05+(1\angle {120}^{\circ })Y_C=0$
$(1\angle {120}^{\circ })Y_C=-0.05=0.05\angle {180}^{\circ }$

Solving for the admittance $Y_C$:
$Y_C=\frac{0.05\angle {180}^{\circ }}{1\angle {120}^{\circ }}=0.05\angle ({180}^{\circ }-{120}^{\circ })=0.05\angle {60}^{\circ }\text{S}$

Finally, we calculate the load impedance $Z_C$:
$Z_C=\frac{1}{Y_C}=\frac{1}{0.05\angle {60}^{\circ }}=20\angle -{60}^{\circ }\Omega$