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 option is 20∠-60°.

Visual Analysis of the Circuit:
From the provided circuit diagram, we can observe the following:
1. A three-phase star-connected source with a neutral node labeled $n$ and phase voltages $E_A$, $E_B$, and $E_C$.
2. A three-phase star-connected load with a neutral node labeled $n^{\prime }$ and phase impedances $Z_A$, $Z_B$, and $Z_C$ connected to the phases respectively.

Step 1: Identify Phase Voltages
The supply is a balanced three-phase system with a line-to-line voltage of $100\sqrt{3}\text{V}$ and phase sequence ABC.
For a star connection, the relationship between the line voltage ($V_L$) and the phase voltage ($V_{\text{ph}}$) is:

$V_{\text{ph}}=\frac{V_L}{\sqrt{3}}=\frac{100\sqrt{3}}{\sqrt{3}}=100\text{V}$

Taking phase A as the reference phasor, the three phase voltages relative to the source neutral $n$ are defined as:

$E_A=100\angle 0°\text{V}$

$E_B=100\angle -120°\text{V}$

$E_C=100\angle 120°\text{V}$

Step 2: Neutral Displacement Voltage Condition
According to Millman's Theorem, the voltage difference between the load neutral $n^{\prime }$ and the source neutral $n$ is given by:

$V_{n^{\prime }n}=\frac{\frac{E_A}{Z_A}+\frac{E_B}{Z_B}+\frac{E_C}{Z_C}}{\frac{1}{Z_A}+\frac{1}{Z_B}+\frac{1}{Z_C}}$

We are given that the voltage difference across nodes $n$ and $n^{\prime }$ is zero ($V_{n^{\prime }n}=0$). This simplifies the numerator of Millman's formula to:

$\frac{E_A}{Z_A}+\frac{E_B}{Z_B}+\frac{E_C}{Z_C}=0$

Step 3: Calculating Phase Currents and Impedance
We are given the load parameters:

$Z_A=10\angle 0°\Omega$

$Z_B=20\angle 60°\Omega$

Substitute the known quantities to find the currents in phases A and B:

$\frac{E_A}{Z_A}=\frac{100\angle 0°}{10\angle 0°}=10\angle 0°=10+j0\text{A}$

$\frac{E_B}{Z_B}=\frac{100\angle -120°}{20\angle 60°}=5\angle -180°=-5+j0\text{A}$

Now, compute their sum:

$\frac{E_A}{Z_A}+\frac{E_B}{Z_B}=10-5=5\text{A}$

For the total sum to be zero, we must have:

$\frac{E_C}{Z_C}=-5=5\angle 180°\text{A}$

Finally, solve for $Z_C$:

$Z_C=\frac{E_C}{5\angle 180°}=\frac{100\angle 120°}{5\angle 180°}=20\angle (120°-180°)=20\angle -60°\Omega$