Windings ‘A’, ‘B’ and ‘C’ have 20 turns each and are wound on the same iron core as shown, along with winding "X' which has 2 turns. The figure shows the sense (clockwise/anti-clockwise) of each of the windings only and does not reflect the exact number of turns. If windings 'A', 'B' and 'C' are supplied with balanced 3-pbase voltages at 50 Hz and there is no core saturation, the no-load RMS voltage (in V, rounded off to 2 decimal places) across winding "X" is ________.

The correct answer is 46 (or 46 V).

1. Understanding the Given Data from the Image:

From the provided diagram:

- Windings A, B, and C each have number of turns $N_A=N_B=N_C=20$.

- Winding X has number of turns $N_X=2$.

- The three-phase balanced AC voltages connected to the windings are visible in the schematic:

$V_A=230\mathrm{\∠}{0}^{\ˆ}\text{\ V}$

$V_B=230\mathrm{\∠}-{120}^{\ˆ}\text{\ V}$

$V_C=230\mathrm{\∠}{120}^{\ˆ}\text{\ V}$

2. Analyzing the Winding Senses and Flux Contributions:

Let us examine the direction of the magnetic flux produced by each winding in the core core limb:

- Looking at winding A: Current entering from the positive terminal wraps over the limb towards the back, creating a magnetic flux in the clockwise direction around the core loop.

- Looking at winding B: The winding is wound in the opposite direction (anti-clockwise sense compared to A), so current entering from its positive terminal produces flux in the anti-clockwise direction.

- Looking at winding C: The connections/winding sense result in flux pointing in the clockwise direction along the main loop.

Thus, the net magnetomotive force (MMF) or net voltage effective across the combined 20-turn primary excitation is:

$V_{\text{net}}=V_A-V_B+V_C$

3. Calculating Net Voltage ($V_{\text{net}}$):

Recall the balanced three-phase identity: $V_A+V_B+V_C=0$, which implies $V_A+V_C=-V_B$.

Substituting this into the expression for net voltage:

$V_{\text{net}}=(V_A+V_C)-V_B=-V_B-V_B=-2V_B$

Taking the magnitude (RMS value):

$\left|V_{\text{net}}\right|=2\×\left|V_B\right|=2\×230=460\text{\ V}$

4. Calculating RMS Voltage across Winding X ($V_X$):

Since the net voltage of 460 V is induced/applied across equivalent 20-turn primary windings, the turns ratio between the total effective primary turns ($N=20$) and winding X ($N_X=2$) gives:

$V_X=V_{\text{net}}\×\frac{N_X}{N}$

$V_X=460\×\frac{2}{20}=460\×0.1=46\text{\ V}$

Thus, the no-load RMS voltage across winding "X" is 46 V.