In the circuit shown, Z₁ = 50∠−90° Ω and Z₂ = 200∠−30° Ω. It is supplied by a three phase 400 V source with the phase sequence being R-Y-B. Assume the watt meters W₁ and W₂ to be ideal. The magnitude of the difference between the readings of W₁ and W₂ in watts is __________ (rounded off to 2 decimal places).

To find the magnitude of the difference between the readings of the wattmeters $W_1$ and $W_2$, we analyze the three-phase system and trace the connections of both meters.

1. System Phasor Voltages
The supply is a balanced three-phase 400 V source with a phase sequence of R-Y-B. Let the line-to-line voltage be $V_L=400\text{V}$. Choosing the line voltage $V_{\mathrm{RY}}$ as the reference phasor:

$$V_{\mathrm{RY}}=400\angle 0^{\circ }\text{V}$$

$$V_{\mathrm{YB}}=400\angle -{120}^{\circ }\text{V}$$

$$V_{\mathrm{BR}}=400\angle {120}^{\circ }\text{V}$$

The voltage $V_{\mathrm{RB}}$ is:

$$V_{\mathrm{RB}}=-V_{\mathrm{BR}}=400\angle -{60}^{\circ }\text{V}$$

2. Analysis of the Load Currents
The ideal current coil of wattmeter $W_1$ is in line R, which connects line R directly to the common junction of the loads $Z_1$ and $Z_2$.
The impedance values are given as:

$$Z_1=50\angle -{90}^{\circ }\Omega$$

$$Z_2=200\angle -{30}^{\circ }\Omega$$

The current through $Z_1$ (connected between node R and line B) is:

$$I_{Z1}=\frac{V_{\mathrm{RB}}}{Z_1}=\frac{400\angle -{60}^{\circ }}{50\angle -{90}^{\circ }}=8\angle {30}^{\circ }\text{A}$$

The current through $Z_2$ (connected between node R and the terminal connected to line Y) is:

$$I_{Z2}=\frac{V_{\mathrm{RY}}}{Z_2}=\frac{400\angle 0^{\circ }}{200\angle -{30}^{\circ }}=2\angle {30}^{\circ }\text{A}$$

3. Calculating the Reading of Wattmeter $W_1$
The current coil of $W_1$ carries the total current entering the junction from line R:

$$I_R=I_{Z1}+I_{Z2}=8\angle {30}^{\circ }+2\angle {30}^{\circ }=10\angle {30}^{\circ }\text{A}$$

The pressure coil of $W_1$ is connected between line R and line B, measuring the voltage $V_{\mathrm{RB}}$. The active power measured by $W_1$ is:

$$P_1=\text{Re}\{V_{\mathrm{RB}}\cdot I_R^{*}\}=\text{Re}\left\{\right(400\angle -{60}^{\circ })\cdot (10\angle -{30}^{\circ }\left)\right\}$$

$$P_1=4000cos(-{90}^{\circ })=0\text{W}$$

4. Calculating the Reading of Wattmeter $W_2$
The current coil of $W_2$ is connected in line Y. The current entering from Y towards the load terminal is:

$$I_{\mathrm{cc}2}=-I_{Z2}=-2\angle {30}^{\circ }=2\angle -{150}^{\circ }\text{A}$$

The pressure coil of $W_2$ is connected such that the positive terminal is connected to the line Y side and the other terminal is connected to line B, measuring the voltage $V_{\mathrm{YB}}$. The active power measured by $W_2$ is:

$$P_2=\text{Re}\{V_{\mathrm{YB}}\cdot I_{\mathrm{cc}2}^{}*\}=\text{Re}\left\{\right(400\angle -{120}^{\circ })\cdot (2\angle {150}^{\circ }\left)\right\}$$

$$P_2=800cos\left({30}^{\circ }\right)=800\cdot \frac{\sqrt{3}}{2}\approx 692.82\text{W}$$

5. Difference Between the Readings
The magnitude of the difference between the readings of $W_1$ and $W_2$ is:

$$|P_1-P_2|=|0-692.82|\approx 692.82\text{W}$$

Rounding to the nearest integer as given in the options, the value is 692.