Question

In the circuit shown, \( Z_1 = 50 \angle -90^\circ \, \Omega \) and \( Z_2 = 200 \angle -30^\circ \, \Omega \). It is supplied by a three-phase 400 V source with the phase sequence being R-Y-B. Assume the wattmeters \( W_1 \) and \( W_2 \) to be ideal. The magnitude of the difference between the readings of \( W_1 \) and \( W_2 \) in watts is \_\_\_\_\_\_ (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{50.png}

Hint:

In the two-wattmeter method, the total power is \( P = W_1 + W_2 \), and the power difference is \( \Delta W = |W_1 - W_2| \).

Answer 1

Step 1: Apply the two-wattmeter method. The two-wattmeter method is used to measure power in a three-phase system. Given: \[ V_{RY} = 400 \angle 0^\circ, \quad V_{YB} = 400 \angle -120^\circ, \quad V_{BR} = 400 \angle -240^\circ. \] According to the given phase sequence, we construct the phasor diagram (\( R \to Y \to B \) phase sequence): \[ V_{BR} = V_{L} \angle -240^\circ. \] \begin{center} \begin{tikzpicture} % Draw the origin \coordinate (O) at (0,0); % Draw vectors \draw[->, thick] (O) -- (3,0) node[midway, above] {$V_{RY} = V_L \angle 0^\circ$}; \draw[->, thick] (O) -- (-1.5,-2.6) node[midway, left] {$V_{YB} = V_L \angle -120^\circ$}; \draw[->, thick] (O) -- (-1.5,2.6) node[midway, left] {$V_{BR} = V_L \angle -240^\circ$}; % Add angles \node at (1,-0.5) {$120^\circ$}; \node at (-1.2,-1.2) {$120^\circ$}; \end{tikzpicture} \end{center} Step 1: Calculate voltages. \[ V_{BR} = V_{BN} - V_{RN} = V_{BN} - V_{RY}. \] From the circuit diagram: \[ I_1 = \frac{V_{RY}}{Z_1} = \frac{400 \angle -60^\circ}{50 \angle -90^\circ} = 8 \angle 30^\circ \, \text{Amp}, \] \[ I_2 = \frac{V_{RY}}{Z_2} = \frac{400 \angle 0^\circ}{200 \angle -30^\circ} = 2 \angle 30^\circ \, \text{Amp}. \] The total current is: \[ I_L = I_1 + I_2 = 8 \angle 30^\circ + 2 \angle 30^\circ = 10 \angle 30^\circ \, \text{Amp}. \] Step 2: Calculate wattmeter readings. For Wattmeter \( W_1 \): \[ W_1 = V_{RY} I_L \cos(\angle V_{RY} - \angle I_L). \] Substitute values: \[ W_1 = 400 \times 10 \times \cos(-60^\circ - 30^\circ) = 400 \times 10 \times \cos(90^\circ). \] Simplify: \[ W_1 = 400 \times 10 \times 0 = 0 \, \text{Watt}. \] For Wattmeter \( W_2 \): \[ W_2 = V_{YB} I_2 \cos(\angle V_{YB} - \angle I_2). \] Substitute values: \[ W_2 = 400 \times 2 \times \cos(-120^\circ - 30^\circ) = 400 \times 2 \times \cos(-150^\circ). \] Simplify: \[ W_2 = 800 \cos(150^\circ) = -800 \cos(30^\circ). \] \[ W_2 = -800 \times \frac{\sqrt{3}}{2} = -692.82 \, \text{Watt}. \] Step 3: Difference between the wattmeter readings. The magnitude of the difference is: \[ |W_1 - W_2| = |0 - (-692.82)| = 692.82 \, \text{Watts}. \] Final Answer: \[ |W_1 - W_2| = 692.82 \, \text{Watts}. \] Step 2: Perform calculations. The readings of \( W_1 \) and \( W_2 \) are calculated, and the difference is found to be: \[ \Delta W = 692 \, \text{W} \, \text{to} \, 693 \, \text{W} \]