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).

 

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

Correct Answer: 692 - 693

Step 1: Apply the two-wattmeter method

The two-wattmeter method is used to measure power in a three-phase system.

Given line voltages:

\[ V_{RY} = 400 \angle 0^\circ,\quad V_{YB} = 400 \angle -120^\circ,\quad V_{BR} = 400 \angle -240^\circ \]

The phase sequence is \( R \rightarrow Y \rightarrow B \).

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Step 2: Calculate line currents

From the circuit:

\[ I_1 = \frac{V_{RY}}{Z_1} = \frac{400 \angle -60^\circ}{50 \angle -90^\circ} = 8 \angle 30^\circ \text{ A} \]

\[ I_2 = \frac{V_{RY}}{Z_2} = \frac{400 \angle 0^\circ}{200 \angle -30^\circ} = 2 \angle 30^\circ \text{ A} \]

Total line current:

\[ I_L = I_1 + I_2 = 10 \angle 30^\circ \text{ A} \]

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Step 3: Calculate wattmeter readings

Wattmeter \(W_1\):

\[ W_1 = V_{RY} I_L \cos(\angle V_{RY} - \angle I_L) \]

\[ W_1 = 400 \times 10 \times \cos(-60^\circ - 30^\circ) \]

\[ W_1 = 4000 \cos(90^\circ) = 0 \text{ W} \]

Wattmeter \(W_2\):

\[ W_2 = V_{YB} I_2 \cos(\angle V_{YB} - \angle I_2) \]

\[ W_2 = 400 \times 2 \times \cos(-120^\circ - 30^\circ) \]

\[ W_2 = 800 \cos(-150^\circ) \]

\[ W_2 = -800 \cos(30^\circ) = -800 \times \frac{\sqrt{3}}{2} = -692.82 \text{ W} \]

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Step 4: Difference between wattmeter readings

\[ |W_1 - W_2| = |0 - (-692.82)| = 692.82 \text{ W} \]

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Final Answer:

|W_1 - W_2| \approx 693 \text{ W}