Sure! Here's the solution to your questions:

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Q1. Two-wattmeter method produces P1 = 1560 W and P2 = 2100 W. Calculate:

• (a) Per-phase average power
  Total power (P) = P1 + P2 = 1560 + 2100 = 3660 W
  Per-phase power = $\frac{P}{3} = \frac{3660}{3} = 1220 \, \text{W}$

• (b) Per-phase reactive power
  Reactive power $Q = \sqrt{3} (P2 - P1) = \sqrt{3} (2100 - 1560) \approx 935.7 \, \text{VAR}$
  Per-phase reactive power = $\frac{Q}{3} = \frac{935.7}{3} \approx 311.9 \, \text{VAR}$

• (c) Power factor
  Power factor $\text{PF} = \frac{P}{\sqrt{P^2 + Q^2}} = \frac{3660}{\sqrt{3660^2 + 935.7^2}} \approx 0.97$

• (d) Phase impedance
  Phase voltage $V_\text{ph} = \frac{V_\text{L}}{\sqrt{3}} = \frac{220}{\sqrt{3}} \approx 127 \, \text{V}$
  Phase current $I_\text{ph} = \frac{P}{3 \cdot V_\text{ph} \cdot \text{PF}} = \frac{3660}{3 \cdot 127 \cdot 0.97} \approx 9.93 \, \text{A}$
  Phase impedance $Z_\text{ph} = \frac{V_\text{ph}}{I_\text{ph}} = \frac{127}{9.93} \approx 12.8 \, \Omega$

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Q2. Three-phase balanced load with Z = 8+j6 Ω, Line Voltage = 208 V. Calculate:

• (a) Wattmeter readings $W_1$ and $W_2$:
  Phase voltage $V_\text{ph} = \frac{V_\text{L}}{\sqrt{3}} = \frac{208}{\sqrt{3}} \approx 120 \, \text{V}$
  Phase current $I_\text{ph} = \frac{V_\text{ph}}{|Z|} = \frac{120}{\sqrt{8^2 + 6^2}} = \frac{120}{10} = 12 \, \text{A}$
  Power factor $\cos\theta = \frac{\text{Re}(Z)}{|Z|} = \frac{8}{10} = 0.8$

  Total power $P = 3 \cdot V_\text{ph} \cdot I_\text{ph} \cdot \text{PF} = 3 \cdot 120 \cdot 12 \cdot 0.8 = 3456 \, \text{W}$
  Reactive power $Q = \sqrt{3} \cdot V_\text{L} \cdot I_\text{ph} \cdot \sin\theta = \sqrt{3} \cdot 208 \cdot 12 \cdot \frac{6}{10} \approx 2598.1 \, \text{VAR}$

  Wattmeter readings:
  $W_1 = \frac{P}{2} + \frac{Q}{\sqrt{3}} \approx \frac{3456}{2} + \frac{2598.1}{\sqrt{3}} \approx 3026 \, \text{W}$
  $W_2 = \frac{P}{2} - \frac{Q}{\sqrt{3}} \approx \frac{3456}{2} - \frac{2598.1}{\sqrt{3}} \approx 430 \, \text{W}$

• (b) Total power (P) and reactive power (Q):
  $P = 3456 \, \text{W}, \, Q = 2598.1 \, \text{VAR}$

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Q3. Phasor diagram for 3-phase star-connected load

In a star connection:

• The phase voltages are $120^\circ$ apart.
• The line voltage is the vector sum of two phase voltages.
  A phasor diagram will have three phase voltages and line voltages represented by vectors.

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Q4. Relationship between line and phase quantities for star connection

• Line voltage $V_\text{L} = \sqrt{3} \cdot V_\text{ph}$
• Line current $I_\text{L} = I_\text{ph}$

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Q5. Advantages of 3-phase system over single-phase system

1. Higher efficiency for the same power rating.
2. Continuous power delivery (less fluctuation).
3. Smaller and lighter machines for the same power output.
4. Better power factor and reduced losses.
5. Easier and more economical transmission of electrical energy.

Let me know if you have further questions!