Page 463 - Basic Electrical Engineering

W = V I cos (30 − ϕ)
1
L L
= 17.3 cos (30 − 60°)

= 17.3 × 0.866 = 15 kW
W = V I cos (30 + ϕ)
2
L L
= 17.3 cos 90°

= 17.3 × 0

= 0

Total power = W + W = 15 + 0 = 15 kW
1
2

Thus, at a load power factor of 0.5, one of the wattmeters has given zero
reading. This has been explained earlier under the effect of change of power

factor on wattmeter readings.

4.9 REVIEW QUESTIONS

A. Short Answer Type Questions

1. What is the difference between a single-phase winding and a three-phase winding ?
2. Draw wave shapes of a three-phase supply.
3. What is the difference between a balanced load and an unbalanced load ?
4. Show that the phasor sum of the three-phase balanced voltages is zero.
5. What do you mean by phase sequence of three-phase voltages?
6. Distinguish between star connection and delta connection of three-phase windings.
7. Derive the relationship between line current, line voltage, phase current, and phase voltage in
case of star and delta connection of three-phase windings.
8. Prove that the power in a three-phase circuit is equal to V I cos ϕ.
L L
9. Distinguish between active power and reactive power in a three-phase system.
10. What is the significance of low power factor of any load on the system?
11. Draw the circuit diagram for measurement of three-phase power with two single-phase
wattmeters.
12. At what value of load power factor the reading of one of the wattmeters, in the two-wattmeter
method of measurement of three-phase power, will be zero?
13. What are the advantages of the three-phase system over the single-phase system?
14. Write the relationship between phase voltage and current in a delta-connected load.
15. Draw the connection diagram for three-phase resistive–inductive balanced load connected
across a three-phase supply. Also draw the phasor diagram showing voltages and currents.