Question

The phasor diagram of a synchronous machine connected to an infinite bus is shown in the figure below. The machine is acting as, 

• A. generator operating at leading p.f.
• B. generator operating at lagging p.f.
• C. motor operating at leading p.f.
• D. motor operating at lagging p.f.

Hint:

For synchronous machines connected to an infinite bus, remember: {E leading V indicates generator operation}, and {over-excitation corresponds to lagging power factor}.

Answer 1

The Correct Option is B

Step 1: Observe the given phasor diagram.
In the phasor diagram, the terminal voltage \(V\) is taken as the reference and is shown along the horizontal axis. The induced emf \(E\) is shown leading the terminal voltage \(V\) by an angle.
Step 2: Identify the operating mode (generator or motor).
For a synchronous machine connected to an infinite bus:
- If the induced emf \(E\) leads the terminal voltage \(V\), the machine is operating as a synchronous generator.
- If the induced emf \(E\) lags behind the terminal voltage \(V\), the machine operates as a synchronous motor.
Since \(E\) is leading \(V\) in the given diagram, the machine is acting as a generator.
Step 3: Determine the power factor nature.
In a synchronous generator:
- If the generator is over-excited, it supplies reactive power to the system and operates at a lagging power factor.
- If the generator is under-excited, it absorbs reactive power and operates at a leading power factor.
The given phasor diagram shows \(E\) significantly leading \(V\), indicating an over-excited condition.
Step 4: Conclusion.
Hence, the synchronous machine is operating as a generator at lagging power factor.

Answer 2

For a synchronous machine tied to an infinite bus, whether it behaves as a generator or a motor, and whether it runs at a leading or lagging power factor, can be read directly off the relative position of the induced emf \(E\) and the terminal voltage \(V\) in the phasor diagram, together with the excitation level. In the given diagram \(E\) is drawn leading \(V\), which is the defining signature of generator action, and \(E\) leads by a large enough angle to indicate over-excitation.

1. Generator operating at leading p.f.: A generator runs at a leading power factor only when under-excited, in which case it absorbs reactive power rather than supplying it. The diagram shows a large lead of \(E\) over \(V\), which is the over-excited condition, not the under-excited one, so this option does not fit.
2. Generator operating at lagging p.f.: Since \(E\) leads \(V\), the machine is generating; and since the lead angle is large (indicating the field current is boosted beyond the level needed for zero reactive power exchange), the machine is over-excited and therefore supplies reactive power to the bus, which corresponds to a lagging power factor as seen from the bus.
3. Motor operating at leading p.f.: A motor would show \(E\) lagging \(V\) instead of leading it, since a motor draws power rather than delivers it. The diagram does not show this, so this option is ruled out.
4. Motor operating at lagging p.f.: This too requires \(E\) to lag \(V\), which again contradicts what the phasor diagram shows.

Reading the diagram this way confirms the machine is delivering power (generator action, \(E\) leading \(V\)) while over-excited (large lead angle), which together mean it is supplying reactive power at a lagging power factor.

Therefore, the correct answer is generator operating at lagging p.f.