The armature current of a synchronous motor has large value for-
Both low and high excitation
A synchronous motor is an AC motor that runs at a constant speed, called synchronous speed, determined by the frequency of the power supply and the number of poles in the motor. Unlike induction motors, synchronous motors require a DC excitation voltage applied to their rotor winding. This excitation significantly impacts the motor's performance, particularly its armature current and power factor.
For a synchronous motor operating at a constant mechanical load (output power), changing the DC excitation voltage applied to the rotor field winding affects the armature current drawn from the AC supply. This relationship is graphically represented by the "V-curve" of the synchronous motor.
The V-curve plots the armature current ($I_a$) on the y-axis against the field excitation current (or excitation voltage, representing the strength of the field) on the x-axis, for a constant motor power output.
Let's analyze the V-curve to understand how armature current changes with excitation:
Therefore, the armature current of a synchronous motor is minimum at normal excitation (unity power factor) and increases as the excitation is either decreased (low excitation, lagging PF) or increased (high excitation, leading PF). This means the armature current has a large value for both low and high excitation levels when compared to the minimum value at medium excitation.
Consider a simplified phasor diagram analysis for constant power ($P$) and terminal voltage ($V$). The power is given by $P = \frac{VE}{X_s} \sin \delta$, where $E$ is the excitation voltage (proportional to excitation), $X_s$ is the synchronous reactance, and $\delta$ is the power angle. Also, $V$, $E$, and the voltage drop $I_a X_s$ form a voltage triangle. To maintain constant power $P$ with constant $V$ and $X_s$, as $E$ changes, $\delta$ and $I_a$ must adjust. A larger $I_a$ is required when $E$ is either much smaller or much larger than the value corresponding to minimum $I_a$ (unity PF operation), causing the motor's power factor to deviate significantly from unity (lagging at low E, leading at high E).
Based on the characteristic V-curve of a synchronous motor operating at a constant load, the armature current is minimal at a specific "normal" or "medium" excitation level where the power factor is unity. Any deviation from this optimal excitation level, either towards low excitation (under-excited, lagging PF) or high excitation (over-excited, leading PF), results in an increase in the armature current. Thus, the armature current has a large value for both low and high excitation compared to the minimum value.
The options provided describe the conditions under which the armature current is large:
Based on our analysis of the V-curve, the armature current is large at both low and high excitation.
| Excitation Level | Motor Condition | Power Factor | Armature Current ($I_a$) |
|---|---|---|---|
| Low (Under-excited) | Acts like a lagging load | Lagging ($\cos \phi < 1$) | Large |
| Medium (Normal) | Minimum $I_a$ | Unity ($\cos \phi = 1$) | Minimum |
| High (Over-excited) | Acts like a leading load (capacitor) | Leading ($\cos \phi < 1$) | Large |
| Concept | Key Point |
|---|---|
| Synchronous Motor | Runs at synchronous speed, requires DC excitation. |
| Armature Current ($I_a$) | AC current drawn from the supply. |
| Excitation | DC current/voltage applied to the rotor field winding. |
| V-Curve | Graph showing $I_a$ vs. Field Excitation at constant load. |
| Minimum $I_a$ | Occurs at unity power factor (normal/medium excitation). |
| Large $I_a$ | Occurs at low excitation (lagging PF) and high excitation (leading PF). |
One of the key advantages of a synchronous motor is its ability to operate at different power factors simply by changing its DC excitation. This makes them valuable for power factor correction in industrial plants.
The magnitude of the armature current is directly related to the reactive power exchange with the grid for a given real power output. For a fixed real power output, a significant reactive power exchange (either consuming or supplying) requires a larger apparent power ($S$) and thus a larger armature current ($I_a$), since $S = V \times I_a$ and $S^2 = P^2 + Q^2$, where $P$ is real power and $Q$ is reactive power.
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