# In a synchronous motor, at no-load condition, and with normal excitation the armature current drawn by a synchronous motor is

### Right Answer is:

Lags the applied voltage V by a small angle

#### SOLUTION

**In a synchronous motor, at no-load condition, and with normal excitation, the armature current drawn by a synchronous motor lags the applied voltage V by a small angle.**

**Synchronous Motor**** Phasor Diagrams**

Consider a normally-excited synchronous motor (i.e. E_{b} = V) supplied with fixed excitation. Since field excitation and speed of the motor = synchronous speed are fixed, the value of E_{b} remains the same.

Let

V = supply voltage/phase

E_{b} = back e.m.f./phase

E_{r} = Net voltage

Z_{s} = synchronous impedance/phase

**(i) Motor on no-load with no losses:-**

- When the synchronous motor is on no-load and has no losses, the center lines of rotor poles and stator poles coincide and the torque angle δ is practically 0°.
- Under this condition, the back e.m.f. E
_{b}, is equal and opposite to the supply voltage V as shown in the phasor diagram in Fig.

- Since net voltage
**E**is zero, the armature current I_{r}= V − E_{b }_{a}= E_{r}/Z_{s}is zero. This is expected because there is neither load nor motor losses. Now, the motor just floats.

**Motor on no-load with losses**

- If the motor is on no-load but it has losses, the torque angle increases by a small angle δ.
- Consequently, the vector E
_{b}(its magnitude is constant as excitation is fixed) falls back (vectors are rotating in an anticlockwise direction) by angle δ as shown in Fig.

- The resultant or net voltage E
_{r}in the stator causes the stator or armature current I_{a}= E_{r}/Z_{s }to have some finite value to supply the no-load losses. - Note that armature current Ia lags behind E
_{r }by a small angle θ = tan^{−1}X_{s}/R_{a}. Since X_{s}>> R_{a}, I_{a}, lags E_{r}by nearly 90° (**∵**θ = tan^{−1}X_{s}/0 = 90°). The phase angle between V and I_{a}is φ_{o}so the no-load power factor is cosφ_{o}.

** No-load input power/phase = VI, cosφ _{o}, **

Thus at no-load, the motor takes a small power VI, cosφ_{o}/phase from the supply to meet the no-load losses while it continues to run at synchronous speed.

**Hence Under no-load conditions with normal excitation, armature current I _{a} drawn by a synchronous motor lags the applied voltage V by a small angle.**

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