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

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. Eb = V) supplied with fixed excitation. Since field excitation and speed of the motor = synchronous speed are fixed, the value of Eb remains the same.

Let

V = supply voltage/phase

Eb = back e.m.f./phase

Er = Net voltage

Zs = 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. Eb, is equal and opposite to the supply voltage V as shown in the phasor diagram in Fig. • Since net voltage Er = V − Eb is zero, the armature current Ia = Er/Zs is zero. This is expected because there is neither load nor motor losses. Now, the motor just floats.

• If the motor is on no-load but it has losses, the torque angle increases by a small angle δ.
• Consequently, the vector Eb (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 Er in the stator causes the stator or armature current Ia = Er/Zs to have some finite value to supply the no-load losses.
• Note that armature current Ia lags behind Er by a small angle θ = tan−1 Xs/Ra. Since Xs >> Ra, Ia, lags Er by nearly 90° ( θ = tan−1 Xs/0 = 90°). The phase angle between V and Ia is φo so the no-load power factor is cosφo.

No-load input power/phase = VI, cosφo

Thus at no-load, the motor takes a small power VI, cosφo/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 Ia drawn by a synchronous motor lags the applied voltage V by a small angle.