Correct Answer - Option 1 : the load torque angle remains the same

**Concept:**

The power in a synchronous motor is given by,

\(P = \frac{{EV}}{{{X_s}}}\sin \delta\)

\(P \propto \left( {\frac{{{V^2}}}{f}} \right)\sin \delta\)

\(T = \frac{P}{{{\omega _0}}};\;\;T \propto \left( {\frac{{{V^2}}}{{{f^2}}}} \right)\sin \delta \)

**Application:**

The terminal voltage and frequency are reduced by 10% i.e. V/f ratio is constant and the torque angle δ is constant.

Therefore, the load torque remains the same.