Question

In a Capacitor-start motor, main winding current is $2.0\text{ pu}$, auxiliary winding current is $2.0\text{ pu}$ and the phase difference is $90^\circ$. Then the starting torque is:

• A. $2.0\text{ pu}$
• B. $4.0\text{ pu}$
• C. $6.0\text{ pu}$
• D. $8.0\text{ pu}$

Hint:

To maximize the starting torque of a single-phase induction motor, the phase angle difference \(\alpha\) should be designed to be as close to \(90^\circ\) as possible, since \(\sin(90^\circ) = 1\) is the maximum value of the sine function.
This is why capacitor-start motors have much higher starting torques compared to split-phase motors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the starting torque of a capacitor-start single-phase induction motor, given the main winding current, auxiliary winding current, and the phase difference between them.

Step 2: Key Formula or Approach:
The starting torque (\(T_{\text{start}}\)) of a split-phase or capacitor-start single-phase induction motor is directly proportional to the product of the currents in the main and auxiliary windings and the sine of the phase angle difference between them:
\[ T_{\text{start}} \propto I_m \cdot I_a \cdot \sin\alpha \] where:
\(I_m\) is the main winding current,
\(I_a\) is the auxiliary winding current, and
\(\alpha\) is the phase difference between the two currents.
In per-unit system representation, we can write the starting torque as:
\[ T_{\text{start}} = I_m \cdot I_a \cdot \sin\alpha \]

Step 3: Detailed Explanation:

• We are given the following values:
- Main winding current, \(I_m = 2.0\text{ pu}\).
- Auxiliary winding current, \(I_a = 2.0\text{ pu}\).
- Phase angle difference, \(\alpha = 90^\circ\).

• Calculate the sine of the phase angle difference:
\[ \sin(90^\circ) = 1.0 \]
• Substitute these values into the starting torque equation:
\[ T_{\text{start}} = 2.0 \times 2.0 \times \sin(90^\circ) \] \[ T_{\text{start}} = 4.0 \times 1.0 = 4.0\text{ pu} \]
• Thus, the starting torque of the motor is \(4.0\text{ pu}\).

Step 4: Final Answer:
The starting torque is 4.0 pu.