Question

A 3-phase star connected slip ring induction motor has the following parameters referred to the stator: \[ R_s = 3 \, \Omega, \, X_s = 2 \, \Omega, \, X_r' = 2 \, \Omega, \, R_r' = 2.5 \, \Omega \] The per phase stator to rotor effective turns ratio is 3:1. The rotor winding is also star connected. The magnetizing reactance and core loss of the motor can be neglected. To have maximum torque at starting, the value of the extra resistance in ohms (referred to the rotor side) to be connected in series with each phase of the rotor winding is \_\_\_\_\_\_ (rounded off to 2 decimal places).

Hint:

For maximum torque at starting, ensure that \( R_r' + R_{ext} = X_r' \), where \( R_{ext} \) is the external resistance added.

Answer 1

Step 1: Condition for maximum torque at starting. To achieve maximum torque at starting, the total rotor resistance must equal the rotor reactance: \[ R_r' + R_{ext} = X_r' \] Step 2: Calculate the extra resistance \( R_{ext} \). Substitute the given values: \[ \frac{R_{2}^{\prime}}{s} = \sqrt{(R_{1})^{2} + (x_{1} + x_{2})^{2}}
= \sqrt{9 + 16}
\frac{R_{2}^{\prime}}{s} = \sqrt{25} = 5
\frac{R_{2}^{\prime}}{s} = 5\] From equation (i): \[ \frac{R_2' + R_{\text{ext}}}{S} = 5 \] To have maximum torque at starting: \[ S = 1 \] Substitute \( S = 1 \): \[ R_2' + R_{\text{ext}} = 5 \] Given \( R_2' = 2.5 \, \Omega \), we calculate \( R_{\text{ext}} \): \[ R_{\text{ext}} = 5 - 2.5 = 2.5 \, \Omega \] Now, \( R_{\text{ext}} \) referred to the rotor side is: \[ R_{\text{ext (referred)}} = \left(\frac{1}{3}\right)^2 R_{\text{ext}} \] Substitute \( R_{\text{ext}} = 2.5 \, \Omega \): \[ R_{\text{ext (referred)}} = \frac{1}{9} \times 2.5 = 0.277 \, \Omega \] Final Answer: The external resistance referred to the rotor side is: \[ R_{\text{ext (referred)}} = 0.277 \, \Omega \]