The rotor of a 4-pole, 50 Hz, slipring induction motor has resistance of 0.25 ohm per phase and runs at 1440 rpm at full load. Calculate the external resistance per phase which must be added to lower the speed to 1200 rpm, torque remaining constant.

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Accepted Answer

No. of poles, [latex]P = 4[/latex]

Supply frequency, [latex]f = 50 Hz[/latex]

Rotor resistance, [latex]R_{2} = 0.25 ohm[/latex]

Full load speed, [latex] N = 1440 rpm [/latex]

Controlled speed, [latex] N^{\prime } = 1200 rpm[/latex]

Synchronous speed, [latex] N_{s} = \frac{120 f}{P} = \frac{120 imes 50}{4} = 1500 rpm[/latex]

Full load slip, [latex] S = \frac{N_{s} - N}{N_{s}} = \frac{1500 - 1400}{1500} = 0.04[/latex]

Let the external resistance per phase added in the rotor circuit to control the speed be r.

[latex]Then total rotor circuit resistance = R_{2} + r[/latex]

Slip at 1200 rpm [latex] S_{2} = \frac{N_{s} - N_{2}}{N_{s}} = \frac{1500 - 1200}{1500} = 0.2[/latex]

For constant torque, [latex] ratio of \frac{R_{2}}{S_{1}} = constant = \frac{R_{2} + r}{S_{2}} [/latex]

[latex] herefore R_{2} + r = \frac{R_{2}}{S} imes S_{2} = \frac{0.25}{0.04} imes 0.2 = 1.25 ohm[/latex]

[latex] herefore r = 1.25 - R_{2} = 1.25 - 0.25 = \mathbf{1 ohm} [/latex]

Question 10.11: The rotor of a 4-pole, 50 Hz, slipring induction motor has r...

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