A 3-phase, 4-pole, 50Hz slip-ring induction motor is rotating at a speed of 1440 rpm at full-load. Its rotor resistance is 0.25 ohm per phase. What external resistance should be added in the rotor circuit to reduce the speed to 1320 rpm, the torque is to be kept the same.

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Accepted Answer

Here, [latex]R_{2} = 0.25 ohm ; f = 50 Hz ; P = 4 ; N_{1} = 1440 rpm ; N_{2} = 1320 rpm[/latex]

Torque developed, [latex] T = \frac{S E_{2S} R_{2}}{\left[R_{2}^{2} + \left(S X_{2S}\right)^{2} \right] } = \frac{S E_{2S} R_{2}}{R_{2}^{2}} \[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \left(Rejecting S X_{2S,} not given\right) [/latex]

[latex] T = \frac{K S }{R_{2}} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \left(Where K is a constant\right) [/latex]

In first case, [latex] T_{1} = \frac{K S_{1} }{R_{2}} [/latex] ...(i)

In second case, [latex] T_{2} = \frac{K S_{2} }{\left(R_{2} + r\right) } [/latex] ...(ii)

[latex] N_{S} = \frac{120 f}{P } = \frac{120 \times 50}{4} = 1500 rpm[/latex]

[latex] S_{1} = \frac{1500 - 1440 }{1500 } = 0.04 ; S_{2} = \frac{1500 - 1320}{1500} =0.12[/latex]

Since [latex]T_{1} = T_{2}[/latex] equating equation (i) and (ii), we get,

[latex] \frac{S_{1}}{R_{2}} = \frac{S_{2}}{R_{2} + r} or R_{2} + r= \frac{S_{2}}{S_{1}} \times R_{z}[/latex]

[latex]\therefore r = \frac{0.12}{0.04} \times 0.25 - 0.25 = \mathbf{0.5 ohm} [/latex]

Question 10.9: A 3-phase, 4-pole, 50Hz slip-ring induction motor is rotatin...

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