Three-Phase Induction Motor Problems

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Three-Phase Induction Motor Problems Problem 1: Torque and Speed Calculations Given: $P = 746 \text{ kW}$, $3\phi$, $50 \text{ Hz}$, $16$-pole. Rotor resistance $R_2 = 0.02 \, \Omega$, reactance $X_2 = 0.15 \, \Omega$ at standstill ($s=1$). Full load speed $N_{FL} = 360 \text{ rpm}$. Synchronous speed $N_s = \frac{120f}{P} = \frac{120 \times 50}{16} = 375 \text{ rpm}$. Full load slip $s_{FL} = \frac{N_s - N_{FL}}{N_s} = \frac{375 - 360}{375} = 0.04$. Slip at maximum torque $s_m = \frac{R_2}{X_2} = \frac{0.02}{0.15} = 0.1333$. (i) Ratio of maximum to full load torque: $T_{max} \propto \frac{1}{2X_2}$ $T_{FL} \propto \frac{s_{FL}}{R_2^2 + (s_{FL}X_2)^2}$ $\frac{T_{max}}{T_{FL}} = \frac{R_2^2 + (s_{FL}X_2)^2}{2s_{FL}R_2 X_2}$ (This formula is simplified for $R_1=0$) A more common formula: $\frac{T_{max}}{T_{FL}} = \frac{s_m^2 + s_{FL}^2}{2s_m s_{FL}}$ (assuming $R_1=0$) Using $s_m = R_2/X_2$: $\frac{T_{max}}{T_{FL}} = \frac{(R_2/X_2)^2 + s_{FL}^2}{2(R_2/X_2)s_{FL}} = \frac{R_2^2 + (s_{FL}X_2)^2}{2s_{FL}R_2 X_2}$ $\frac{T_{max}}{T_{FL}} = \frac{(0.02)^2 + (0.04 \times 0.15)^2}{2 \times 0.04 \times 0.02 \times 0.15} = \frac{0.0004 + 0.000036}{0.00024} = \frac{0.000436}{0.00024} \approx 1.816$. (ii) Speed at maximum torque $N_m = N_s(1 - s_m) = 375(1 - 0.1333) = 375 \times 0.8667 \approx 325 \text{ rpm}$. (iii) For maximum starting torque, $s=1$. We need $s_m = 1$. $s_m = \frac{R_{2,new}}{X_2} = 1 \implies R_{2,new} = X_2 = 0.15 \, \Omega$. Resistance to be added $R_{add} = R_{2,new} - R_2 = 0.15 - 0.02 = 0.13 \, \Omega$. Problem 2: Starting Torque with Added Resistance Given: $4$-pole, $50 \text{ Hz}$, $3\phi$. $T_{max} = 162.8 \text{ N-m}$ at $N_m = 1365 \text{ rpm}$. Rotor resistance $R_2 = 0.2 \, \Omega$/phase. Synchronous speed $N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm}$. Slip at maximum torque $s_m = \frac{N_s - N_m}{N_s} = \frac{1500 - 1365}{1500} = \frac{135}{1500} = 0.09$. At maximum torque, $s_m = \frac{R_2}{X_{2s}}$, where $X_{2s}$ is standstill reactance. $X_{2s} = \frac{R_2}{s_m} = \frac{0.2}{0.09} = 2.222 \, \Omega$. Starting torque $T_{st} \propto \frac{R_{2,total}}{R_{2,total}^2 + X_{2s}^2}$ (at $s=1$). Maximum torque $T_{max} \propto \frac{1}{2X_{2s}}$. We want starting torque $T_{st} = \frac{1}{2} T_{max}$. $\frac{T_{st}}{T_{max}} = \frac{2X_{2s} R_{2,total}}{R_{2,total}^2 + X_{2s}^2} = \frac{1}{2}$. $4X_{2s} R_{2,total} = R_{2,total}^2 + X_{2s}^2$. $R_{2,total}^2 - 4X_{2s} R_{2,total} + X_{2s}^2 = 0$. Substitute $X_{2s} = 2.222$: $R_{2,total}^2 - 4(2.222) R_{2,total} + (2.222)^2 = 0$. $R_{2,total}^2 - 8.888 R_{2,total} + 4.937 = 0$. Solve quadratic equation for $R_{2,total}$: $R_{2,total} = \frac{-(-8.888) \pm \sqrt{(-8.888)^2 - 4(1)(4.937)}}{2(1)}$ $R_{2,total} = \frac{8.888 \pm \sqrt{79.00 - 19.748}}{2} = \frac{8.888 \pm \sqrt{59.252}}{2}$ $R_{2,total} = \frac{8.888 \pm 7.697}{2}$. Two possible values for $R_{2,total}$: $R_{2,total,1} = \frac{8.888 + 7.697}{2} = \frac{16.585}{2} = 8.2925 \, \Omega$. $R_{2,total,2} = \frac{8.888 - 7.697}{2} = \frac{1.191}{2} = 0.5955 \, \Omega$. Resistance to be inserted $R_{add} = R_{2,total} - R_2$. $R_{add,1} = 8.2925 - 0.2 = 8.0925 \, \Omega$. $R_{add,2} = 0.5955 - 0.2 = 0.3955 \, \Omega$. (Typically the smaller positive resistance is chosen if possible, or context might specify which value makes sense.) Problem 3: Performance Parameters Given: $V_L = 440 \text{ V}$, $3\phi$, $50 \text{ Hz}$, $4$-pole, Y-connected. $N_{FL} = 125 \text{ rpm}$. Rotor impedance $Z_2 = (0.4 + j4) \, \Omega$ at standstill. $R_2 = 0.4 \, \Omega$, $X_2 = 4 \, \Omega$. Turn ratio $K = \frac{E_2}{E_1} = 0.8$. Synchronous speed $N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm}$. Full load slip $s_{FL} = \frac{1500 - 125}{1500} = \frac{1375}{1500} \approx 0.9167$. (This is a very high slip, indicating it might be a special application or a mistake in problem statement, but we proceed with it). Per phase voltage $V_1 = \frac{440}{\sqrt{3}} = 254.03 \text{ V}$. Rotor impedance referred to stator $Z_2' = (\frac{1}{K})^2 Z_2 = (\frac{1}{0.8})^2 (0.4 + j4) = (1.25)^2 (0.4 + j4) = 1.5625 (0.4 + j4) = (0.625 + j6.25) \, \Omega$. So $R_2' = 0.625 \, \Omega$, $X_2' = 6.25 \, \Omega$. (i) Full load torque $T_{FL}$: Rotor current $I_2' = \frac{V_1}{\sqrt{(R_1 + R_2'/s_{FL})^2 + (X_1 + X_2')^2}}$. Assuming $R_1, X_1$ are negligible for now as they are not given. $I_2' \approx \frac{V_1}{\sqrt{(R_2'/s_{FL})^2 + (X_2')^2}} = \frac{254.03}{\sqrt{(0.625/0.9167)^2 + (6.25)^2}} = \frac{254.03}{\sqrt{(0.6818)^2 + (6.25)^2}} = \frac{254.03}{\sqrt{0.4648 + 39.0625}} = \frac{254.03}{\sqrt{39.5273}} = \frac{254.03}{6.287} \approx 40.41 \text{ A}$. Rotor copper loss per phase $P_{cu2} = (I_2')^2 R_2'/s_{FL} = (40.41)^2 \times (0.625/0.9167) = 1632.9 \times 0.6818 \approx 1113.3 \text{ W}$. Total rotor input power $P_{in,rotor} = 3 \times P_{cu2} = 3 \times 1113.3 = 3339.9 \text{ W}$. Mechanical power developed $P_m = P_{in,rotor}(1-s_{FL}) = 3339.9 (1-0.9167) = 3339.9 \times 0.0833 = 278.2 \text{ W}$. Torque $T_{FL} = \frac{P_m}{\omega_m} = \frac{P_m}{2\pi N_{FL}/60} = \frac{278.2}{2\pi \times 125/60} = \frac{278.2}{13.09} \approx 21.25 \text{ N-m}$. (ii) Rotor current $I_2'$ (referred to stator) is $40.41 \text{ A}$. Actual rotor current $I_2 = I_2'/K = 40.41/0.8 = 50.51 \text{ A}$. Full load rotor Cu-loss (total) $= 3 I_2^2 R_2 = 3 \times (50.51)^2 \times 0.4 = 3 \times 2551.26 \times 0.4 = 3061.5 \text{ W}$. (iii) Power output $P_{out} = P_m - P_{fw} = 278.2 - 500 = -221.8 \text{ W}$. (Negative output indicates the motor cannot sustain this load given the losses, or the $N_{FL}$ value is for a different operating point, or this motor is very inefficient at this slip). Assuming it's a calculation exercise, we proceed. (iv) Maximum torque $T_{max}$ and speed at which it occurs: Slip at max torque $s_m = \frac{R_2'}{X_2'} = \frac{0.625}{6.25} = 0.1$. Speed at max torque $N_m = N_s(1-s_m) = 1500(1-0.1) = 1350 \text{ rpm}$. Rotor current at max torque $I_{2,m}' = \frac{V_1}{\sqrt{(R_2'/s_m)^2 + (X_2')^2}} = \frac{254.03}{\sqrt{(0.625/0.1)^2 + (6.25)^2}} = \frac{254.03}{\sqrt{(6.25)^2 + (6.25)^2}} = \frac{254.03}{6.25\sqrt{2}} = \frac{254.03}{8.838} \approx 28.74 \text{ A}$. Power input to rotor at max torque $P_{in,rotor,m} = 3 (I_{2,m}')^2 \frac{R_2'}{s_m} = 3 \times (28.74)^2 \times \frac{0.625}{0.1} = 3 \times 825.99 \times 6.25 = 15487.3 \text{ W}$. Max mechanical power $P_{m,max} = P_{in,rotor,m}(1-s_m) = 15487.3 (1-0.1) = 15487.3 \times 0.9 = 13938.6 \text{ W}$. Max torque $T_{max} = \frac{P_{m,max}}{\omega_m} = \frac{13938.6}{2\pi \times 1350/60} = \frac{13938.6}{141.37} \approx 98.6 \text{ N-m}$. (v) Starting current ($s=1$): $I_{2,st}' = \frac{V_1}{\sqrt{(R_2'/1)^2 + (X_2')^2}} = \frac{254.03}{\sqrt{(0.625)^2 + (6.25)^2}} = \frac{254.03}{\sqrt{0.3906 + 39.0625}} = \frac{254.03}{\sqrt{39.4531}} = \frac{254.03}{6.281} \approx 40.44 \text{ A}$. Actual starting current $I_{2,st} = I_{2,st}'/K = 40.44/0.8 = 50.55 \text{ A}$. (vi) Starting torque ($s=1$): $P_{in,rotor,st} = 3 (I_{2,st}')^2 R_2' = 3 \times (40.44)^2 \times 0.625 = 3 \times 1635.49 \times 0.625 = 3066.5 \text{ W}$. $T_{st} = \frac{P_{in,rotor,st}}{\omega_s} = \frac{3066.5}{2\pi \times 1500/60} = \frac{3066.5}{157.08} \approx 19.52 \text{ N-m}$. Problem 4: Voltage Reduction for Torque at Half Speed Given: $R_2 = 0.015 \, \Omega$, $X_2 = 0.09 \, \Omega$. Full load slip $s_{FL} = 3\% = 0.03$. Synchronous speed $N_s$. Full load speed $N_{FL} = N_s(1-s_{FL}) = N_s(1-0.03) = 0.97 N_s$. We want full load torque $T_{FL}$ at half full load speed $N_{new} = \frac{1}{2} N_{FL} = \frac{1}{2} (0.97 N_s) = 0.485 N_s$. New slip $s_{new} = \frac{N_s - N_{new}}{N_s} = \frac{N_s - 0.485 N_s}{N_s} = 0.515$. Torque $T \propto \frac{s V^2 R_2}{R_2^2 + (sX_2)^2}$. Since the torque is the same: $T_{FL,normal} = T_{FL,new}$. $\frac{s_{FL} V_{normal}^2 R_2}{R_2^2 + (s_{FL}X_2)^2} = \frac{s_{new} V_{new}^2 R_2}{R_2^2 + (s_{new}X_2)^2}$. $\frac{s_{FL}}{R_2^2 + (s_{FL}X_2)^2} V_{normal}^2 = \frac{s_{new}}{R_2^2 + (s_{new}X_2)^2} V_{new}^2$. $V_{new}^2 = V_{normal}^2 \frac{s_{FL}}{s_{new}} \frac{R_2^2 + (s_{new}X_2)^2}{R_2^2 + (s_{FL}X_2)^2}$. $V_{new} = V_{normal} \sqrt{\frac{s_{FL}}{s_{new}} \frac{R_2^2 + (s_{new}X_2)^2}{R_2^2 + (s_{FL}X_2)^2}}$. Substitute values: $s_{FL} = 0.03$, $s_{new} = 0.515$. $R_2 = 0.015$, $X_2 = 0.09$. $s_{FL}X_2 = 0.03 \times 0.09 = 0.0027$. $s_{new}X_2 = 0.515 \times 0.09 = 0.04635$. $R_2^2 = (0.015)^2 = 0.000225$. $(s_{FL}X_2)^2 = (0.0027)^2 = 0.00000729$. $(s_{new}X_2)^2 = (0.04635)^2 = 0.002148$. $V_{new} = V_{normal} \sqrt{\frac{0.03}{0.515} \frac{0.000225 + 0.002148}{0.000225 + 0.00000729}}$. $V_{new} = V_{normal} \sqrt{\frac{0.03}{0.515} \frac{0.002373}{0.00023229}}$. $V_{new} = V_{normal} \sqrt{0.05825 \times 10.215} = V_{normal} \sqrt{0.595}$. $V_{new} = V_{normal} \times 0.771$. Percentage reduction in stator voltage $= (1 - 0.771) \times 100\% = 22.9\%$. Power factor: Not enough information (stator resistance $R_1$ and stator reactance $X_1$ are not given). Power factor calculation requires the total equivalent impedance. Problem 5: Power and Torque Calculations Given: $P_{out} = 18.65 \text{ kW}$, $4$-pole, $50 \text{ Hz}$, $3\phi$. Friction and windage losses $P_{fw} = 2.5\% \text{ of } P_{out} = 0.025 \times 18.65 = 0.46625 \text{ kW}$. Full load slip $s_{FL} = 4\% = 0.04$. Synchronous speed $N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm}$. Mechanical power developed $P_m = P_{out} + P_{fw} = 18.65 + 0.46625 = 19.11625 \text{ kW}$. (a) Rotor Cu loss $P_{cu2} = \frac{s}{1-s} P_m = \frac{0.04}{1-0.04} \times 19.11625 = \frac{0.04}{0.96} \times 19.11625 = 0.041667 \times 19.11625 \approx 0.7965 \text{ kW}$. (b) Rotor input $P_{in,rotor} = P_m + P_{cu2} = 19.11625 + 0.7965 = 19.91275 \text{ kW}$. (Alternatively, $P_{in,rotor} = P_m / (1-s) = 19.11625 / 0.96 = 19.91275 \text{ kW}$). (c) Shaft torque $T_{sh} = \frac{P_{out}}{\omega_{FL}}$. Full load speed $N_{FL} = N_s(1-s_{FL}) = 1500(1-0.04) = 1500 \times 0.96 = 1440 \text{ rpm}$. $\omega_{FL} = \frac{2\pi N_{FL}}{60} = \frac{2\pi \times 1440}{60} = 150.8 \text{ rad/s}$. $T_{sh} = \frac{18.65 \times 1000}{150.8} \approx 123.67 \text{ N-m}$. (d) Gross electromagnetic torque $T_g = \frac{P_m}{\omega_s}$. $\omega_s = \frac{2\pi N_s}{60} = \frac{2\pi \times 1500}{60} = 157.08 \text{ rad/s}$. $T_g = \frac{19.11625 \times 1000}{157.08} \approx 121.7 \text{ N-m}$. Problem 6: Torque and Slip Relationships Given: $V_L = 400 \text{ V}$, $4$-pole, $3\phi$, $50 \text{ Hz}$. Rotor resistance $R_2 = 0.01 \, \Omega$, reactance $X_2 = 0.1 \, \Omega$. Stator to rotor turns ratio $a = 4$. Synchronous speed $N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm}$. Rotor impedance referred to stator $R_2' = a^2 R_2 = 4^2 \times 0.01 = 0.16 \, \Omega$. $X_2' = a^2 X_2 = 4^2 \times 0.1 = 1.6 \, \Omega$. (Assuming phase voltage $V_1 = V_L/\sqrt{3}$ for Y-connection, or $V_L$ for delta, typically line values are used for per phase calculations with referred values). Assuming $V_1 = 400/\sqrt{3} = 230.94 \text{ V}$. (a) Maximum torque $T_{max}$ and corresponding slip $s_m$: $s_m = \frac{R_2'}{X_2'} = \frac{0.16}{1.6} = 0.1$. $T_{max} = \frac{3}{\omega_s} \frac{V_1^2}{2X_2'}$ (assuming $R_1=0$, $X_1$ is lumped into $X_2'$). $\omega_s = \frac{2\pi \times 1500}{60} = 157.08 \text{ rad/s}$. $T_{max} = \frac{3}{157.08} \frac{(230.94)^2}{2 \times 1.6} = \frac{3}{157.08} \frac{53333.1}{3.2} = \frac{3}{157.08} \times 16666.6 = 0.01909 \times 16666.6 \approx 317.9 \text{ N-m}$. (b) Full load slip $s_{FL}$ and power output $P_{out}$, if $T_{max} = 2 T_{FL}$. $\frac{T_{max}}{T_{FL}} = \frac{s_m^2 + s_{FL}^2}{2s_m s_{FL}} = 2$. $s_m^2 + s_{FL}^2 = 4s_m s_{FL}$. $s_{FL}^2 - 4s_m s_{FL} + s_m^2 = 0$. Substitute $s_m = 0.1$: $s_{FL}^2 - 4(0.1)s_{FL} + (0.1)^2 = 0$. $s_{FL}^2 - 0.4s_{FL} + 0.01 = 0$. Solve quadratic for $s_{FL}$: $s_{FL} = \frac{-(-0.4) \pm \sqrt{(-0.4)^2 - 4(1)(0.01)}}{2(1)}$ $s_{FL} = \frac{0.4 \pm \sqrt{0.16 - 0.04}}{2} = \frac{0.4 \pm \sqrt{0.12}}{2} = \frac{0.4 \pm 0.3464}{2}$. Two possible slips: $s_{FL,1} = \frac{0.4 + 0.3464}{2} = \frac{0.7464}{2} = 0.3732$. $s_{FL,2} = \frac{0.4 - 0.3464}{2} = \frac{0.0536}{2} = 0.0268$. Typically, full load slip is small, so $s_{FL} = 0.0268$. Full load torque $T_{FL} = \frac{T_{max}}{2} = \frac{317.9}{2} = 158.95 \text{ N-m}$. Mechanical power developed $P_m = T_{FL} \omega_s (1-s_{FL}) = 158.95 \times 157.08 \times (1-0.0268) = 158.95 \times 157.08 \times 0.9732 \approx 24324 \text{ W}$. Power output $P_{out}$ (assuming no friction/windage losses given, or they are negligible) $\approx P_m = 24324 \text{ W}$. Problem 7: Load Resistance, Voltage, and Current Given: Maximum torque at $s_m = 12\% = 0.12$. Equivalent secondary resistance $R_2' = 0.08 \, \Omega$/phase. Gross power output $P_m = 9000 \text{ W}$. We know $s_m = \frac{R_2'}{X_2'}$, so $X_2' = \frac{R_2'}{s_m} = \frac{0.08}{0.12} = 0.6667 \, \Omega$. At operating slip $s = s_m = 0.12$. Mechanical power $P_m = 3 \frac{V_1^2}{(R_1+R_2'/s)^2 + (X_1+X_2')^2} R_2' \frac{1-s}{s}$. (Assuming $R_1, X_1$ negligible unless given). $P_m = 3 I_2'^2 R_2' \frac{1-s}{s}$. $9000 = 3 I_2'^2 \times 0.08 \times \frac{1-0.12}{0.12} = 3 I_2'^2 \times 0.08 \times \frac{0.88}{0.12} = 3 I_2'^2 \times 0.08 \times 7.333 = 1.76 I_2'^2$. $I_2'^2 = \frac{9000}{1.76} = 5113.6$. $I_2' = \sqrt{5113.6} \approx 71.51 \text{ A}$. (This is the current at $s=0.12$) Equivalent load resistance $R_L$: The equivalent resistance representing the mechanical load is $R_m = R_2' \frac{1-s}{s}$. $R_{L,eq} = R_2' \frac{1-s}{s} = 0.08 \times \frac{1-0.12}{0.12} = 0.08 \times \frac{0.88}{0.12} = 0.08 \times 7.333 = 0.5866 \, \Omega$. Equivalent load voltage $V_L$: Voltage across the equivalent load resistance. $V_{L,eq} = I_2' R_{L,eq} = 71.51 \times 0.5866 \approx 41.96 \text{ V}$. Current at this slip: $I_2' = 71.51 \text{ A}$. Problem 8: Starting to Full Load Current Ratio (Star-Delta Starter) Given: $P_{out} = 10 \text{ kW}$, $V = 400 \text{ V}$, $3\phi$. Full load power factor $\cos\phi_{FL} = 0.85$. Full load efficiency $\eta_{FL} = 0.88$. Blocked rotor current $I_{sc} = 40 \text{ A}$ at $400 \text{ V}$. Star-delta starter. Input power at full load $P_{in,FL} = \frac{P_{out}}{\eta_{FL}} = \frac{10000}{0.88} = 11363.6 \text{ W}$. Full load line current $I_{FL} = \frac{P_{in,FL}}{\sqrt{3} V \cos\phi_{FL}} = \frac{11363.6}{\sqrt{3} \times 400 \times 0.85} = \frac{11363.6}{588.8} \approx 19.3 \text{ A}$. Starting current with direct online (DOL) starting: $I_{st,DOL} = I_{sc} = 40 \text{ A}$. For a star-delta starter, the starting line current is $1/\sqrt{3}$ of the direct-on-line starting current. $I_{st,Y-\Delta} = \frac{1}{\sqrt{3}} I_{st,DOL} = \frac{1}{\sqrt{3}} \times 40 = 23.09 \text{ A}$. Ratio of starting to full load current = $\frac{I_{st,Y-\Delta}}{I_{FL}} = \frac{23.09}{19.3} \approx 1.196$. Problem 9: Starting Torque and Current (Auto-Transformer Starter) Given: Star-delta starter takes $I_{st,Y-\Delta} = 1.8 I_{FL}$ and develops $T_{st,Y-\Delta} = 0.35 T_{FL}$. For an auto-transformer starter with tapping $x = 75\% = 0.75$. Starting current with auto-transformer: $I_{st,auto} = x^2 I_{st,DOL}$. We know $I_{st,Y-\Delta} = \frac{1}{\sqrt{3}} I_{st,DOL}$, so $I_{st,DOL} = \sqrt{3} I_{st,Y-\Delta} = \sqrt{3} \times 1.8 I_{FL} = 3.117 I_{FL}$. $I_{st,auto} = (0.75)^2 \times 3.117 I_{FL} = 0.5625 \times 3.117 I_{FL} \approx 1.75 I_{FL}$. Starting torque with auto-transformer: $T_{st,auto} = x^2 T_{st,DOL}$. We know $T_{st,Y-\Delta} = \frac{1}{3} T_{st,DOL}$ for star-delta starter (torque is proportional to $V^2$, and phase voltage is $1/\sqrt{3}$ of line voltage in star, so $V_{phase}^2 = (1/\sqrt{3})^2 V_{line}^2 = 1/3 V_{line}^2$). So $T_{st,DOL} = 3 T_{st,Y-\Delta} = 3 \times 0.35 T_{FL} = 1.05 T_{FL}$. $T_{st,auto} = (0.75)^2 \times 1.05 T_{FL} = 0.5625 \times 1.05 T_{FL} \approx 0.59 T_{FL}$. Problem 10: Double Cage Rotor Torque Ratio Given: Inner cage impedance $Z_i = (0.4 + j2) \, \Omega$. Outer cage impedance $Z_o = (2 + j0.4) \, \Omega$. $R_i = 0.4$, $X_i = 2$. $R_o = 2$, $X_o = 0.4$. Torque $T \propto \frac{R}{R^2 + (sX)^2}$ for a given slip $s$ and voltage. For a double cage rotor, total torque is the sum of torques from inner and outer cages. (i) At standstill ($s=1$): Torque from inner cage $T_i \propto \frac{R_i}{R_i^2 + X_i^2} = \frac{0.4}{(0.4)^2 + (2)^2} = \frac{0.4}{0.16 + 4} = \frac{0.4}{4.16} \approx 0.09615$. Torque from outer cage $T_o \propto \frac{R_o}{R_o^2 + X_o^2} = \frac{2}{(2)^2 + (0.4)^2} = \frac{2}{4 + 0.16} = \frac{2}{4.16} \approx 0.48077$. Ratio of torques $\frac{T_o}{T_i} = \frac{0.48077}{0.09615} \approx 5$. (ii) At $s = 5\% = 0.05$: Torque from inner cage $T_i \propto \frac{R_i}{R_i^2 + (sX_i)^2} = \frac{0.4}{(0.4)^2 + (0.05 \times 2)^2} = \frac{0.4}{0.16 + (0.1)^2} = \frac{0.4}{0.16 + 0.01} = \frac{0.4}{0.17} \approx 2.353$. Torque from outer cage $T_o \propto \frac{R_o}{R_o^2 + (sX_o)^2} = \frac{2}{(2)^2 + (0.05 \times 0.4)^2} = \frac{2}{4 + (0.02)^2} = \frac{2}{4 + 0.0004} = \frac{2}{4.0004} \approx 0.4999$. Ratio of torques $\frac{T_o}{T_i} = \frac{0.4999}{2.353} \approx 0.212$. Problem 11: Double Cage Equal Torques Slip Given: Outer cage impedance at standstill $Z_o = (2 + j1.2) \, \Omega$. Inner cage impedance at standstill $Z_i = (0.5 + j3.5) \, \Omega$. $R_o = 2$, $X_o = 1.2$. $R_i = 0.5$, $X_i = 3.5$. For equal torques $T_o = T_i$: $\frac{R_o}{R_o^2 + (sX_o)^2} = \frac{R_i}{R_i^2 + (sX_i)^2}$. $\frac{2}{2^2 + (s \times 1.2)^2} = \frac{0.5}{0.5^2 + (s \times 3.5)^2}$. $\frac{2}{4 + 1.44s^2} = \frac{0.5}{0.25 + 12.25s^2}$. Cross-multiply: $2(0.25 + 12.25s^2) = 0.5(4 + 1.44s^2)$. $0.5 + 24.5s^2 = 2 + 0.72s^2$. $24.5s^2 - 0.72s^2 = 2 - 0.5$. $23.78s^2 = 1.5$. $s^2 = \frac{1.5}{23.78} \approx 0.06308$. $s = \sqrt{0.06308} \approx 0.251$. The slip at which the two cages develop equal torques is approximately $0.251$ or $25.1\%$.