1. **Problem statement:** A 3-phase 440V 50Hz 4-pole slip ring induction motor runs at 50% slip at full load. Rotor impedance at standstill is $1 + j5$ Ohm per phase. We add an external resistance of 5 Ohm per phase in the rotor circuit and want to find the new speed of the motor while keeping torque constant. 2. **Given data:** - Supply voltage, $V = 440$ V (line-to-line, but not needed for speed calculation) - Frequency, $f = 50$ Hz - Number of poles, $P = 4$ - Full load slip, $s_1 = 0.5$ - Rotor standstill impedance, $Z_r = 1 + j5$ Ohm - External resistance added, $R_{ext} = 5$ Ohm 3. **Step 1: Calculate synchronous speed $N_s$** The synchronous speed of a motor is given by: $$N_s = \frac{120f}{P}$$ Substitute values: $$N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm}$$ 4. **Step 2: Calculate rotor speed at full load slip $s_1$** Rotor speed $N_r$ is: $$N_r = N_s (1 - s)$$ At full load slip $s_1 = 0.5$: $$N_{r1} = 1500 \times (1 - 0.5) = 750 \text{ rpm}$$ 5. **Step 3: Understand torque-slip relation** Torque $T$ is proportional to rotor current and rotor resistance. For constant torque, the slip changes when external resistance is added. 6. **Step 4: Calculate rotor resistance at standstill** Rotor resistance $R_2 = 1$ Ohm (real part of $Z_r$) Rotor reactance $X_2 = 5$ Ohm 7. **Step 5: Calculate new slip $s_2$ with external resistance** Total rotor resistance with external resistance: $$R_2' = R_2 + R_{ext} = 1 + 5 = 6 \text{ Ohm}$$ Torque $T$ is proportional to: $$T \propto \frac{s R_2'}{R_2'^2 + (s X_2)^2}$$ Since torque is constant, equate torque at slip $s_1$ and $s_2$: $$\frac{s_1 R_2}{R_2^2 + (s_1 X_2)^2} = \frac{s_2 R_2'}{R_2'^2 + (s_2 X_2)^2}$$ Substitute values: $$\frac{0.5 \times 1}{1^2 + (0.5 \times 5)^2} = \frac{s_2 \times 6}{6^2 + (s_2 \times 5)^2}$$ Calculate left side denominator: $$1 + (2.5)^2 = 1 + 6.25 = 7.25$$ Left side: $$\frac{0.5}{7.25} = 0.06897$$ Right side denominator: $$36 + 25 s_2^2$$ Equation: $$0.06897 = \frac{6 s_2}{36 + 25 s_2^2}$$ Cross multiply: $$0.06897 (36 + 25 s_2^2) = 6 s_2$$ $$2.483 + 1.724 s_2^2 = 6 s_2$$ Rearranged: $$1.724 s_2^2 - 6 s_2 + 2.483 = 0$$ 8. **Step 6: Solve quadratic for $s_2$** Use quadratic formula: $$s_2 = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1.724 \times 2.483}}{2 \times 1.724}$$ Calculate discriminant: $$36 - 4 \times 1.724 \times 2.483 = 36 - 17.13 = 18.87$$ Square root: $$\sqrt{18.87} = 4.34$$ Calculate roots: $$s_2 = \frac{6 \pm 4.34}{3.448}$$ Two solutions: $$s_2 = \frac{6 + 4.34}{3.448} = \frac{10.34}{3.448} = 3.0$$ (not possible, slip must be less than 1) $$s_2 = \frac{6 - 4.34}{3.448} = \frac{1.66}{3.448} = 0.48$$ 9. **Step 7: Calculate new rotor speed $N_{r2}$** $$N_{r2} = N_s (1 - s_2) = 1500 \times (1 - 0.48) = 1500 \times 0.52 = 780 \text{ rpm}$$ **Final answer:** The speed of the motor when an external resistance of 5 Ohm per phase is added, keeping torque constant, is approximately **780 rpm**.