A 40 kW, 3-phse slip ring induction motor of negligible stator impedance runs at a speed0.96 times synchronous speed at rated torque. The slip at maximum torque is 4 times the fullload value. If the rotor resistance of the motor is is increased by 5 times, determine,(i) The speed, power output and rotor ohmic losses at rated torque(ii) The speed corresponding to maximum torque.Neglect mechanical losses.

This problem involves the use of the slip and torque equations for a 3-phase slip ring induction motor. Here are the steps to solve it:

(i) The speed, power output and rotor ohmic losses at rated torque

1. Calculate the slip at full load. The slip (s) is given by 1 - (rotor speed / synchronous speed). Given that the rotor speed is 0.96 times the synchronous speed, the slip at full load is 1 - 0.96 = 0.04.

2. The power output at full load is given as 40 kW.

3. The rotor ohmic losses at full load can be calculated using the formula: P_loss = s * P_out, where P_out is the power output and s is the slip. Substituting the given values, P_loss = 0.04 * 40 kW = 1.6 kW.

(ii) The speed corresponding to maximum torque

1. The problem states that the slip at maximum torque is 4 times the full load value. So, s_max = 4 * 0.04 = 0.16.

2. The speed at maximum torque can be calculated using the formula: N_max = N_s * (1 - s_max), where N_s is the synchronous speed. However, the synchronous speed is not given in the problem.

3. If we assume that the synchronous speed at full load is the base speed (which is a common assumption), then N_s = 1 (per unit). So, N_max = 1 * (1 - 0.16) = 0.84 (per unit).

4. If the rotor resistance is increased by 5 times, the slip at maximum torque will also increase by 5 times (since the slip is directly proportional to the rotor resistance). So, the new slip at maximum torque is s_max_new = 5 * 0.16 = 0.8.

5. The new speed at maximum torque is N_max_new = N_s * (1 - s_max_new) = 1 * (1 - 0.8) = 0.2 (per unit).

Please note that the actual speeds (in rpm) would depend on the actual synchronous speed, which is not given in the problem. The speeds calculated here are in per unit (p.u.), relative to the synchronous speed.

This problem involves the use of the slip and torque equations for a 3-phase slip ring induction motor. Here are the steps to solve it:

(i) The speed, power output and rotor ohmic losses at rated torque:

1. First, we need to calculate the slip at full load. The slip (s) is given by the formula s = (Ns - N) / Ns, where Ns is the synchronous speed and N is the actual speed. Given that the motor runs at 0.96 times the synchronous speed at full load, the slip at full load is s = (1 - 0.96) = 0.04.

2. The power output (Pout) at full load is given by the formula Pout = Pinput * (1 - s), where Pinput is the input power. Given that the input power is 40 kW, the power output at full load is Pout = 40 kW * (1 - 0.04) = 38.4 kW.

3. The rotor ohmic losses (Prot) at full load are given by the formula Prot = Pinput * s. So, the rotor ohmic losses at full load are Prot = 40 kW * 0.04 = 1.6 kW.

(ii) The speed corresponding to maximum torque:

1. The problem states that the slip at maximum torque is 4 times the full load value. So, the slip at maximum torque is s_max = 4 * 0.04 = 0.16.

2. The speed at maximum torque (N_max) is given by the formula N_max = Ns * (1 - s_max). So, the speed at maximum torque is N_max = 1 * (1 - 0.16) = 0.84 times the synchronous speed.

Note: The rotor resistance is increased by 5 times, but this does not affect the speed at rated torque or maximum torque, as these speeds are determined by the slip, which is a function of the motor's electrical characteristics and load, not its resistance. The increased resistance will, however, increase the rotor's ohmic losses and decrease the motor's efficiency.