Question:
A balanced three-phase supply is given to a \(30\ \text{kW}\), \(4\)-pole, \(400\ \text{V}\), \(50\ \text{Hz}\), wound rotor induction motor with Y-connected stator and rotor windings. The motor is driving a constant torque load. With shorted sliprings, the machine runs at \(1476\ \text{rpm}\).
When an external non-inductive resistance of \(0.27\ \Omega\) per phase is connected in series in the rotor circuit, the steady-state speed drops to \(1404\ \text{rpm}\).
Neglecting rotational losses, the actual per phase rotor winding resistance is \(\Omega\) (Round off to two decimal places)
A balanced three-phase supply is given to a \(30\ \text{kW}\), \(4\)-pole, \(400\ \text{V}\), \(50\ \text{Hz}\), wound rotor induction motor with Y-connected stator and rotor windings. The motor is driving a constant torque load. With shorted sliprings, the machine runs at \(1476\ \text{rpm}\).
When an external non-inductive resistance of \(0.27\ \Omega\) per phase is connected in series in the rotor circuit, the steady-state speed drops to \(1404\ \text{rpm}\).
Neglecting rotational losses, the actual per phase rotor winding resistance is \(\Omega\) (Round off to two decimal places)
When an external non-inductive resistance of \(0.27\ \Omega\) per phase is connected in series in the rotor circuit, the steady-state speed drops to \(1404\ \text{rpm}\).
Neglecting rotational losses, the actual per phase rotor winding resistance is \(\Omega\) (Round off to two decimal places)
Show Hint
For a constant torque load near synchronous speed, slip divided by rotor resistance stays the same before and after adding external resistance.
Updated On: Jul 20, 2026
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Verified By Collegedunia
Correct Answer: 0.09
Solution and Explanation
Step 1: Find the synchronous speed.
For a \(4\)-pole machine on a \(50\ \text{Hz}\) supply,
\[ N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm} \]
Step 2: Find the slip with shorted sliprings.
The rotor speed here is \(1476\ \text{rpm}\), so
\[ s_1=\frac{N_s-N_1}{N_s}=\frac{1500-1476}{1500}=\frac{24}{1500}=0.016 \]
Step 3: Find the slip after adding external resistance.
The new speed is \(1404\ \text{rpm}\), so
\[ s_2=\frac{N_s-N_2}{N_s}=\frac{1500-1404}{1500}=\frac{96}{1500}=0.064 \]
Step 4: Use the constant torque condition.
The load is a constant torque load, so the torque produced is the same before and after the extra resistance is added. Near synchronous speed the slip is small, so the rotor leakage reactance term in the torque expression can be ignored next to resistance, and torque becomes proportional to slip divided by rotor resistance:
\[ T\propto\frac{s}{r_2} \]
Since torque stays the same in both cases,
\[ \frac{s_1}{r_2}=\frac{s_2}{r_2+R_{ext}} \]
Step 5: Substitute the known numbers.
\[ \frac{0.016}{r_2}=\frac{0.064}{r_2+0.27} \]
Cross multiply:
\[ 0.016\,(r_2+0.27)=0.064\,r_2 \]
\[ 0.016\,r_2+0.00432=0.064\,r_2 \]
Step 6: Solve for the rotor resistance.
\[ 0.00432=0.064\,r_2-0.016\,r_2=0.048\,r_2 \]
\[ r_2=\frac{0.00432}{0.048}=0.09\ \Omega \]
Final Answer:
The actual per phase rotor winding resistance is
\[ \boxed{0.09\ \Omega} \]
For a \(4\)-pole machine on a \(50\ \text{Hz}\) supply,
\[ N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm} \]
Step 2: Find the slip with shorted sliprings.
The rotor speed here is \(1476\ \text{rpm}\), so
\[ s_1=\frac{N_s-N_1}{N_s}=\frac{1500-1476}{1500}=\frac{24}{1500}=0.016 \]
Step 3: Find the slip after adding external resistance.
The new speed is \(1404\ \text{rpm}\), so
\[ s_2=\frac{N_s-N_2}{N_s}=\frac{1500-1404}{1500}=\frac{96}{1500}=0.064 \]
Step 4: Use the constant torque condition.
The load is a constant torque load, so the torque produced is the same before and after the extra resistance is added. Near synchronous speed the slip is small, so the rotor leakage reactance term in the torque expression can be ignored next to resistance, and torque becomes proportional to slip divided by rotor resistance:
\[ T\propto\frac{s}{r_2} \]
Since torque stays the same in both cases,
\[ \frac{s_1}{r_2}=\frac{s_2}{r_2+R_{ext}} \]
Step 5: Substitute the known numbers.
\[ \frac{0.016}{r_2}=\frac{0.064}{r_2+0.27} \]
Cross multiply:
\[ 0.016\,(r_2+0.27)=0.064\,r_2 \]
\[ 0.016\,r_2+0.00432=0.064\,r_2 \]
Step 6: Solve for the rotor resistance.
\[ 0.00432=0.064\,r_2-0.016\,r_2=0.048\,r_2 \]
\[ r_2=\frac{0.00432}{0.048}=0.09\ \Omega \]
Final Answer:
The actual per phase rotor winding resistance is
\[ \boxed{0.09\ \Omega} \]
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