Question:hard
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
Show Solution
Correct Answer: 0.09
Solution and Explanation
Step 1: Get the two slips from the two speeds.
The synchronous speed is $N_s=120f/P=120(50)/4=1500$ rpm. With shorted rings the speed is $1476$ rpm, so
\[ s_1=\frac{1500-1476}{1500}=0.016 \]
After the extra $0.27\ \Omega$ per phase is added, the speed falls to $1404$ rpm, so
\[ s_2=\frac{1500-1404}{1500}=0.064 \]
Step 2: Turn the constant torque condition into a ratio.
For a torque-slip curve close to synchronous speed, slip is proportional to rotor resistance for a fixed torque, since the reactance drop is small there. So instead of writing $s/r$ as equal on both sides, write the ratio of the two slips directly against the ratio of the two total rotor resistances:
\[ \frac{s_1}{s_2}=\frac{r_2}{r_2+R_{ext}} \]
Step 3: Plug in the slip ratio.
\[ \frac{0.016}{0.064}=\frac{r_2}{r_2+0.27} \]
\[ 0.25=\frac{r_2}{r_2+0.27} \]
Step 4: Cross multiply and isolate $r_2$.
\[ 0.25(r_2+0.27)=r_2 \]
\[ 0.25r_2+0.0675=r_2 \]
\[ 0.0675=r_2-0.25r_2=0.75r_2 \]
Step 5: Solve.
\[ r_2=\frac{0.0675}{0.75}=0.09\ \Omega \]
\[ \boxed{r_2=0.09\ \Omega} \]
The synchronous speed is $N_s=120f/P=120(50)/4=1500$ rpm. With shorted rings the speed is $1476$ rpm, so
\[ s_1=\frac{1500-1476}{1500}=0.016 \]
After the extra $0.27\ \Omega$ per phase is added, the speed falls to $1404$ rpm, so
\[ s_2=\frac{1500-1404}{1500}=0.064 \]
Step 2: Turn the constant torque condition into a ratio.
For a torque-slip curve close to synchronous speed, slip is proportional to rotor resistance for a fixed torque, since the reactance drop is small there. So instead of writing $s/r$ as equal on both sides, write the ratio of the two slips directly against the ratio of the two total rotor resistances:
\[ \frac{s_1}{s_2}=\frac{r_2}{r_2+R_{ext}} \]
Step 3: Plug in the slip ratio.
\[ \frac{0.016}{0.064}=\frac{r_2}{r_2+0.27} \]
\[ 0.25=\frac{r_2}{r_2+0.27} \]
Step 4: Cross multiply and isolate $r_2$.
\[ 0.25(r_2+0.27)=r_2 \]
\[ 0.25r_2+0.0675=r_2 \]
\[ 0.0675=r_2-0.25r_2=0.75r_2 \]
Step 5: Solve.
\[ r_2=\frac{0.0675}{0.75}=0.09\ \Omega \]
\[ \boxed{r_2=0.09\ \Omega} \]
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