Question

A 30 kW, 3-phase induction motor has full load slip of 3%, the stator losses amounts to 5% of the input and mechanical losses 1.5% of output. If the current in each rotor phase is 45 A, then the resistance per phase of the rotor is ________$\Omega$.(Answer upto three decimal)

• A. 0.155

Answer 1

The correct option is A 0.155

$\text{Given, Slip, s = 0.03}$

$\text{Per phase output power,}$
$P_0=\frac{30}{3}=10kW$

$\text{Now, Mechanical losses}$ $=\frac{10\times 1.5}{100}=0.15kW$

$\text{Mechanical developed power = 10.15~kW}$

$\text{Rotor copper loss = s}\times \text{(air gap power)}$

$\text{or Rotor copper loss =}\frac{s(\text{Mechanical developed power )}}{1-s}$

$P_{cu\left(rotor\right)}=\frac{0.03}{0.97}\times 10.15=0.3139kW$

$I^{2}R=0.3139kW$

$\text{Rotor resistance / phase,}$

$R=\frac{0.03139\times 1000}{45\times 45}=0.155\Omega$