Question

A 60 kW, 250 V shunt motor takes 16 A when running at 1440 rpm. The resistance of the armature and field are 0.2 $\Omega$ and 125 $\Omega$ respectively. The efficiency of the motor when taking 152 A is

• A. 77.73%
• B. 85.17%
• C. 87.17%
• D. 81.45%

Answer 1

The correct option is A 77.73%

$\text{Field current,}{I}_f=\frac{250}{125}=2\text{A}$

$\text{No load armature current,}$ $I_{a0}=16-2=14$ A

$\text{Constant losses,}$ $P_K=(250\times 14-(14{)}^2\times 0.2)+250\times 2=3960.8W$

$I_a=152-2=150$ A

$P_L=I_a{}^2{R}_a+P_K=(150{)}^2\times 0.2+3960.8$
$=8.461kW$

$P_{in}=250\times 152=38kW$

$\therefore$ $\text{Efficiency,}$ $\eta =\frac{38-8.461}{38}\times 100=77.73\%$