Question

A 200 V shunt motor takes 10 A when running on no load. At higher loads, the brush drop is 2 V and at light loads, it is negligible. The stray load loss at a line current of 100 A is 50% of the no load loss. Calculate the efficiency at a line current of 100 A if armature and field resistances are 0.2 and 100 ohm respectively.

• A. 74.48%
• B. 81.2%
• C. 78.14%
• D. 66.12%

Answer 1

The correct option is A 74.48%

$\text{At no load,}$
$\text{No load loss}=200\times 10=2000W$

$I_f=\frac{200}{100}=2A$

$I_0=10-2=8A$

$P_{io}+P_{wf}=200\times 8-(8{)}^2\times 0.2=1587.2W$

$\text{At load,}{I}_a=100-2=98A$

$I_a^2{R}_a=(98{)}^2\times 0.2=1920.8W$

$\text{Stray load loss}=0.5\times 2000=1000W$

$P_L=(I_a^2{R}_a+V_b{I}_a+P_{st})+(P_{io}+P_{wf}+P_{sh})$

$=(1920+2\times 98+1000)+(1587.2+200\times 2)=5104W$

$\text{Efficiency ,}\eta =\frac{P_{in}-P_L}{P_{in}}$

$=\frac{200\times 100-5104}{200\times 100}=0.7448$

$\%\eta =74.48\%$