A 230 V, 1800 rpm, long shunt, cumulative compound motor has a shunt field resistance of 460 $\Omega$, series field resistance of 0.2 $\Omega$ and armature circuit resistance of 1.3 $\Omega$. The full-load line current is 10.5 A. The rotational loss is 5% of the power developed. Then the efficiency (in percentage) of the motor at full load is________.

1. 84.57

The correct option is A 84.57
$\text{At full load:}$


$\text{Armature current,}{I}_a=10.5-\frac{230}{460}=10\text{A}$

$\text{Back emf,}{E}_b=V-I_a(R_a+R_{se})$
$=230-10(1.3+0.2)$
$E_b=215Volt$

$\therefore \text{Power developed in armature}=E_b{I}_a$
$=215\times 10=2150Watt$

$\text{Given that, rotational losses is 5\% of power developed}$
$P_{output}=0.95\times P_{developed}$
$=0.95\times 2150$
$=2042.5Watt$

$\therefore \text{Efficiency,}$ $\eta =\frac{P_{output}}{P_{input}}=\frac{2042.5}{230\times 10.5}=0.8457\left(or\right)84.57\%$