A dc motor run by a source of emf 200V draws a current of 10A . If...

Let us assume that the source voltage is $E$, the back emf developed in the armature coil is $e$, the armature resistance is $R$ and the armature current is $i$.

The efficiency of the motor is

$η = \dfrac{ie}{iE} = \dfrac{e}{200} = 0. 4$

$⇒ e = 80V$

The armature current is

$i = \dfrac{E - e}{R}$

$∴ 10 = \dfrac{200 - 80}{R}$

$∴ R = 12Ω$