Q. An electric motor runs a DC source of emf 200V and draws a current of 10A. If the efficiency is 40%, then the resistance of the armature is
A) 5$\Omega$
B) 12$\ \Omega$
C) 120$\ \Omega$
D) 160$\ \Omega$
Explanation
(b) Here, $\eta = \frac{e}{E} \times 100$
$\ \ \ \ \ \ \ \ \ \ \ \frac{40}{100} = \frac{e}{E}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ e = \frac{2E}{5} = \frac{2 \times 200}{5} = 80\ V$
But $\ \ \ \ \ \ \ \ \ I = \frac{E - e}{R}$
$\therefore\ \ \ \ \ \ \ \ \ \ 10 = \frac{200 - 80}{R}$
$\ \ \ \ \ \ \ \ 10R = 120$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ R = 12\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.