Q. A 6V cell with 0.5$\ \Omega$ internal resistance, a10V cell with 1$\ \Omega$ internal resistance and a 12$\ \Omega$ external resistance are connected in parallel. The current (in ampere) through the 10V cell is
A) 0.60
B) 2.27
C) 2.87
D) 5.14
Explanation
(c) In closed loop ABGFEHA
$10 - i_{2} \times 1 + i_{1} \times 0.5 - 6 = 0$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.5i_{1} - i_{2} = - 4$
In closed loop BCDEB
$\left( i_{1} + i_{2} \right) \times 12 + i_{2} \times 1 - 10 = 0$
$12i_{i} + 13i_{2} = 10$
From Eqs. (i) and (ii)
$i_{2} = 2.87A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.