Question

A $6\, V$ cell with $0.5\, \Omega$ internal resistance, a $10\, V$ cell with $1\, \Omega$ internal resistance and a $12\, \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10\, V$ cell is

• A. 0.60
• B. 2.27
• C. 2.87
• D. 5.14

Answer 1

Answer: (C)

In closed loop $ABGFEHA$,
$10-i_{2} \times 1+i_{1} \times 0.5-6=0$
$0.5 i_{1}-i_{2}=-4\,\,\,(i)$
In closed loop $B C D E B$,
$\left(i_{1}+i_{2}\right) \times 12+i_{2} \times 1-10=0$
$12 i_{1}+13 i_{2}=10\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$i_{2}=2.87 \,A$