Question

A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs $E_1=2V$ and $E_2=1.5V$ and internal resistances $r_1=1\Omega$ and $r_2=2\Omega$ respectively, with polarities as shown in the figure.

(IIT-JEE 1997)

• A. $E=\frac{5}{6}V$
• B. E = 1.2 V
• C. r = 1.5 Ω
• D. $r=\frac{2}{3}\Omega$

Answer 1

The correct option is A

$E=\frac{5}{6}V$

Simple,
All you have to do is resolve the cell combination.
$E=\frac{E_1{r}_2-E_2{r}_1}{r_1+r_2}=\frac{2\times 2-1.5\times 1}{2+1}$
$=\frac{2.5}{3}=\frac{5}{6}V$