Question

Two cells $A$ and $B$ of e.m.f $2V$ and $1.5V$ respectively, are connected as shown in figure through an external resistance $10\Omega$. The internal resistance of each cell is $5\Omega$. The potential difference $E_A$ and $E_B$ across the terminals of the cells $A$ and $B$ respectively are

• A. $E_A=2.0V,E_B=1.5V$
• B. $E_A=2.12V,E_B=1.375V$
• C. $E_A=1.875V,E_B=1.625V$
• D. $E_A=1.875V,E_B=1.375V$

Answer 1

Answer: (C)

The figure can be redrawn as,

The current through the circuit
$i=\frac{\text{ net emf }}{\text{ effective resistance }}$
$=\frac{2-1.5}{5+5+10}=\frac{0.5}{20}$
$=\frac{1}{40}=0.025A$
The terminal potential difference of the batteries
$V_A={\epsilon }_A-i{r}_A=2-0.025\times 5$
$=2-0.0125=1.875V$
and $V_B={\epsilon }_B+i{r}_B$
$=1.5+0.025\times 5$
$=1.5+0.0125=1.625V$