Question

Two cells $A$ and $B$ of e.m.f $2\,V$ and $1.5\,V$ 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. $E_A = 2.0\, V,\, E_B = 1.5\, V$
• B. $E_A = 2.12\, V,\, E_B = 1.375\, V$
• C. $E_A = 1.875\, V,\, E_B = 1.625\, V$
• D. $E_A = 1.875\, V,\, E_B = 1.375\, V$

Answer 1

Answer: (C)

The figure can be redrawn as,

The current through the circuit
$i =\frac{\text { net emf }}{\text { effective resistance }}$
$=\frac{2-1.5}{5+5+10}=\frac{0.5}{20}$
$=\frac{1}{40}=0.025\, A$
The terminal potential difference of the batteries
$V_{A} =\varepsilon_{A}-i r_{A}=2-0.025 \times 5$
$=2-0.0125=1.875\, V$
and $V_{B} =\varepsilon_{B}+ir_{B}$
$=1.5+ 0.025 \times 5$
$=1.5+0.0125=1.625\, V$