Question

The emf of a Daniel cell is $1.08V.$ When the terminals of the cells are connected to the resistance of $3\Omega$, the potential difference across the terminals is found to be $0.6V$. Then the internal resistance of the cell is

• A. $1.8\Omega$
• B. $2.4\Omega$
• C. $3.24\Omega$
• D. $0.2\Omega$

Answer 1

The correct option is B $2.4\Omega$

$\Rightarrow E-Ir=0.6$
$Ir=0.48$
$E=I(R+r)$
$\frac{1.08}{3+r}=I$
$\Rightarrow \frac{1.08r}{3+r}=0.48$
$1.08r=0.48r+0.48\times 3$
$0.6r=0.48\times 3$
$r=\frac{4.8\times 3}{6}$
$r=2.4\Omega$