Rankers Physics

Combination of Batteries: Practice Problem & Solution

For a cell terminal P.D. is $2.2 \text{ V}$ when circuit is open and reduces to $1.8 \text{ V}$ when cell is connected to a resistance of $R = 5 \Omega$. Determine internal resistance of cell ($r$): (2002)
$\frac{10}{9} \Omega$
$\frac{9}{10} \Omega$
$\frac{11}{9} \Omega$
$\frac{5}{9} \Omega$

Solution Explained:

To solve this problem, we apply the core principles of Combination of Batteries. Understanding the underlying formula is key to arriving at the correct answer below:

Internal resistance is given by $r = R(\frac{E}{V} - 1)$. Substituting $E = 2.2 \text{ V}$, $V = 1.8 \text{ V}$, and $R = 5 \Omega$, we get $r = 5(\frac{2.2}{1.8} - 1) = 5(\frac{0.4}{1.8}) = 5(\frac{2}{9}) = \frac{10}{9} \Omega$.

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