Combination of Batteries: Practice Problem & Solution
A cell can be balanced against $110\text{ cm}$ and $100\text{ cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10\text{ }\Omega$. Its internal resistance is: (2008)
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:
The open-circuit balancing length is $l_1 = 110\text{ cm}$ and closed-circuit balancing length is $l_2 = 100\text{ cm}$.
Using internal resistance formula $r = R \left(\frac{l_1}{l_2} - 1\right)$:
$r = 10 \cdot \left(\frac{110}{100} - 1\right) = 10 \cdot 0.1 = 1.0\text{ ohm}$.
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