Combination of Resistors: Practice Problem & Solution
A set of '$n$' equal resistors, of value '$R$' each, are connected in series to a battery of emf '$E$' and internal resistance '$R$'. The current drawn is $I$. Now, the '$n$' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes $10 I$. The value of '$n$' is (2018)
Solution Explained:
To solve this problem, we apply the core principles of Combination of Resistors. Understanding the underlying formula is key to arriving at the correct answer below:
Current in series: $I = \frac{E}{nR + R}$. Current in parallel: $10I = \frac{E}{R/n + R}$. Dividing the two equations gives $10 = \frac{nR + R}{R/n + R} = \frac{n(n+1)R}{(n+1)R} = n$. Thus, $n = 10$.
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