Question

First, a set of n equal resistors of 10 \(\Omega\) each are connected in series to a battery of emf 20 V and internal resistance 10 \(\Omega\). A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is _________.

Hint:

When setting up circuit problems, always account for the internal resistance of the battery by adding it to the total external resistance of the circuit. This is a common point where errors are made.

Answer 1

Correct Answer: 20

Step 1: Understanding the Question:
We have 'n' identical resistors connected to a battery first in series and then in parallel. We are given the relationship between the currents in the two cases and need to find the number of resistors, 'n'.
Step 2: Key Formula or Approach:
We will use Ohm's law for the complete circuit, \(I = \frac{\mathcal{E}}{R_{ext} + r_{int}}\). 1. Calculate the total external resistance for the series combination (\(R_{series}\)). 2. Calculate the total external resistance for the parallel combination (\(R_{parallel}\)). 3. Write the expressions for the current in both cases (\(I_{series}\) and \(I_{parallel}\)). 4. Use the given relation \(I_{parallel} = 20 \times I_{series}\) to solve for 'n'.
Step 3: Detailed Explanation:
Given values: Resistance of each resistor, \(R = 10 \, \Omega\). EMF of the battery, \(\mathcal{E} = 20\) V. Internal resistance of the battery, \(r_{int} = 10 \, \Omega\).
Case 1: Series Connection
The equivalent resistance of n resistors in series is \(R_{series} = nR = n \times 10 = 10n \, \Omega\). The current in the series circuit is: \[ I_{series} = \frac{\mathcal{E}}{R_{series} + r_{int}} = \frac{20}{10n + 10} = \frac{20}{10(n+1)} = \frac{2}{n+1} \] Case 2: Parallel Connection
The equivalent resistance of n resistors in parallel is \(R_{parallel} = \frac{R}{n} = \frac{10}{n} \, \Omega\). The current in the parallel circuit is: \[ I_{parallel} = \frac{\mathcal{E}}{R_{parallel} + r_{int}} = \frac{20}{\frac{10}{n} + 10} = \frac{20}{\frac{10 + 10n}{n}} = \frac{20n}{10(1+n)} = \frac{2n}{n+1} \] Relating the Currents
We are given that the current increased 20 times, which means \(I_{parallel} = 20 \times I_{series}\). \[ \frac{2n}{n+1} = 20 \times \left( \frac{2}{n+1} \right) \] Since \(n \ge 1\), the denominator \((n+1)\) is not zero and can be cancelled from both sides. \[ 2n = 20 \times 2 \] \[ 2n = 40 \] \[ n = 20 \] Step 4: Final Answer:
The value of n is 20.