Question

When an external resistance of 5 \(\Omega\) is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 \(\Omega\) resistor is replaced by a 2 \(\Omega\) resistor, a current of 0.5 A flows through it. The internal resistance of the cell is \dots \(\Omega\).

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
The current \( I \) in a closed circuit containing a cell of emf \( E \), internal resistance \( r \), and external resistance \( R \) is given by Ohm's law.
: Key Formula or Approach:
\( I = \frac{E}{R + r} \implies E = I(R + r) \).
Step 2: Detailed Explanation:
From the first case:
\( E = 0.25 \cdot (5 + r) \quad \dots(1) \).
From the second case:
\( E = 0.5 \cdot (2 + r) \quad \dots(2) \).
Since the emf \( E \) is constant for the cell, equate (1) and (2):
\[ 0.25(5 + r) = 0.5(2 + r) \]
Divide both sides by 0.25:
\[ 5 + r = 2(2 + r) \implies 5 + r = 4 + 2r \]
\[ r = 1\ \Omega \].
Step 3: Final Answer:
The internal resistance of the cell is 1 \(\Omega\).