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Step-by-Step Solution
1. Understanding the Problem
Two identical cells are connected to an external resistance of 5\,\Omega . Whether these two cells are connected in series or in parallel, the current in the external circuit remains the same. We need to find the internal resistance r of each cell.
2. Series Connection
When the cells are in series, their emfs add up, giving a total emf of 2\epsilon . The total internal resistance becomes 2r . The current through the external resistance in series is:
i_s = \frac{2\epsilon}{5 + 2r} .
3. Parallel Connection
When the cells are in parallel, the effective emf remains \epsilon . However, two identical internal resistances r in parallel combine to give \frac{r}{2} . Thus, the current in the external circuit in parallel is:
i_p = \frac{\epsilon}{5 + \frac{r}{2}} .
4. Equating the Currents
According to the problem statement, i_s = i_p . Therefore:
\frac{2\epsilon}{5 + 2r} = \frac{\epsilon}{5 + \frac{r}{2}} .
We can cancel \epsilon on both sides (assuming \epsilon \neq 0 ).
5. Simplifying the Equation
After canceling \epsilon , we get:
\frac{2}{5 + 2r} = \frac{1}{5 + \frac{r}{2}} .
Cross-multiplying leads to:
2 \Bigl(5 + \frac{r}{2}\Bigr) = 5 + 2r .
Simplify step by step:
10 + r = 5 + 2r .
Rearranging:
10 - 5 = 2r - r .
5 = r .
6. Final Answer
Hence, the internal resistance of each cell is:
r = 5\,\Omega .
Reference Figure
Below is a representative figure of series and parallel connections of cells:
