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Step-by-Step Solution
Step 1: Understand the Series Combination
When five identical cells (each of emf $5\text{ V}$ and internal resistance $1 \Omega$) are connected in series, their total emf adds up, and their total internal resistance also adds up.
So, for series connection:
Total emf, $E_{\text{series}} = 5 \times 5 = 25\text{ V}$.
Total internal resistance, $r_{\text{series}} = 5 \times 1 = 5\ \Omega$.
If $R$ is the external resistance, then the current in the series circuit, denoted by $i_1$, is given by:
$$ i_1 = \frac{E_{\text{series}}}{r_{\text{series}} + R} = \frac{25}{5 + R}. $$
Step 2: Understand the Parallel Combination
When the same five cells are connected in parallel:
The effective emf remains $5\text{ V}$, because all cells in parallel have the same emf.
The total internal resistance, $r_{\text{parallel}}$, for $n$ identical resistors of resistance $r$ in parallel is given by $\frac{r}{n}$. Here, $n=5$ and $r=1\ \Omega$, so
$$r_{\text{parallel}} = \frac{1}{5}\ \Omega.$$
Hence, the current in the parallel circuit, denoted by $i_2$, is:
$$ i_2 = \frac{5}{R + \frac{1}{5}}. $$
Step 3: Equate the Two Currents
According to the condition given in the problem, the current in the series arrangement ($i_1$) must be equal to that in the parallel arrangement ($i_2$). So we set:
$$ \frac{25}{5 + R} = \frac{5}{R + \frac{1}{5}}. $$
Step 4: Solve for $R$
Cross-multiply to find $R$:
$$ 25\Bigl(R + \frac{1}{5}\Bigr) = 5 \bigl(5 + R\bigr). $$
Expand both sides:
$$ 25R + 5 = 25 + 5R. $$
Rearrange terms to isolate $R$:
$$ 25R - 5R = 25 - 5, $$
$$ 20R = 20, $$
$$ R = 1\ \Omega. $$
Step 5: State the Final Answer
Thus, the value of the external resistance $R$ for which the currents in both the series and parallel combinations are the same is
$$ \boxed{1\ \Omega}. $$
