A cell with an e.m.f of 1.5 V is connected in series with an external resistance of 5 . If the current flowing through the circuit is 0.25 A, calculate the internal resistance of the cell.

The electromotive force (e.m.f) of the cell is $E = 1.5 \text{ V}$.
The external resistance connected in series is $R = 5 \Omega$.
The current flowing through the circuit is $I = 0.25 \text{ A}$.
We need to find the internal resistance, $r$.

The total resistance in the circuit is the sum of the external resistance and the internal resistance, $R_{total} = R + r$.
According to Ohm's law for a circuit with internal resistance, the current $I$ is given by:

Multiply both sides by $(5 + r)$:

Distribute 0.25:

Subtract 1.25 from both sides:

Divide by 0.25:

The internal resistance is $1 \Omega$. This corresponds to option A.

The final answer is $\text{A. } 1 \Omega$.