Given,
The current, I=0.25 A
The total charge flowing in the circuit is q=6812 C.
The electric current can be defined as the time rate of flow of the charges.
Thus, the electric current is given by,
[tex]I=\frac{q}{t}[/tex]Where t is the time duration.
[tex]\begin{gathered} 0.25=\frac{6812}{t} \\ t=\frac{6812}{0.25} \\ =27248\text{ s} \end{gathered}[/tex]Thus the batteries will fry out after 27248 s
