A cell of emf 1 . 8 volts gives a current of 17   A when directly connected to an ammeter of resistance…

Let the Internal resistance of the cell is $r,$the ammeter connected in series showing a reading of $17 \mathrm{A}$ is the current flowing in the circuit having resistance, $0.06 \mathrm{\Omega }$

Therefore, $E=Ir+IR$

$$\begin{gathered} \Rightarrow r=\frac{E-IR}{I}=\frac{1.8-\left(17\times 0.06\right)}{17}=0.0458 \mathrm{\Omega } \\ \Rightarrow r=0.0458 \mathrm{\Omega }=0.046 \mathrm{\Omega } \end{gathered}$$