We have following data of current in first wire:

Current \(\displaystyle{I}_{{1}}={\frac{{{Q}_{{1}}}}{{{t}_{{1}}}}}={4}{A}\)

The current in a second wire that delivers twice as much charge in half the time can be calculate as \(\displaystyle{I}_{{2}}\)

In general we have:

\(\displaystyle{I}={\frac{{{Q}}}{{{t}}}}\)

If \(\displaystyle{Q}_{{2}}={2}{Q}_{{1}}\) and \(\displaystyle{t}_{{2}}={\frac{{{t}_{{1}}}}{{{2}}}}\), we get:

\(\displaystyle{I}_{{2}}={\frac{{{Q}_{{2}}}}{{{t}_{{2}}}}}={\frac{{{2}{Q}_{{1}}}}{{{\frac{{{t}_{{1}}}}{{{2}}}}}}}={4}{\frac{{{Q}_{{1}}}}{{{t}_{{1}}}}}\)

\(\displaystyle{I}_{{2}}={16}{A}\)

Current \(\displaystyle{I}_{{1}}={\frac{{{Q}_{{1}}}}{{{t}_{{1}}}}}={4}{A}\)

The current in a second wire that delivers twice as much charge in half the time can be calculate as \(\displaystyle{I}_{{2}}\)

In general we have:

\(\displaystyle{I}={\frac{{{Q}}}{{{t}}}}\)

If \(\displaystyle{Q}_{{2}}={2}{Q}_{{1}}\) and \(\displaystyle{t}_{{2}}={\frac{{{t}_{{1}}}}{{{2}}}}\), we get:

\(\displaystyle{I}_{{2}}={\frac{{{Q}_{{2}}}}{{{t}_{{2}}}}}={\frac{{{2}{Q}_{{1}}}}{{{\frac{{{t}_{{1}}}}{{{2}}}}}}}={4}{\frac{{{Q}_{{1}}}}{{{t}_{{1}}}}}\)

\(\displaystyle{I}_{{2}}={16}{A}\)