A wire carries a 4 A current. What is the current in a second wire that delivers

A wire carries a 4 A current. What is the current in a second wire that delivers twice as much charge in half the time?

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James Kilian
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:
$$\displaystyle{I}_{{2}}={\frac{{{Q}_{{2}}}}{{{t}_{{2}}}}}={\frac{{{2}{Q}_{{1}}}}{{{\frac{{{t}_{{1}}}}{{{2}}}}}}}={4}{\frac{{{Q}_{{1}}}}{{{t}_{{1}}}}}$$
$$\displaystyle{I}_{{2}}={16}{A}$$