# Two batteries in parallel: why they deliver “twice the current for the same time”?

I am reading "Make Electronics - 2nd edition" by Charles Platt, and I am looking at batteries in series in parallel right now. There's a sentence I really can't understand:

Figure 1-73. Batteries in parallel, powering the same load as before, will run it for for about twice as long. Alternatively, they can provide twice the current for the same time as a single battery.

What puzzles me is the last part: if the V stays the same, how can the battery provide twice the current for the same time? Are we talking about an implicit resistance drop (but in that case I wouldn't need the second battery, would I?), or what, exactly?

• The battery does not provide twice the current, two batteries provides the same current. Overall current is twice as you have two batteries instead of one. – Bence Kaulics Oct 26 '16 at 13:50
• There should be another thing highlighted in the book - that batteries themselves are having resistance, and connecting those two may cause, at some unfortunate circumstance, to have one being discharged through another if latter will appear having defect or being "less changed" than former one. Thus practice may differ with idealistic theory because theory does not take some conditions into account. – Anonymous Oct 26 '16 at 14:12
• The "internal resistance" of each battery matters; when you put them in parallel, the internal resistances are effectively in parallel, and thus halved. – pjc50 Oct 26 '16 at 14:17
• Oh, and "provide twice the current" should not be read as "will always output twice the current" - current draw is still determined by the circuit connected. – pjc50 Oct 26 '16 at 14:18

Batteries' charge, $Q$, equals to the product of current drawn and the duration: $Q = I \cdot t$ or $\Delta Q = I \cdot \Delta t$.
Think of it like this: For parallel batteries, $\Delta Q$ is same for each, but $I$ is half per each (Because they both supply the same load with equal -half- contribution at the same time). Yields $\Delta t_{new} = \frac{\Delta Q}{(I/2)} = 2\cdot\Delta t$.
Now we have total charge of $2Q$. If you keep $\Delta t$ constant, $\Delta I_{new} = 2I$.