Why does a weakened 9V battery show lower voltage but a weakened 1.5V battery doesn't?
A ""typical"" ""1.5v"" Alkaline C ""Battery"" is a single Cell (so not really a battery), ranging from ~1.6v at full charge, averaging 1.5v to 1.3v for most of it's life, eventually rapidly dropping to ~0.9/0.8v.

A ""typical"" ""9v"" Alkaline Battery (notice, no quotes) is composed of multiple 1.5v cells in series. So even in a perfect scenario, where all 6 cells discharge at an equal rate, the lower voltage of each cell is multiplied. The same drop from 1.5v to 1.4v (under load) for a C battery would mean 0.1v * 6 or 0.6v for a 9v Battery.

THEN we have the problem that a 9v battery is designed for higher voltage and a long life via low current draw. A C battery, on the other hand, is designed for high current draw and higher capacity. For comparison, a typical 9v is ~400mAh while a C is ~8000mAh. That is 20 times the capacity. The tradeoff? Lower voltage, higher weight and size.

For clarification, electrical semantics dictate that A Battery is a device consisting of multiple cells, while a cell is a device that converts chemical to electrical energy.

And this ALL varies based on manufacturer and quality of the battery and battery chemistry. Some Alkaline 9Vs are composed of AAAA batteries, while other's have fat stack types inside. Carbon Zinc are weaker, Lithium stronger, etc etc.