002 How can a 2V EMF outsine a 3V EMF?


While a battery provides the EMF (E) to push current through the external circuit, it also comes with its own internal resistance (r). The amount of EMF spent to push current through its own internal resistance is directly proportional to the current passing through the battery (Ir). So the amount of EMF left to push current through the external circuit, called the terminal potential difference (Vt), decreases whenever the battery is made to pump out larger current.

In this video, as we connect more and more bulbs (in parallel) across the battery, we draw larger and larger current from the battery. This must lead to a larger and larger potential difference across the internal resistance of the battery, which implies a smaller and smaller terminal pd, which dims the bulbs.

There is also a simple explanation why the 2V battery could “outshine” the 3V. The 2V battery may have a lower EMF, but by having a smaller internal resistance, it provided a higher terminal pd to the external circuit.


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