A 3 V flashlight battery has an internal resistance of 1 Ω when new, and 6 Ω when old. Imagine this is a resis?
A 3 V flashlight battery has an internal resistance of 1 Ω when new, and 6 Ω when old. Imagine this is a resistor in series with the battery mounted INSIDE the battery itself, which describes the chemical degradation of the battery with age. Calculate the voltage across a 13 Ω lamp (the load) when the battery is new. Express the answer with one decimal place.
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The total load across the "perfect" 3V battery = 13+ 1 Ω = 14 Ω
So the current through the load is 3V / 14Ω = 0.214 A
So the voltage across the lamp is V = 0.214A * 13Ω = 2.8 V (1 d.p.)
[2.8V is the battery's terminal voltage on load. The internal volt drop across the 1Ω resistance is 3 - 2.8V = 0.2V = 1Ω * 0.214 A )
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