A 8 V flashlight battery has an internal resistance of 0.6 Ω when new, and 5 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 14 Ω lamp (the load) when the battery is old. Express the answer with one decimal place.
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Verified answer
The total load across the "perfect" 8V battery = 14+ 5 Ω = 19 Ω
So the current through the load is 8V / 19Ω = 0.421 A
So the voltage across the lamp is V = 0.421A * 14Ω = 5.9 V (1 d.p.)
[5.9V is the battery's terminal voltage on load. The internal volt drop across the 5Ω resistor is 8 - 5.9V = 2.1V = 5 * 0.421 A )
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