A bank of 80 1.5-volt flashlight batteries in series is prepared. The voltage di

Question

A bank of 80 1.5-volt flashlight batteries in series is prepared. The voltage difference across the bank is 120 V. A 120-volt bulb that operates at 25 watts glows at normal brightness when connected across the bank of batteries. The 25-watt bulb is then removed. A 120-volt bulb that usually operates at 500 watts glows only dimly when connected across the same bank of batteries. Explain.

I'm thinking of this formula: P=I^2R, so when power increases, resistivity increases, thus decreasing brightness. However, I'm not sure if this is correct. I read somewhere that brighter bulbs have a higher resistivity? But I don't know how that makes sense.

Explanation / Answer

That's not quite correct. Since power is he equal to the product of voltage and current (P = EI) and voltage is constant at 120V, the higher power bulb must have higher current flowing through it. Therefore its resistance must be less ... the resistance (at operating temperature) of a 500 watt bulb is much less than that of a 25 watt bulb.

The reason the 500 W bulb is dimmer is the internal resistance of the batteries. Basically you have a bulb in series with 80 resistors as well as 80 voltage sources -- although individually the resistance of each battery isn't all that high, collectively it is high enough to affect the light.

Look at it this way: the 25 watt bulb has resistance of 576 and the 500 watt bulb only 29 (in each case calculate from P = E2/R). So if each battery has resistance of say, one ohm, the 80 ohms affect the 25 watt bulb very little but the 500 watt bulb a lot.

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