When four identical light bulbs are connected in parallel and this combination is connected to an...

When four identical light bulbs are connected in parallel and this combination is connected to an ideal battery, the bulbs each glow with a certain brightness until the battery runs down after six hours. [Recall that an ideal battery is one with no internal resistance.] If two of the light bulbs are removed so that there are just two light bulbs in parallel and then this combination is connected to a new (but otherwise identical) battery, Question 6 options:

the remaining two bulbs will each glow with the same brightness as each bulb did before and the battery will last for six hours.

the remaining two bulbs will each glow more brightly than each bulb did before and the battery will still last for six hours.

the remaining two bulbs will each glow more dimly than each bulb did before and the battery will last for twelve hours.

the remaining two bulbs will each glow with the same brightness as each bulb did before and the battery will last for twelve hours.

the remaining two bulbs will each glow more dimly than each bulb did before and the battery will last for 24 hours.

the remaining two bulbs will each glow more brightly than each bulb did before and the battery will last for three hours.

Expert Solution

Consider that each bulb has resistance r.
First case.
When 4 bulbs are connected in parallel,
Net resistance, 1/R1 = 1/r + 1/r + 1/r + 1/r = 4/r
R1 = r/4
Take the voltage in the battery as V.
Current drawn, I1 = V/(r/4) = 4V/r
Power = VI1 = 4V²/r
Battery lasted for 6 hrs,
Total energy the battery have = 4V²/r x 6 = 24V²/r

Second case.
Now only two bulbs are connected in parallel.
Net resistance, 1/R2 = 1/r + 1/r
R2 = r/2
Voltage of the battery is same as V
Current drawn by the bulbs, I2 = V/(r/2) = 2V/r,
Since I2 = 1/2I1, bulbs will glow dimly than before

Total energy of the battery is same for both the cases
Take that the battery lasted for t hrs
Energy delivered in t hrs
= Power x time
= VI2 x t = (2V²/r)t

Equating the energy in both the cases,
24V²/r = (2V²/r)t
t = 12 hrs
the remaining two bulbs will each glow more dimly than each bulb did before and the battery will last for twelve hours.

genius_generous answered 2 years ago

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