Brighter is Mightier

A kind boy with name Steven who wants to study can sit in two rooms A A and B B . Room A A has a bulb A A rated at 60 W 60W and Room B B has a bulb B B which is rated at 100 W 100W . Both the bulbs are connected in a series ,with a battery of V V volts. He wants to study in a brighter room. In which room should he go? After going in the room the room will be locked for some hours.

Anyone B Data inadequate A

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1 solution

Steven Chase
Apr 13, 2020

This is a nice little exercise, and the result is a bit counter-intuitive, perhaps. The analysis is shown in the code below. As it turns out, the bulb with the smaller nominal power output is the brighter bulb when the two are in series.

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import math

V = 120.0           # Applied voltage from wall outlet

PAnom = 60.0        # Nominal powers at rated voltage
PBnom = 100.0

# P = (V**2.0)/R

RA = (V**2.0)/PAnom # Resistances
RB = (V**2.0)/PBnom

Rtot = RA + RB   # Total series resistance

I = V/Rtot       # Current through bulbs

PA = (I**2.0)*RA # Actual power in Bulb A
PB = (I**2.0)*RB # Actual power in Bulb B

print PA
print PB

#23.4375     Bulb A is brighter, despite the lower nominal rating
#14.0625

@Steven Chase Exactly. Can these question will come in problem of week??

A Former Brilliant Member - 1 year, 1 month ago

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Brilliant used to do problems of the week, which were taken from the community section. They have since moved to the "daily problems" format, which features problems written by the staff.

Steven Chase - 1 year, 1 month ago

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So can my problems move to daily problems??

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member Probably not, since those are generally written by staff members

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase Sir please can you post more questions on Electricity and magnetism in which calculus can be done in wolfram,today. I am getting bored living in home only .

A Former Brilliant Member - 1 year, 1 month ago

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