# Finding out how long a light bulb will light [closed]

A cylindrical wire used to form a light bulb filament has radius 3.7 micrometers and length 1.7 cm. The resistivity of the wire is 5.25 * 10^-5 ohm meters. The light bulb is connected to a 12V battery.

With the given information; resistance equals 20.75 ohm meters. Current will be 0.58 amps and power will be 6.94 W.

The question I do not understand the meaning of is, if the battery has a total stored charge of 0.5 A hr, and produces a constant potential difference until discharged, how long will the light bulb light?

Can anyone explain to me what the question means?

• The total energy stored in a battery is its voltage $V$ times the current it produces $I$ times the length of time it can produce that current $t$ i.e. $E = VIt$. Since $VI$ is just the power (in Watts) the energy is just power times time. In this case you're told the battery capacity is $0.5$A hours and this means the total energy stored in the battery is $12$ volts times $0.5$ amps time $3600$ seconds. Divide this by the power drawn by your bulb and you have the time the battery can power the bulb. Aug 25, 2016 at 5:25
• @JohnRennie Thank you very much sir! I would really appreciate it if you could explain to me why it is necessary to divide (E = V It) by the power that is drawn by the bulb. Thank you again Aug 25, 2016 at 5:50
• @JohnRennie Also if you could explain why the units of time is in seconds. Aug 25, 2016 at 6:02
• @Utsav Time is in seconds because the SI unit for time is seconds. The reason JohnRennie says divide energy by power is that energy / power = energy / (energy / time) = time, which is what you were after. Aug 25, 2016 at 6:31

You've done all the hard work and now there's one little part left.

You're being asked, given a certain battery capacity in amp hours and a certain current in amps flowing out of the battery, for how many hours (or seconds) will the current flow? The assumption is that as long as the current is flowing the light will be on and as soon as the capacity of the battery is drained the light will go off.

Explanation

You know the power the bulb will use. You know the voltage of the battery, you know the total charge store in the battery in the form "amp hours" (A hr). Amp hours are an engineering term for charge and are equivalent to coulombs.

1 amp flowing for 1 hour = 1 coulomb / second flowing for 3600 seconds = 3600 C

A battery capacity of 1 amp hour means that:

• a current of 1 amp can flow for 1 hour
• a current of 2 amps can flow for 0.5 hours
• a current of 0.25 amps can flow for 4 hours

Notice that in each of these the current (amps) * time (hours) = capacity (amp hours): 1 * 1 = 2 * 0.5 = 0.25 * 4 = 1

A constant potential difference across the light means that the current will also be constant. In the real world as batteries discharge the voltage across their terminals decreases. This drop in voltage is due to internal reistance of the battery increasing as chemical reactions take place inside the battery but that's outside the scope of this question. If the driving voltage wasn't contant the current would also change which makes the problem harder.

• @Utsav You're welcome. Please feel free to click the "up button" that appears at the top left of each answer if you feel it is useful. Doing that will help other people find useful information. You can do this for any answer to any question no matter who asked and it doesn't have to be the best answer, just a good answer. As the person who asked the question you have an extra privilege to mark an answer as "accepted" if believe the answer satisfies your question but you can do this only once. Edit: and I can see that you did that as I was typing this comment. Thanks. :) Aug 25, 2016 at 6:25