*Very* basic battery question

[coder1]

Ok. I am about as ignorant as my grandmother when it comes to basic electrical understanding, so bear with me on this question.

I have a 6-volt batery for one of my flashlights (large flashlight) Also in my alarm clock there is a 9-volt battery. Now, the 6-volt is a much larger battery in size and weight. The 9-volt is... well you all know what a little 9-volt looks like. Well here is my question? Does the smaller 9-volt just put out more voltage & ampers for a shorter amount of time compared to the 6-volt that would put out less voltage & ampers, but for a extended amount of time. For instance if I hooked up the larger lattern-flashlight that requires a 6-volt to the little 9-volt would it be extremley bright but for a very short time?

Ok, thanks for any insight in this simple subject that thats has seemed to elude me over the years.

 

[InverseOfNeo]

You are about halfway on the right track. There is also this thing called load resistance or internal resistance of a battery. Without going into a lot of engineering explanation about that, I will just say its how much load a battery can power. A small everyday 9V battery will not be able to make the lantern work because the lantern exerts too much load on it. Thus a bigger battery is needed because for the most part, the bigger a battery the bigger the internal resistnace and the more load it can power. Same reason why we dont use 8 C or D cell batteries in our car. The load is much too great for them and would last less than an hour.

 

[OrganizedChaos]

yes it would but it dosen't work like that. power is measured in watts. 1 volt x 1 amp x 1hour =1watt/hour. voltage is detemined by the number of cells in the battery with each cell in this case being 1.5v. current is determined by the load placed across a voltage, in this case its a light bulb. batteries something called an amperehour rating. your avrage 9 volts we'll say has a raing of .130AH. that means it can sustain a .130 amp load for 1 hour before its dead. your bigger 6v battery contains 4 1.5v cells in series. each cell being about the sice of 2 C cells placed end to end. the 6v will have a AH rating of about 12AH many times larger than that of the 9volt.

ex. a light bulb that could kill a 6v battery in 1 hour would be have a resistance of .5 ohms. 6/12=.5 . this same lightbulb placed across a 9v would draw 50% more current (remember voltage X current =wattage) thats 18 amps of load on a battery with a .130 AH rating. T=C/I so that nine volt would last for .007 hours.

to top that off nothing i just said works in the real world. there are many more variables to account for such as the interneal resistance of the battery, peukerts number, tempreture and manufacturing variances and to top it all off batteries aern't very linear in high drain applications

so in the end you'll end you have a 50% brighter light for maybe 10 seconds


/disclaimer not an EE

 

[her209]

Using a plumbing analogy:

Think of V as water pressure.
Think of I as as water flow.
Think of R as the diameter of pipe.

And remember that water flows from a point of higher water pressure to lower water pressure.

 

[Ornery]

Electricity/Water Analogy With Diagrams

The 9V battery "squirts the juice" out at a higher pressure than the 6V battery, but can't do it for nearly as long. Its power reserve could be spent in just a few seconds, compared to the hours the 6V would last. When the 9V is used in the correct application, a load with very low resistance or short duration, it can last for months.

 

[AyashiKaibutsu]

Voltage is the difference between the two ends. Resistance is how hard it is for electricty to travel. Amperage is the amount of power that goes across it (determined by the voltage and resistance). V/R=A. Basicly electricty will flow from the high end to the low end. Resistance is how hard it is for the electricty to travel across the two ends. Amperage is the end result of electricty transfered. For those who dislike analogies.

 

[coder1]

Thanks for everyones help