Using a conventional flashlight bulb and a 1.5 v. battery, a bulb burns at a certain brightness. Using a 9 v. battery, it doesn’t burn as bright. At least, that’s what a student of mine is reporting, and I don’t understand what would cause that to occur. I would not be surprised if the bulb burnt out with the greater force of the 9 v. battery, but I don’t know why it would glow less bright. She said that a text she was using predicted that would happen. What is the basic straight dope on this? thanks. xo C.
The current the 9v can put out is very small and the battery just can’t produce enough and maintain 9V. An interesting experiment for your student, have them measure the voltage drop across the lightbulb.
I guess I need a more basic course in electricity, but I don’t see why this should be. The old trick of putting a 9 v. battery on your tongue seems to indicate a good bit of current is flowing. Why would a battery with more electromotive force produce less current? Why would the voltage drop from one side of the bulb to the other?
The battery could be kaput. It’s not producing 9 V of EMF. If you licked the battery this particular student is using, you wouldn’t get so a zetz. (I find electricity much easier to understand in Yiddish, don’t you?)
Because of Ohm’s law: V=IR. The voltage drop across a resistor (such as a bulb) equals the current times the resistance. The bulb is using up energy (turning it from electrical potential energy into light and heat.)
Oy vey is mir. That should read: “you wouldn’t get so much of a zetz.”
Because that is the way it is designed. Otherwise it wouldn’t last long enough to make it effective.
The internal resistance of a 9V battery is 1200 mOhm.
The internal resistance of a D cell is 126 mOhm.
The 9v simply cannot supply as much current as a D-cell.
Ok, but my basic intuition is correct, yes? A higher voltage battery SHOULD cause a bulb to burn brighter. I really knew that the resistance of the bulb would cause the voltage to drop somewhat. I don’t know why I included that in my OP. But are you saying that unless the battery is shot, the higher voltage battery should produce more light? Jeez, this would be so much more easy if we were all just sitting around talking. Hey, you guys want to come over this afternoon?
Crap. That should be 1700 mOhm.
Ok, let me get this straight. My intuition was wrong. A 9 v. battery produces only a small amount of current because of its internal resistance. It comes out fairly forcefully because of the 9 v. but there’s not much of it. Therefore, it doesn’t have enough current to light a bulb very brightly, whereas a 1.5 v. battery, because of it’s lower internal resistance, produces a greater amount of current, enough to light the bulb brighter, even though the current is coming out of the battery with less force. Do I have that correct?
IIRC, it might help to think of current as “mass” and voltage as “speed,” although this is not really very accurate (voltage is potential). Some electronic things need the “speed,” while others need the “mass.” So a flashlight can give you a bright light for quite a while with two cells (3 volts) or four cells (6 volts) in series; it needs the large current because the light and heat produced give off a lot of energy. Meanwhile, a digital phone message machine doesn’t need that much current (it’s all solid state semiconductors), so it can run off of a 9 volt battery. Just the comparative heft of the two things would seem to indicate so. (Also, the “D” is a cell, while the “9” is a battery–a collection of cells.)
A little Law of Ohm shows that the 1.5v D cell can source just under 12 amps to a dead short and the 9v batter can source a bit over 5 amps to a short. Juggling that into watts: 212 watts for the D cell and 252 watts for the 9 volt on the assumed dead short.
The datasheets from Duracell show that a 9v battery will die in very short order if called on to deliver even one watt. The D-cell will deliver a watt for three to four times as long.
Unless someone adds some scientific rigor to this, I’m inclined to say that the 9v batter in question is kaput.
I am fairly certain your average 9V battery is actually just 6 1.5V batteries in series. They could deliver no more current than any single 1.5V could. For more current, you need to put batteries in parallel, not in series.
No you need a course in electrical engineering. In the basic courses you learn with ‘ideal’ power sources and superconductor wires, it’s only in advanced classes that you factor in that there are real world limitations like batteries have a limited ability to supply power and wires do have resistance.
I need a course in electrical engineering to understand the answer to my question? hmmmm Things are out of whack.