Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Stby on 23/01/2008 00:07:11

Title: Loss of current in a bulb?
Post by: Stby on 23/01/2008 00:07:11: Hello!

I'm new to teaching conceptual physics and electronics has come up and the kids are asking questions that I'm having a hard time answering.

In a simple circuit (battery, wire, bulb) Ohm's law indicates that the current is going to be the same no matter where the ammeter is placed in the circuit. If the current is flowing from the battery through the bulb and then back to the battery again, what is "missing" when the current comes back out of the bulb if it isn't the current or voltage?

Are we trying to mix an ideal situation that can be found on paper when drawing a circuit where there is no acknowledgment of loss of energy and the real world?

Thanks!!
***Stby
Title: Loss of current in a bulb?
Post by: another_someone on 23/01/2008 05:23:45: The loss of energy comes from the loss of charge in the battery.

The current represents a flow of of charge (a current) that is going from the negative terminal of the battery to the positive, and as it does this, the available charge in the battery is depleted.

You also have a drop in voltage across the bulb - since the sum of all the voltages in the circuit must come to zero - so, ignoring the slight resistance of the wire, the entire voltage of the battery much be matched by an equivalent drop in voltage across the bulb.
Title: Loss of current in a bulb?
Post by: Pumblechook on 23/01/2008 15:13:44: Power = voltage x current.

A resistor (of suitable resistance in each case) with only 10 mA flowing through it but 2000 Volts across it will get hotter than one with 1 A flowing with 12 Volts across it.

Try measuring the resistance of a bulb and then calculte the resistance by measuring the current and voltage with the bulb fully lit.
Title: Loss of current in a bulb?
Post by: daveshorts on 23/01/2008 15:41:57: I am not sure if this is the confusion, but the resistance of a bulb is by no means constant, when a metal heats up normally its resistance increases, so if you measure the resistance of a bulb cold it will be a lot less than when it is running and hot.

When an electron moves through the bulb it's electric potential energy reduces, which is measured in volts. You can think of it as on the negative side of the battery the electrons are slightly compressed together and because they repel one another this takes energy, this energy can be released by letting the electrons move apart slightly as they move through the bulb.

Your multimeter measures the difference in potential (potential energy per unit of charge) between two places. If we measure the potential from the + end of the battery on the -ve side there is -1.5V of potential, but if you measure between the + and the + there is 0V of potential, so as the charge moves through the bulb is looses potential, and therefore potential energy which ends up being turned into light.
Title: Loss of current in a bulb?
Post by: techmind on 23/01/2008 19:04:20: Quote from: Stby on 23/01/2008 00:07:11
Hello!

I'm new to teaching conceptual physics and electronics has come up and the kids are asking questions that I'm having a hard time answering.

In a simple circuit (battery, wire, bulb) Ohm's law indicates that the current is going to be the same no matter where the ammeter is placed in the circuit. If the current is flowing from the battery through the bulb and then back to the battery again, what is "missing" when the current comes back out of the bulb if it isn't the current or voltage?

Are we trying to mix an ideal situation that can be found on paper when drawing a circuit where there is no acknowledgment of loss of energy and the real world?

Thanks!!
***Stby
It is the voltage that is "depleted" (if you measure always with reference to the same point, eg the negative terminal of the battery) when it comes out of the "other side" of the lightbulb. Energy = current X voltage ; current is the same everywhere, but the voltage has dropped, so energy has been used somewhere (in the bulb).
Title: Loss of current in a bulb?
Post by: Pumblechook on 23/01/2008 20:09:42: Split hairs maybe.

Energy = Voltage x Current x Time.
Title: Loss of current in a bulb?
Post by: lyner on 23/01/2008 22:09:10: What is 'conceptual physics'?
How does it differ from the Physics in School textbooks?
A good textbook will explain the concept of electrical potential.
There is a good analogy with the gravitational potential which is gradually transferred as a ball rolls downhill. The ball doesn't get any less but the energy it has, by virtue of its position, is transferred to kinetic and thermal energy.
Title: Loss of current in a bulb?
Post by: lyner on 24/01/2008 18:06:56: One Joule of energy is transferred when the potential of one Coulomb of charge changes by one volt.
Or; 1 Joule is 1 Volt per Coulomb
Title: Loss of current in a bulb?
Post by: Stby on 24/01/2008 18:19:19: Thank you everyone, I was hoping there would be an easy answer so I'm going to have to digest what everyone here is saying. :)

Conceptual Physics covers the same topics that a Physics class would cover but they are covered at a ~conceptual~ level - there is less math involved. It works well as a high school freshman course to get the basics covered so that when a Physics course is taken later in school where the equations come in to play the foundation is already solid.