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Name: Andrea
Status: other
Grade: 9-12
Country: United Kingdom
Date: N/A 


Question:
In a series DC circuit where there are two equal bulbs how do electrons know to save some energy for the second bulb?



Replies:
Andrea, let me give you an analogy that perhaps might help. Imagine your light bulbs are like a ball rolling down a smooth, hard ramp. After rolling a little while, the ball picks up some speed, but then the ball hits a section of soft sand. The ball slows down, but it still keeps rolling. Then some more smooth ramp, then another section of sand. The ramp is like the wire, and the sand is like the bulb. There is a voltage drop across a light bulb because some of the energy stored in the voltage is converted to heat and light -- in the same way that the sand takes some of the energy of the rolling ball (although the sand does not light up). The nature of the bulb (and the nature of the sand) is what determines how much energy is used (different wattage bulbs might be like softer or harder sand). In the same way the ball is simply following the rules of physics when it rolls, so too is the electrical current just following the rules of physics. Now, suppose the bulb were to use more energy than the cord could supply. Then, you would see a difference in the second bulb. But it is not that the electricity "knows" anything, it is just that it is following the path set forth in front of it (the wire, and the bulb).

Hope this helps,
Burr Zimmerman


Andrea,

Before the switch is closed, the electrons are evenly distributed through the wires and the bulbs. When the switch is first turned on, electrons throughout the circuit start to feel electric field within the circuit. They all begin to move. Electrons find moving through a bulb more difficult than moving through wire: they build up at the entrance of the bulb. This happens at the entry of all the bulbs. The shift in electron distribution gives electrons the extra push they need to get through the bulbs. If too much energy is lost in the first bulb, there will be a greater buildup at the second bulb. This reorganization causes energy and current distributions to quickly balance out. When total energy lost during one cycle through the bulbs balances the energy gained from the battery, there will be no more change in buildup of electrons. This readjustment occurs while the bulbs are going through the process of turning on.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College


Hi Andrea

To answer your question directly: The electrons that are enter the circuit do not see two resistors. They see the total resistance of the circuit which is the sum of the two series resistors. In your case, I think you are assuming that both bulbs have the same amount of resistance But here is an example where the resistances are different to demonstrate how the electrons work in the circuit.

(referring to the attached drawing)

Since there is only one path, the (positive) current entering at T1 must equal the current leaving at T2 and the voltage drops across R1 and R2 must equal the voltage applied across the terminals.

So if R1 = 5 ohms and R2 = 25 ohms, the total resistance is 30 ohms, and let us say that the applied voltage from T1 to T2 is 60 Volts Using Voltage (V) = Current (I) times Resistance (R), the current I passing through both resistors is 60 Volts / 30 Ohms = 2 Amps. So 2 Amps passing through R1 at 5 ohms gives a voltage drop of 10 Volts 2 Amps passing through R2 of 25 ohms gives a voltage drop of 50 Volts Giving a total voltage drop of 60 Volts.

Negative current made up of electrons give the same result, they just flow from T2 to T1 and the voltage polarities are as shown.

Sincere regards,

Mike Stewart


The bulbs do not "have to know" how to divide up the electric current. That happens "automatically" -- following various modifications of
Ohm's Law: E (voltage) = I (electrical current) x R (electrical resistance).

That is: E = I x R

Perhaps an example from the flow of water through a hose or pipe will help clarify the issue. In the case of water: P (pressure drop) = Q (water amount) x L ( pipe resistance to water flow). That is: P = Q x L .

Notice the direct analogy between the flow of electrons and the flow of water: P ~ E; Q ~ I; L ~ R

In both examples the fluids (i.e. water amount "Q", and electrical current "I") behave like an incompressible fluids. The flow of the electrical current, or water. The "I" or the "Q" does not have to "know" anything. Also consider a water fall. The water does not have to "know" that it is going to tumble off the falls. Gravity demands it. The river does not have to "save" any part of the water for various parts of the river. It all obeys the laws of fluids and gravity.

Vince Calder



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