Voltage drop in a series circuit

[eggsbenedict]

I am trying to help a student understand the concept of voltage drop but I'm not sure I fully grasp the concept myself.

From Conceptual Physics 9th edition

"The total voltage impressed across a series circuit divides among the individual electrical devices in the circuit so that the sum of the voltage drops across each individual device is equal to the total voltage supplied by the source. This follows from the fact that the amount of energy used to move each unit of charge through the entire circuit equals the sum of the energies used to move that unit of charge through each electrical device in turn."

How do I explain the reason that the voltage drop of the resistors must add up to the source voltage to a student? I am really having a hard time coming up with good analogies or examples.


Ok, another thing I am finding confusing. The textbook my school uses shows a diagram with a battery connected to a circuit that has one light bulb with a charge's path traced. As the charge moves towards the light bulb from the positive terminal of the battery it shows a bar graph at several points with decreasing PE. Is that because the distance from the positive terminal is increasing? It then shows the charge going through the light bulb and coming out on the other side with no PE and going back to the battery to 'get charged up'. This is confusing me. By not having an PE that means the charge could do no additional work. But does that have any effect on it's attraction to the negative terminal?

 

[wheelatharm]

A classic analogy for current and voltage is a mountain stream.

Consider the voltage to be the drop in elevation. A high elevation with a supply of energy, like from a battery, or a pond, can induce a flow of energy through mass carrying KE, like a in stream, or in the case of a circuit, current. Don't think of it exactly as a charge flow though, there are differences!

The amount of current is analogous to the amount of water flow.

The analogy for a series of 3 resistors would be where the stream steps down in a series of 3 sets of rapids. The total elevation is equal to the total voltage drop. The voltage drops of each rapids is not the same, but less than the total, and will add up to the total.
The more the stream is twisted and rocky, the more it resists the flow, and less flow is given for the same height when it is this way. In the same way, a high resistance limits current.
In between the rapids are pools that do not resist the flow much, these are wires and connectors. Like in pools the height does not change much, so there is little voltage drop.
To increase flow, you must increase elevation, voltage, or widen the stream, decrease resistance.

The 2nd part of ur ? confuses me.

 

[wheelatharm]

I you tell me the page number of the diagram of the 2nd part, I can explain, but might take a day or 2

 

[R Power]

As the charge moves towards the light bulb from the positive terminal of the battery it shows a bar graph at several points with decreasing PE. Is that because the distance from the positive terminal is increasing?

Remember, charges flowing in a circuit never come from the battery, they come from the conducting wire itself(free electrons). So there is no point that PE is decreasing because of increasing distance from battery.

This is what actually happens:- Whenever a battery powers a light bulb, the battery spreads electrical energy into space. That EM field energy is then grabbed firmly by the wires and guided by them. The field energy flows parallel to the wires, and eventually it dives into the lightbulb filament. There it drives the metal's population of movable charges forward, against the resisting force of electrical "friction." Electrons in the metal momentarily speed up before colliding with tungsten atoms. In this way the electrical energy gets converted into thermal energy. As a whole, an electric circuit is like a duct for electrical energy, but this duct has no walls.

"Try to explain your student physics instead of electronics."

 

[K^2]

^Gah. Energy flowing through space from a battery? I can see an urge to simplify the problem, but you shouldn't be using nonsense to explain something.

A battery does generate an electric field through the conductor. That electric field pushes electrons in the conductor. The potential difference between two points is the distance along the current path times the average electric field along that path. So the potential must continuously decrease as you move along the circuit from positive terminal to the negative. In places where current flows easily, electric field required to drive current is low, and so potential decreases slowly. In places where current experiences a lot of resistance, the electric field driving the current is higher, and so the potential drops faster.

Potential drop along the wire should be almost zero, assuming no short circuit, and so potential drops across loads such as lamps and resistors will make up almost the entire potential difference between the two terminals.

 

[esspeajay]

First of all dc current flows from the negative to the positive not positive to negative. in ac, it ebbs and flows creating expanding and contracting electro-magnetic fields around the conductors. In the U.S this happens 60 times a second, or 60hz. which is why a digital clock made for the U.S will not work in Europe where they operate on 50hz thereby losing about 10 min an hour. Second, voltage drops through any resistance including the wire that carries the current. If you siphon water through a hose into, let's say, four glass jugs, some water would be lost in the hose. The rest would divide equally between the four jugs. if you limit the waters ability to flow through the mouth of (resistance), let's say, two of the jugs so that it could only accept half of the water, the other two jugs would fill twice as full. to get the same amount in each of the four jugs at the same rate of flow you need to double the pressure (voltage) So, if you have 2-50 ohm resistances and 2-100 ohm resistances in series, with 120 volts applied to the circuit, disregarding the voltage dropped in the conductor, which would of course vary with the overall distance of the conductor, you would realize a voltage drop of 20 volts dropped at each of the 50 ohm resistances and 40 volts at each of the 100 ohm resistances. Add those up, and it will of course equal 120v.

An Electrician