PAPER ASSIGNMENT #1: ELECTRIC CIRCUITS Due at the beginning of class Saturday, February 9, 2008

Transcription

1 PHYS Foundations of Science II By Richard Matthews PAPER ASSIGNMENT #1: ELECTRIC CIRCUITS Due at the beginning of class Saturday, February 9, 2008 Part I; Outline of the important elements of the current model for the behavior of electric circuits. A. Electric circuits. 1. Materials. a. Battery b. Wire (conductor) c. Bulb (resistor) battery bulb wire switch d. Switch (to allow current to flow or not to flow) 2. Complete closed circuit. a. Battery with positive and negative terminals. b. Conductor connects both the positive and negative terminals. c. Coated or insulated wire must have bare, metal wire exposed. d. Direction is not determined, nor important. battery Resistors (obstacles in the path of electric current). a. Indicators of electric flow (light bulb). b. Switch (opens / closes electric flow) c. nichrome wire battery bulb switch

2 4. Requirements met in order for bulb to light. (Battery, bulb, and wire connection). a. Operational definition. Wire should have at least two exposed or bare ends. Battery with positive and negative ends with reasonable charges. The bulb should be compatible to its energy source. Bulb should have two wires or inner filament connected to the base and side of metal bulb. Lighting the bulb using the copper or external wire, touch and connect an exposed end to the side of the metal part of the bulb. This can be accomplished by wrapping the exposed wire around the metal end of the battery. Take the other end of the exposed end of the wire and connect it to the negative side of the battery. Now touch the base of the bulb to the positive end of the battery. The bulb should now light. b. Conductors and Insulators of electricity. Conductors- metal, metal alloys, lead and water. Insulators- plastic, rubber, styrofoam, glass, and porcelain. Early in the course, an investigation was performed to observe what materials were defined as conductors of electricity and which are the insulators. Those that were the conductors were then categorized by whether electricity was a continuous current flow, if the indicator bulb was brightly lit or dim. Finally, whether the indicator bulb would light at all. The follow are the results:

3 Objects Stayed lit Brightly lit Dim Did not light penny nail screw (steel) aluminum plastic nichrome paper rubber pencil (wood) glass lead porcelain water (not tested) Upon conclusion of this investigation, it justified the statement mentioned above; that metals, metal alloys, and lead are conductors of electricity. Water was not attempted. Copper wires and later, nichrome wires would be used in most of the circuit models constructed in class. 5. Circuit Diagrams. In order to draw circuit diagrams correctly, the following symbols were introduced and used frequently in class as well as on the homework assignments. This allowed me to recreate, on paper, actual electric current models that were both constructed and demonstrated with in class. It in turn helped me see and understand the concept of a series circuit and a parallel circuit and the relationships between indicator bulbs when more than one was placed in the circuit in terms of their brightness.

4 battery current path indicator bulb switch (open circuit) node resistor wire 6. The following circuits were constructed, demonstrated with and discussed. a. Open b. Closed c. Short d. Series e. Parallel f. Parallel within a series (components) A open closed bulb A shorts out when switch is closed series parallel parallel w/in a series

5 B. Electric current flow. 1. Requirements met in order for bulb to light. (inside the bulb) a.the bulb (an indicator) filament glass housing support post screw threads knob / rivet base b. Lighting the bulb. To light a bulb using a battery, copper wire and bulb- the low voltage must be in a complete circuit motion where all parts or components are operational. The inside of the metal part of the bulb must have two components (the base, soldered area and the aluminum side). The wire only needs to touch one of These components while the other touches the battery. Lighting of the bulb is not going to matter if the battery is flipped, with bulb touching the negative end, or if the bulb is turned on its side.

6 2. Movement through the circuit. a. Battery terminals. In a closed circuit, current is a continuous flow from the positive battery terminal to the negative battery terminal is round trip. In other words it leaves the battery and returns to the battery s opposite end (+ and / or -) and is not used up. Direction is not determined. The conservation of current is Kirchhoff s first rule; energy is conserved. This is true for all closed circuits. b. Current through the resistor(s). In the models and demonstrations performed in class, small light bulbs were used to indicate that electricity moved through the circuit network. In a two bulb, series circuit, current will pass through each bulb lighting them equally. The bulbs in this network will be dimmer than a similar network with just one bulb. So, it is safe to suggest that for each bulb added in a series network the dimmer the bulbs will become until they eventually don t light. Thus as resistance increases, the current flow decreases. For current to pass through a parallel circuit network, it must split at a junction that we later referred to as a node. By arranging bulbs in a parallel circuit network, current is increased, thus lighting bulbs equally. The bulbs in a two bulb, parallel circuit network will be brighter than the two bulbs in a series circuit network. This is so because, although each network has the same number of bulbs, the current is equal to or greater than the resistance in the parallel network. c. Pathways (branches), obstacles and nodes (conservation of current). Current will flow freely through a closed network with no obstacles. Obstacles in the pathway, such as bulbs, will use some

7 of the current as it passes through it. When current comes to a junction or node, current will split and travel through all branches. The total current out of a node is equal to the total current into the node. So the algebraic sum of the currents at a node is zero. Because the current splits one bulb will receive more current than the other due to the separate pathways. As branches are added, and / or removed current is unchanged, thus bulb brightness is not affected. d. Independent and dependent branches in the network. Independent = no change when adding or removing bulbs and / or branches. Bulbs in a parallel circuit network are Independent of one another. The bulbs will show no significant change. Dependent = effects occurring on other bulbs such as its brightness. e. Linear resistors. Nichrome wire- in conducting experiments in class, it was discovered that, in series circuits, when the length of nichrome wire was increased the resistance was also increased; however the current through the entire circuit network was decreased. The length of the resistor wire is directly proportional to resistance and inversely proportional to current flow. As linear resistors are added in parallel circuits, the current flow is increased as observed in class. When there exists multiple branches in a parallel circuit network with varying numbers and lengths of nichrome wire, the current can be distributed equally by using the quantitative method Req = R/N (the length of resistance over the total number of branches. Linear resistance is having a resistance essentially independent of the current through them.

8 Part II; Brightness of the indicator bulbs within circuits 1 and 2 (independently of each other). A B C F H G I J D E (circuit 2) (circuit 1) In circuit one, the ranking of bulb brightness is based on current flow from the battery s positive terminal end to its negative terminal end and the positioning of the bulbs within the components. In this closed, series circuit network, electric current will go through all of the resistor bulbs, thus lighting each one. The brightness of all the bulbs depends on the amount of current going through them, as one bulb s brightness is determined by another bulb s position within the network. Although this circuit network s bulbs are arranged in a series, there exist two parallel components within. Because of this type of arrangement, and not knowing the direction of current flow, the current will split before one of the components. It will come back together between the components before splitting again. Upon returning to the battery s opposite terminal the current will come back together again. The brightness of each bulb will depend on how much current travels through each resistor bulb. As current moves around the network and toward a junction node, the current will enter and split. The amount of electric current that goes into the node will be the same amount coming out. That is, the current will now be traveling through different branches with less current prior to the node. The current of the two branches together will equal the networks total current. Because the network s top component has one bulb on one half the side and two bulbs in a series on the other half, more current will go to the path with the single bulb. Circuit one s ranking is as follows: Bright > A, > (D and E- same), > B, > C, > Dim.

9 In circuit two s closed, parallel circuit network, electric current will go through all of the resistor bulbs, and light each one. Their brightness will depend on the amount of current going them. The ranking of brightness is also based on current flow from the battery s positive terminal end to its negative terminal end and the arrangement of additional bulbs through the branches as well as within the component. Generally, in parallel circuit networks of multiple branches, each with a single bulb, the indicator bulbs will light equally as an equal amount of current goes through each bulb. The parallel branches are independent of each other, and as resistance is increased or decreased in one branch it will not affect another. So, if some of these parallel branches had more or less bulbs than other branches, all of the bulbs within the entire network would light equally. In this network, both branches have multiple bulbs. The branch closest to the battery is arranged in a two bulb series. The branch furthest from the battery is arranged in a two bulb, parallel component that is in series with a single bulb. As direction is not determined, current will travel through the network, splitting at a node at the fist branch and continuing to the second branch. Since the second branch has alternate paths with the component and one bulb in series, current will be increased through this branch, thus making bulb H the brightest in this network as bulb H receives more of the network s current. Next, bulbs F and G are arranged in a series and therefore will light equally as they share the current through this branch. Finally, the bulbs in the component, bulbs I and J, are in series with bulb H. So they are sharing the current through that branch with H. Since they are in parallel with each other, the current will split at a node upon going through and lighting them making them the dimmest in this circuit network. Circuit two s ranking is as follows: Bright > H, > (F and G- same), > (I and J- same) > Dim.