Electric Circuits - Physics
Electric Circuits
[pic]
Electric Current
In the scenario below, how can we get a flow of electric charge from the wall socket to the lamp?
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Could any common material be used to create a flow? Why or why not?
No, only materials that conduct electric charge can be used.
Current ( I ) – the rate at which charge pass through a wire.
• Units = C/s = Amperes (A)
Although many practical applications and devices are based on the principles of static electricity, electricity did not become an integral part of our daily lives until scientists learned to control the movement of electric charge, known as current.
1. What is the current in a wire in which 600. C of charge pass a point every 4.0 minutes?
[pic]
2. How many charge carriers flow through a circuit if a 10.0 A current is allowed to flow for 20. seconds?
[pic]
|Table of Common Currents |
|Computer chip |10-12 – 10-6 A |
|Electron beam of a television |10-3 A |
|Current dangerous to a human |10-2 – 10-1 A |
|Household light bulb |1 A |
|Car starter motor |200 A |
|Lightning |104 A |
Drift Velocity
When you turn on a light switch, the light comes on almost immediately. Does this mean that electrons flow very rapidly from the battery to the bulb? No. Actually, it takes electrons over an hour to travel one meter through a copper wire. How can this be explained?
Because the signal for charges to begin moving (the electric field) travels at nearly the speed of light.
[pic] [pic]
Types of Current
|Direct Current (DC) |Alternating Current (AC) |
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|[pic] |[pic] |
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|DC – charges move in only one direction. |AC – the motion of charges continuously changes in the forward and reverse |
| |directions. |
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|[pic] |[pic] |
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Conditions Necessary for an Electric Current
1. Complete circuit
• a closed path along which charged particles move.
2. Potential between two points in the circuit
• May be supplied by a battery (has positive and negative terminals)
[pic]
3. Conductive material for which charge can move.
• Metal are good conductors because their electrons move readily.
4. Resistance
• If resistance was not present the circuit would overheat and burn out.
Resistance
Resistance ( R ) – the opposition that a device or conductor present to the flow of electric current.
• Unit – Ohm (Ω)
Factors that Affect the Resistance of a Conductor
|Factor |Less Resistance |More Resistance |
|Length |[pic] |[pic] |
|Cross – Sectional Area |[pic] |[pic] |
|Material |[pic] |[pic] |
|Temperature |[pic] |[pic] |
Resistivity ( ρ ) – a characteristic of a material that depends on its electronic structure and temperature.
• Unit - Ω•m
• Good conductors have low resistivities and good insulators have high resistivities.
1. To make a wire with the lowest resistance, what material would you use? Highest resistance?
Lowest (Silver) / Highest (Nichrome)
2. What is the resistance of a nichrome wire 2.0 meters long with a cross-sectional area of
6.4 × 10 8 m2?
[pic]
3. What would be the resistance of an identical copper wire twice as long as the one in the above question?
[pic]
4. Determine the resistance of a 4.00 meter length of copper wire having a diameter of 2.00 millimeters. Assume a temperature of 20º C.
[pic]
Ohm’s Law
Ohm’s Law – the ratio of the potential difference to the current is always a constant for a given conductor.
1. A student measures a current of 0.10 ampere flowing though a lamp connected by short wires to a 12.0 volt source. What is the resistance of the lamp?
[pic]
2. A wire has a resistance of 0.300 ohms and is connected directly to both ends of a 1.5 volt battery. What is the current in the wire?
[pic]
Graphs
[pic]
A Simple Circuit Model
What is the purpose of a battery?
|[pic] |A Water Analogy |
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What is a cell?
A cell (dry or wet), release energy in a chemical reaction occurring inside the cell
How is a battery different from a cell?
A battery is just two or more cells wired together.
[pic]
What is the difference between a D cell and an AA cell considering that both have a potential difference of 1.5V?
A "D-cell" has a rating of 1.5 volts which means that for every Coulomb of charge that moves from the negative side of the cell to the positive side will do 1.5 Joules worth of work. A "AA-cell" also has a rating of 1.5 volts so each Coulomb of charge that moves from one side to the other can and will do 1.5 Joules worth of work. The difference between the D-cell and the AA-cell is that the D-cell has more Coulombs worth of charge, so it will last longer. As a result of having more charge, the D-cell has more energy and can do more work, but it will still do work at the same rate (or has the same power) as the AA-cell.
Energy Transformation
|[pic] |Describe the energy transferals and transformations |[pic] |
| |that take place in these circuits. | |
Electric Power
Power ( P ) – the rate of conversion of electrical energy.
• Units – Watts (W)
1. How much power is dissipated by a mini light bulb connected to a 1.5 volt battery drawing a current of 28 mA?
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2. What is the resistance of the light bulb from question #1?
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3. What is the resistance of the motor shown?
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4. How much power does it draw from the battery?
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5. How fast could the motor raise a 2.0 kg weight?
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Electrical Energy
How much energy is required to power a 60. Watt light bulb for 4.5 hours?
W = Pt = (60. W)(16200 s) = 9.7 x 105 J
The Cost of Electrical Energy
[pic][pic]
Convert 1 kW•h into joules.
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A 1320 W space heater runs all night (~ 8.0 hrs). Estimate how much this costs is the price $.01249 per kilowatt-hour.
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Summary
|Quantity |Symbol |Unit |Formula |
|Resistance |R |Ohm (Ω) |[pic] |
|Potential Difference (Voltage) |V |J/C = Volts (V) |V=IR |
|Charge |q |Coulomb (C) |q=It, [pic] |
|Current |I |C/s = Amperes (A) |[pic], [pic] |
|Energy |E or W |Joules (J) |[pic] |
|Power |P |Watts (W) |[pic] |
Drawing Circuit Schematic Diagrams
Circuit schematic – a diagram of an electric circuit using standard symbols for the circuit elements.
Steps for Drawing Schematic Diagrams
1. Begin by drawing the symbol for the battery or other source of electric energy (such as a cell or generator).
2. Draw a wire coming out from the positive terminal of the power source.
• Draw wires as straight lines (use a ruler if needed).
3. When the path of the current reaches a resistor or other device, draw the appropriate symbol with values.
4. Follow the current path until you reach the negative terminal of the battery.
5. Check your work to make sure that you have included all parts and that there are complete paths for the current to flow.
|[pic] |[pic] |
Use the appropriate Circuit Symbols on your Reference Tables to draw a schematic diagram of the circuit shown below.
|[pic] |[pic] |
Battery Terminals
Label the positive and negative terminals on the battery and cell symbols below.
[pic] [pic]
Two Types of Current
|Electron Flow |Conventional Current |
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|[pic] |[pic] |
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|Negative charges tend to move from lower potential to the higher potential. |Positive charges tend to move from higher potential to lower potential. |
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|- → + |+ → - |
[pic]
The direction of a current in an electric circuit can be defined as either of these directions. In terms of mathematical treatment it is convenient to treat the current as flowing from positive to negative, that is, as conventional current.
| |Electron Flow |Conventional Current |
|Motion of Charge Carrier |[pic] |[pic] |
Ammeters and Voltmeters in a Circuit
[pic]
Voltmeter – a device used to measure the potential difference across a circuit.
• Connect outside the direct path of the current (parallel connection).
Ammeter – a device used to measure the current through a circuit.
• Connect in the direct path of the current (series connection).
Use the appropriate Circuit Symbols on your Reference Tables to draw a schematic diagram of the circuit above.
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|[pic] |
Draw a circuit diagram to include a 60.0 V battery, an ammeter, and a resistance of 12.5 Ω in series. Indicate the ammeter reading the direction of the conventional current.
[pic]
If a battery supplies a potential difference of 9.0 volts and the light bulb has a resistance of 150 Ω, what would be the readings on each meter if the switch were open and if it were closed?
|[pic] |Meter |Reading when Open |Reading when Closed |
| |[pic] |0 V |9.0 V |
| |[pic] |0 A |.060 A |
Resistors
Resistor – a device used in a circuit to limit current flow or provide a potential drop.
|Picture |Symbol |
|[pic] |[pic] |
Variable Resistor – a coil of resistance wire whose effective resistance can be varied by sliding a contact point.
• As more of the coil is used in a circuit, the resistance of the circuit increases, the current decreases.
• Also called a rheostat, or a potentiometer.
|Picture |Symbol |
|[pic] |[pic] |
A variable resistor set to 100. ohms is connected to a voltage source of 12.0 V.
What is the current in the circuit?
[pic]
If the resistor is then set to 200. ohms, what is the new current?
[pic]
If a lamp is connected to the circuit. What will happen to its brightness if the resistance in the circuit is increased? Why?
The brightness of the lamp will decrease because the current will decrease as a result of the increasing resistance.
Resistors in Series Circuits
Series Connection – a circuit in which all parts are connected end to end to provide a single path for the current.
|[pic] |[pic] |
Since there is only one path for the current in a series circuit, the current is ___the same________ through each resistor.
[pic]
In an electric circuit the increase in voltage provided by the power source is ______equal_______ to the _____sum_____ of the voltage drops across the resistors.
[pic]
|[pic] |Equivalent → |[pic] |
Equivalent resistance – the single resistance that could replace the several resistors in a circuit.
1. Three resistors, with resistances of 4.0 ohms, 6.0 ohms, and 8.0 ohms respectively, are connected in series to an applied potential difference of 36 volts. Determine
[pic]
|Known(s) |Unknown(s) |
| | |
|R1= 4.0 Ω |Req |
|R2=6.0 Ω |I1, I2, I3 |
|R3=8.0 Ω |V1, V2, V3 |
|V = 36 V | |
| | |
a. The equivalent resistance.
Req = R1 + R2 + R3
Req = 4.0 Ω + 6.0 Ω + 8.0 Ω = 18.0 Ω
b. The current running through each resistor.
[pic]
c. The potential drop across each resistor.
V1 = I1R1 = (2.0 A) (4.0 Ω) = 8.0 V
V2 = I2R2 = (2.0 A) (6.0 Ω) = 12 V
V3 = I3R3 = (2.0 A) (8.0 Ω) = 16 V
V = V1 + V2 + V3 = 8.0 V + 12 V + 16 V = 36 V
d. The resistor that dissipates the most power.
The third resistor will dissipate the most power, because it has the greatest potential difference.
e. The resistor that receives the most energy.
The third resistor will receive the most power, because it has the greatest power.
f. Which location if replaced by a light bulb would shine the brightest.
The third bulb will shine the brightest because is have the greatest power (due to a greater voltage drop).
2. In the circuit shown to the right, a voltage of 6 V pushes charge through a single resistor of 2 Ω. According to Ohm’s law, the current in the resistor (and therefore the whole circuit) is
[pic]
3. If a second identical lamp is added to the circuit, as shown on the left, determine the equivalent resistance as well as the current in the circuit.
[pic]
4. Does current flow THROUGH a resistor, or ACROSS a resistor?
5. Does voltage establish THROUGH a resistor, or ACROSS a resistor?
6. Circuits a and b are identical with all bulbs rated at equal wattage (therefore equal resistance). The only difference between the circuits is that Bulb 5 has a bypass, as shown.
[pic]
a. In which circuit is the current the greatest? ______b (less resistance)________
b. In which circuit are all three bulbs equally bright? _______ a_______
c. What bulb(s) are the brightest? _______ 4 & 6_______
d. What bulb(s) are the dimmest? ______ 5_(electricity takes path of least resistance)____
e. What bulbs have the largest voltage drops across them? ______4 & 6____
f. What circuit dissipates more power? _____ b (less resistance)_____
g. What circuit produces more light? ______b (more power)________
h. If Bulb 3 were unscrewed (or burned out) in circuit a, what would happen to the other bulbs in the circuit? _All the bulbs will go out (the path connecting terminals of the voltage source will break and current will cease.)
Resistors in Parallel Circuits
Parallel Connection – a circuit in which there are several paths for current flow.
|[pic] |[pic] |
Since there multiple paths for the current in a parallel circuit, the current is ___different________ through each resistor. The _____sum_____ of the current in the branches is _____ equal_____ to the total current from the source.
[pic]
[pic]
The drop in potential difference for each path is ____the same______ .
[pic]
Using the equation I = I1 + I2 + I3 + … and ohm’s law [pic] , we find the equivalent resistance to be:
* Note – the equivalent resistance is always less than the resistance of any branch.
1. Three resistors of 4.0 ohms, 6.0 ohms, and 12 ohms are connected parallel to an applied potential difference of 12 volts. Calculate
|Known(s) |Unknown(s) |
| | |
|R1= 4.0 Ω |Req |
|R2=6.0 Ω |I1, I2, I3 |
|R3=12 Ω |V |
|V = 12 V | |
| | |
[pic]
a. The equivalent resistance.
[pic]
b. The potential difference across each resistor.
The potential difference across each branch of the circuit is the same as the applied potential difference.
12 V
c. The current through each resistor.
[pic]
d. The resistor that dissipates the most power.
The first resistor will dissipate the most power, because it has the greatest current.
e. The resistor that receives the most energy.
The first resistor will receive the most energy, for it has the greatest current and lowest resistance.
f. Which location if replaced by a light bulb would shine the brightest.
The first (4.0 Ω) light bulb would shine the brightest, because it has the greatest current.
2. In the circuit shown below, there is a voltage drop of 6 V across each 2 Ω resistor. Determine
a. The current in each resistor.
[pic]
b. The equivalent resistance.
[pic]
c. Fill in the current in the eight blank spaces in the view of the same circuit shown above.
[pic]
3. Cross out the circuit below that is not equivalent to the circuit above.
[pic]
4. A 50. Ω, a 100. Ω and a 150. Ω resistor are to be connected in a circuit. What type of combination will give the highest resistance? The least resistance? Calculate each.
[pic]
Circuit Segments
Each segment below is connected to a 120 V voltage source. Find the equivalent resistance of each circuit segment shown below. Then, compare the currents through each resistor, the voltage drop across each resistor, and the power dissipated by each resistor.
|Segment |Equivalent Resistance |Current |Voltage Drop |Power |
|1 | |40. Ω |3.0 A |60. V |180 W |
| |[pic] | | | | |
|2 |[pic] |10. Ω |I1 = 6.0 A |120 V |P1 = 720 W |
| | | |I2 = 6.0 A | |P2 = 720 W |
|3 | |60. Ω |2.0 A |V1 = 40. V |P1 = 80. W |
| |[pic] | | |V2 = 80. V |P2 = 160 W |
|4 |[pic] |13 Ω |I1 = 6.0 A |120 V |P1 = 720 W |
| | | |I2 = 3.0 A | |P2 = 360 W |
|5 | |180 Ω |.67 A |V1 = 20. V |P1 = 13 W |
| |[pic] | | |V2 = 40. V |P2 = 27 W |
| | | | |V3 = 60. V |P3 = 40 W |
|6 |[pic] |16 Ω |I1 = 4.0 A |120 V |P1 = 480 W |
| | | |I2 = 2.0 A | |P2 = 240 W |
| | | |I3 = 1.3 A | |P3 = 160 W |
Junctions
Junction – a point where two or more current paths join.
|Junction A |Junction B |Junction C |
|[pic] |[pic] |[pic] |
The Junction Rule - The total current directed into a junction must equal the total current directed out of the junction.
Try these following problems on your own.
|Junction D |Junction E |Junction F |
|[pic] |[pic] |[pic] |
Adding Extra Resistors
A single light bulb is connected to a battery as shown. Then, a second light bulb is added to the circuit. Predict what will happen to the circuit’s characteristics if this second bulb is added in series or in parallel to the first bulb.
|In Series | |In Parallel |
|[pic] |→ |[pic] | |[pic] |→ |[pic] |
|Power of 1st light bulb |decrease | |Power of 1st light bulb |same |
|Total power of circuit |decrease | |Total power of circuit |increase |
|Current through 1st light bulb|decrease | |Current through 1st light bulb|same |
|Total current through circuit |decrease | |Total current through circuit |increase |
|Voltage Drop across 1st light |decrease | |Voltage Drop across 1st light |same |
|bulb | | |bulb | |
|Total Voltage of circuit |same | |Total Voltage of circuit |same |
|Resistance of 1st light bulb |same | |Resistance of 1st light bulb |same |
|Total Resistance of circuit |increases | |Total Resistance of circuit |decrease |
Light Bulbs in Series and Parallel
Two light bulbs use the standard 110 V light socket. One is rated at 60 W and the other at 100 W.
a) Which bulb carries the greater current?
Power is proportional to current, therefore the bulb with higher power carries greater current.
b) Which bulb has a higher resistance?
Power is inversely proportional to resistance, therefore the bulb with lower power has a higher resistance.
c) Which bulb is brighter?
The bulb with higher power (100 W) provides more brightness.
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Thomas Edison
[pic]
George Westinghouse
In 1887 direct current (DC) was king. At that time there were 121 Edison power stations scattered across the United States delivering DC electricity to its customers. But DC had a great limitation -- namely, that power plants could only send DC electricity about a mile before the electricity began to lose power. So when George Westinghouse introduced his system based on high-voltage alternating current (AC), which could carry electricity hundreds of miles with little loss of power, people naturally took notice. A "battle of the currents" ensued. In the end, Westinghouse's AC prevailed. The major advantage that AC electricity has over DC is that AC voltages can be transformed to higher or lower voltages. This means that the high voltages used to send electricity over great distances from the power station could be reduced to a safer voltage for use in the house. This is done by the use of a transformer.
Potential Difference (V)
[pic]
Potential Difference (V)
Current (A)
Current (A)
Resistance for
Non-Ohmic Material
Resistance for
Ohmic Material
Georg Simon Ohm
(1787-1854)
[pic]
Ohm’s main interest was current electricity, which had recently been advanced by Alessandro Volta’s invention of the battery. Ohm made only a modes[pic] Ž‘’¨©«»¼ " # $ % & ? Ž ? ? ’ “ ¸ × üøéøåμΪ¼¤ž?€|xjb^V^Vt living and as a result his experimental equipment was primitive. Despite this, he made his own metal wire, producing a range of thickness and lengths of remarkable consistent quality. The nine years he spent at the Jesuit’s college, he did considerable experimental research on the nature of electric circuits. He took considerable pains to be brutally accurate with every detail of his work. In 1827, he was able to show from his experiments that there was a simple relationship between resistance, current and voltage.
Use the table of Resistivities at 200 C on your Reference Tables to answer the following questions.
Resistance
Current
Potential Difference (Water moves from high pressure to low pressure.)
DEMONSTRATION – fill groove of ruler from end to end with marbles. Add one more marble at one end, and the last marble should roll off immediately. (Each marble had a very small drift speed, yet the information traveled very quickly)
Ohm’s law is a behavior that is valid for only certain materials.
No loss in energy occurs as the charge moves though the wire (A to B). But when the charge moves though the filament of the light bulb (B to C), which has a higher resistance than the wire has, it loses electrical potential energy due to collisions. This electrical energy is converted into internal energy, and the filament warms up. When the charge first returns to the battery’s terminal (D) the electrical potential energy is zero, and the battery must do work on the charge. As the charge moves between terminals of the battery (D to A), it electrical potential energy increases by q∆V.
[pic]
PE increases
PE decreases
PE remains the same
PE remains the same
Recall that heat travels from hot to cold, and air from high pressure to low pressure, until both areas reach the same temperature or pressure.
If a wire connected to the ground and the other end placed in contact with the sphere of a Van de Graff generator charged to a high potential, a surge would flow through the wire. The flow would be brief, for the sphere would quickly reach a common potential with the ground.
Dry Cell – Magnesium Dioxide
Wet – Nickel Cadmium (cell phone)
| | |V | |I |R |P |
|R1 |= |12 |• |3 |4 |36 |
|R2 |= |12 |• |2 |6 |24 |
|R3 |= |12 |• |1 |12 |12 |
|RT |= |12 |• |6 |2 |72 |
= + 1/R +
| | |V | |I |R |P |
|R1 |= |8 |• |2 |4 |16 |
|R2 |= |12 |• |2 |6 |24 |
|R3 |= |16 |• |2 |8 |32 |
|RT |= |36 |• |2 |18 |72 |
+ = + +
∆E = W = Pt
A simple circuit can be described in words. It can also be depicted by photographs or artist’s drawings of the parts. Most frequently, however, a diagram of an electric circuit is drawn using standard symbols for the circuit elements.
The black lines symbolize the conducting path provided by the wires, clips, and sockets.
Digital Multimeter
Common uses for variable resistor: Dimmer switch, volume control, idle screw (adjusts speed of motor)
The sum of the drops in PE is equal to the total PE from the top to the bottom.
[pic]
[pic]
30. Ω
[pic]
Ptotal = 10 W
If R = 1Ω for each bulb, than v = 2.25 V; and [pic]
Ptotal = 6.75 W
If R = 1Ω for each bulb, than v = 1.5 V; and [pic]
As resistance increases, current increases.
[pic]
[pic]
[pic]
6
6
3
3
3
3
3
6
The light intensity for each lamp is unchanged as other lamps are introduced (or removed). Only the total resistance and total current in the total circuit is changes, which is to say, the current in the battery changes. As lamps are introduced, more paths are available between the battery terminals, which effectively decreases total circuit resistance. This decreased resistance is accompanied by an increased current. Although changes of resistance and current occur for the circuit as a whole, no changes occur in any individual branch in the circuit. →
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