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Electric Circuits IB 12 In the electric circuit shown below, energy is transferred from the battery to the light bulb by charges that move through a conducting wire because of a potential difference set up in the wire by the battery. The circuit shown contains a typical 9-volt battery. a) What is the emf of the circuit? b) How much energy does one coulomb of charge carry around the circuit? Schematic c) How much energy do two coulombs of charge carry around the circuit? d) How much energy does each coulomb of charge have at point B? e) How much energy does each coulomb of charge have at point C? f) What is VB? What is VC? mark potentials at each spot g) What is ΔVBC? What is ΔVCD? What is ΔVDA? Electric Current Formula: Units: Type: I = Δq/Δt A (ampere) = C/s Scalar Unofficial definition: rate of flow of electric charge Official Definition of One Ampere (1 A) of current – a fundamental unit One ampere is the amount of current flowing in each of two infinitely-long parallel wires of negligible cross-sectional area separated by a distance of one meter in a vacuum that results in a force of exactly 2 x 10 -7 N per meter of length of each wire. Short form – Current is defined in terms of the force per unit length between parallel current-carrying conductors. Closed circuit: complete pathway Open circuit: incomplete pathway for current – Short circuit: circuit with little to no for current break in circuit – infinite resistance resistance – extremely high current – overheating sketch sketch 1 Resistance IB 12 Resistance: ratio of potential difference applied Formula: Units: Type: across a piece of material to the current through the material R=V/I ohm (Ω) = V/A scalar For a wire conductor: Formula: A short fat cold wire is the best conductor R = ρL/A A long hot skinny wire has the most resistance Power: energy per unit time Unit: W = J/s Type: scalar Mechanical Power: Electrical Power: Alternate Formulas: P = E/t = (Δq V)/Δt Substitute V = IR P = W/t = F s/t = F v P=IV P = I (IR) = I2 R P = (V/R)V = V2/ R Meters in a circuit Ammeter: measures current Placement: Must be placed in series to allow current to flow through it Circuit must be broken to insert ammeter Ideal ammeter: has zero resistance so it will not affect current flowing through it Schematic diagram Voltmeter: measures potential difference Placement: Must be placed in parallel to measure potential difference between two points circuit does not to be broken Ideal voltmeter: has infinite resistance so it will not allow any current to flow through it and disrupt circuit 2 Series and Parallel Circuits IB 12 Characteristic Series Circuit Parallel Circuit Number of pathways one More than one for current Same everywhere – same for all Current Current splits – shared among devices devices Potential Difference Voltage shared among devices – Same for all devices (Voltage) voltage splits Overall resistance high low Power low high VT = V1 + V2 + … VT = V1 = V2 = . . . Formulas: IT = I1 = I2 = … IT = I1 + I2 + . . . RT = R1 + R2 + … 1/RT = 1/R1 + 1/R2 + . . . PT = P1 + P2 + . . . PT = P 1 + P2 + . . . Series Circuits Parallel Circuits Voltage Ratio Current Ratio Power Ratio Power Ratio 3 Analyzing Circuits IB 12 Determine the current through and the voltage drop across each resistor in each circuit below. 1. 2. 3. 4. Potential Divider: Resistors in series act as a “potential (voltage) divider.” They split the potential of the source between them. 5. A 20Ω device requires 40 V to operate properly but no 40 V source is available. In each case below, determine the value of added resistor R1 that will reduce the voltage of the source to the necessary 40V for device R 2. (A) (B) (C) (D) 6. A mini light bulb is rated for 0.60 W at 200 mA and is placed in series with a variable resistor. Only a 9.0 volt battery is available to power it. To what value should the variable resistor be set to power the bulb correctly? Bulb needs only 3 V Bulb has resistance of 15 Ω at rated power Added resistance should be 30 ohms 4 The Use of Sensors in Circuits IB 12 1. Light-Dependent Resistor (LDR) or Light Sensor: A photo-conductive cell made of semiconducting material whose resistance decreases as the intensity of the incident light increases. Automatic light switch Describe how the LDR activates the light switch. As ambient light decreases, resistance of LDR increases Potential difference across LDR increases Switch needs minimum PD to turn on When light intensity drops to desired level, PD is high enough to turn on switch 2. Negative Temperature Coefficient (NTC) Thermistor or Temperature Sensor: A sensor made of semiconducting material whose resistance decreases as its temperature increases. Fire alarm Describe how the NTC thermistor activates the fire alarm. As external temperature increases, resistance of NTC decreases Potential difference across R2 increases Switch needs minimum PD to turn on When temperature increases to desired level, PD is high enough to turn on switch 3. Strain Gauge or Force Sensor: A long thin metal wire whose resistance increases as it is stretched since it becomes longer and thinner. Describe how the strain gauge can measure the strain put on a section of an airplane body. As strain increases, resistance of strain gauge increases Potential difference across R2 decreases Voltmeter can read change in voltage which can be used to determine amount of strain on part 5 Combination Series-Parallel Circuits IB 12 1. Determine the current through and the voltage drop across each resistor. 2. The battery has an emf of 12 V and negligible internal resistance and the voltmeter has an internal resistance of 20 kΩ. Determine the reading on the voltmeter. 3. A cell with negligible internal resistance is connected to three resistors as shown. Compare the currents in each part of the circuit. 6 IB 12 4. Determine the current through and the voltage drop across each resistor. 5. A battery with emf E and negligible internal resistance is connected in a circuit with three identical light bulbs. a) Determine the reading on the voltmeter when the switch is open and when it is closed. b) State what effect closing the switch has on the current through each bulb and the brightness of each bulb. 7 Ohm’s Law IB 12 Resistance: ratio of potential difference applied across a piece of material Formula: to the current through the material R=V/I Relationship: Ohm’s Law: for a conductor at constant temperature, the current flowing VαI through it is proportional to the potential difference across it Ohmic Device: a device that obeys Ohm’s law for a wide range of potential differences Meaning: a device with constant resistance Example: resistor 1. On the axes at right, sketch the I-V characteristics for a resistor. Resistance: a) R = V/I at any point b) related to slope of graph (Reciprocal = resistance) 2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the resistor. R = V/I R = 7.5 Ω b) If the voltage is doubled, what is the new current? V = IR for resistor Resistance is constant so double current 8 IB 12 Non-Ohmic Device: a device that does not obey Ohm’s law Meaning: resistance is not constant Example: filament lamp 1. On the axes at right, sketch the I-V characteristics for a filament lamp. Resistance: a) R = V/I at any point b) as current increases, wire filament heats up and resistance increases c) Resistance is NOT related to the slope d) except in initial linear region 2. A flashlight bulb is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the bulb. R = V/I R = 7.5 Ω b) If the voltage is doubled, what is the new current? V = IR for bulb but resistance is not 7.5 ohms any more – R increases with T so less than double current 3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z. X: resistance increases - ratio V/I increases Y: resistance is constant – ratio V/I is constant Z: resistance decreases – ratio V/I decreases 9 Using a Potentiometer to Measure I-V Characteristics IB 12 Potentiometer: a type of variable resistor with three contact points Common use: as a potential divider to measure the I-V characteristics of a device The schematic shows how a potentiometer can be used as a potential divider to measure the I-V characteristics of a filament lamp. It is placed in parallel with the lamp and the slider (center contact point) effectively splits the potentiometer into two separate resistors AB and BC. By moving the slider, the ratio of the voltage drops across the resistors AB and BC is varied. Redraw the schematic with an ammeter and a voltmeter correctly placed to measure the I-V characteristics of the filament lamp. Comment on the circuit characteristics as the slider is moved from A to B to C. Slider at A: Slider at B: Slider at C: 10 IB 12 Internal Resistance of Batteries A 6 volt battery is connected to a variable resistor and the current in the circuit and potential difference across the terminals of the battery are measured over a wide range of values of the resistor. The results are shown in the table. Predicted Actual Voltage across Resistance (Ω) Current (A) Current (A) battery (V) 2000 0.003 0.003 6.00 200 0.03 0.03 5.99 20 0.3 0.29 5.85 2 3 2.4 4.80 0.2 30 8.8 1.71 0.02 300 11.5 0.23 0.002 3000 12.0 0.02 0.0002 30000 12.0 0.00 Why does the current seem to be limited to a maximum of 12.0 amperes and why does the voltage across the battery not remain constant at 6.0 volts? The battery has some internal resistance. As the external resistance decreases, more and more of the energy supplied by the battery is used up inside the battery. Electromotive force (emf): total energy per unit charge supplied by the battery Symbol: ε or E Units: V = J/C Terminal Voltage (Vterm): potential difference across the terminals of the battery Ideal Behavior: Vterm always equals emf since no internal resistance Real Behavior: 1) Think of battery as internal E and tiny internal resistor r 2) Vterm only equals the emf when no current is flowing 3) E is split between R and r 4) When R>>r, Vterm ≈ emf 5) As R decreases, Vr increases and VR decreases 11 IB 12 Relationship between emf and terminal voltage Treat internal resistance as a series resistor ε = I RT ε = I (R + r) ε = IR + Ir Note that in the absence of internal resistance, ε = Vterm 1. A resistor is connected to a 12 V source and a switch. With the switch open, a voltmeter reads the potential difference across the battery as 12 V yet with the switch closed, the voltmeter reads only 9.6 V and an ammeter reads 0.40 A for the current through the resistor. Sketch an appropriate circuit diagram and calculate the internal resistance of the source. 2. Discuss the expected I-V characteristics for this battery and how they can be experimentally determined. R can be adjusted from 0 to its max value A graph of Vterm vs. I can be drawn Specific equation of graph can be compared to math model to derive internal resistance Emf = Vterm + Ir Vterm = -Ir + emf so slope = -r and y-intercept = emf 12

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posted: | 9/23/2011 |

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