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Unformatted text preview: On the Brightness of Bulbs
Resistance Blackbody Radiation Ohm's Law UCSD: Physics 8; 2006 Review: What makes a bulb light up? The critical ingredient is closing a circuit so that current is forced through the bulb filament more on filaments and what is physically going on later The more the current, the brighter the bulb The higher the voltage, the brighter the bulb Power "expended" is P = VI this is energy transfer from chemical potential energy in the bulb to radiant energy at the bulb Spring 2006 2 UCSD: Physics 8; 2006 Bulb Design Basics
Tungsten Filament Sealed Bulb Electrical contacts
120 W bulb at 120 V must be conducting 1 Amp (P = VI) Bulb resistance is then about 120 Ohms (V = IR) Spring 2006 3 UCSD: Physics 8; 2006 What makes the bulb light up? Bulb contains a very thin wire (filament), through which current flows The filament presents resistance to the current electrons bang into things and produce heat a lot like friction Filament gets hot, and consequently emits light gets "red hot" Spring 2006 4 UCSD: Physics 8; 2006 Everything is Aglow All objects emit "light" Though almost all the light we see is reflected light, emitted by some other object The color and intensity of the emitted radiation depend on the object's temperature Not surprisingly, our eyes are optimized for detection of light emitted by the sun Spring 2006 5 UCSD: Physics 8; 2006 Color Temperature
We often try to make artificial light sources resemble sunlight Object You Heat Lamp Candle Flame Bulb Filament Sun's Surface Temperature ~ 30 C 300 K ~ 500 C 770 K ~ 1700 C 2000 K ~ 2500 C 2800 K ~ 5500 C 5800 K Color Infrared (invisible) Dull red Dim orange Yellow Brilliant white The hotter it gets, the "bluer" the emitted light The hotter it gets, the more intense the radiation
Spring 2006 6 UCSD: Physics 8; 2006 The "Blackbody" Spectrum Spring 2006 7 UCSD: Physics 8; 2006 Blackbody spectra on logarithmic scale Sun peaks in visible band (0.5 microns), light bulbs at 1 m, we at 10 m. (note: 0C = 273K; 300K = 27C = 81F)
Spring 2006 8 UCSD: Physics 8; 2006 Bulbs aren't black! Blackbody??!! Black in this context just means reflected light isn't important Hot charcoal in a BBQ grill may glow bright orange when hot, even though they're black Sure, not everything is truly black, but at thermal infrared wavelengths (250 microns), you'd be surprised your skin is 90% black (absorbing) even white paint is practically black metals are still shiny, though This property is called emissivity: radiated power law modified to P = A T4, where have added , a dimensionless number between 0 (perfectly shiny) and 1.0 (perfectly black) , recall, is 5.67 10-8 in MKS units, T in Kelvin Why do we use aluminum foil? Spring 2006 9 UCSD: Physics 8; 2006 What Limits a Bulb's Lifetime Heated tungsten filament drives off tungsten atoms heat is, after all, vibration of atoms: violent vibration can eject atoms occasionally Tradeoff between filament temperature and lifetime Brighter/whiter means hotter, but this means more vigorous vibration and more ejected atoms "Halogen" bulbs scavenge this and redeposit it on the filament so can burn hotter Eventually the filament burns out, and current no longer flows no more light! How "efficient" do you think incandescent bulbs are? Ratio between energy doing what you want vs. energy supplied Efficiency = (energy emitted as visible light)/(total supplied) Spring 2006 10 UCSD: Physics 8; 2006 Predicting Brightness in Bulb Networks This is a very instructive (and visual) way to learn about the behavior of electronics, how current flows, etc. The main concept is Ohm's Law: V = IR voltage = current resistance We've already seen voltage and current before, but what's this R? R stands for resistance: an element that impedes the flow of current measured in Ohms () Remember the bumper-cars nature of a bulb filament? Electrons bounce off of lattice atoms this constitutes a resistance to the flow of current
Spring 2006 11 UCSD: Physics 8; 2006 Interpretation of Ohm's Law The best way to think about Ohm's law is: when I have a current, I, running through a resistance, R, there will be a voltage drop across this: V = IR "voltage drop" means change in voltage Alternative interpretations: when I put a voltage, V, across a resistor, R, a current will flow through the resistor of magnitude: I = V/R if I see a current, I, flow across a resistor when I put a voltage, V, across it, the value of the resistance is R = V/I Ohm's Law is key to understanding how current decides to split up at junctions try to develop a qualitative understanding as well as quantitative Spring 2006 12 UCSD: Physics 8; 2006 Bulbs in Series Each (identical) light bulb presents a "resistance" to the circulating electrical current Adding more bulbs in series adds resistance to the current, so less current flows
_ + A Which bulb is brighter? WHY? B
Spring 2006 13 + _ Bulbstravaganza
Exploration of Circuits & Ohm's Law UCSD: Physics 8; 2006 Reminder: Ohm's Law There is a simple relationship between voltage, current and resistance: V is in Volts (V) I is in Amperes, or amps (A) R is in Ohms () V=IR
Ohm's Law V I R How much voltage is supplied ... If a 12 Volt battery..
Spring 2006 15 UCSD: Physics 8; 2006 Conductors are at Constant Voltage Conductors in circuits are idealized as zeroresistance pieces so V = IR means V = 0 (if R = 0) Can assign a voltage for each segment of conductor in a circuit 1.5 V 1.5 V 3.0 V 0V
batteries in parallel add energy, but not voltage
Spring 2006 0V
batteries in series add voltage
16 UCSD: Physics 8; 2006 Power Dissipation How much power does a bulb (or resistor) give off? P = VI but V = IR so P = I2R and P = V2/R are both also valid Bottom line: for a fixed resistance, power dissipated is dramatic function of either current OR voltage Spring 2006 17 UCSD: Physics 8; 2006 How about multiple resistances? Resistances in series simply add Voltage across each one is V = IR V = 3.0 Volts Total resistance is 10 + 20 = 30 So current that flows must be I = V/R = 3.0 V / 30 = 0.1 A What are the Voltages across R1 and R2? Spring 2006 18 UCSD: Physics 8; 2006 Parallel resistances are a little trickier.... Rule for resistances in parallel: 1/Rtot = 1/R1 + 1/R2 10 Ohms 10 Ohms 5 Ohms Can arrive at this by applying Ohm's Law to find equal current in each leg. To get twice the current of a single10 , could use 5 . Spring 2006 19 UCSD: Physics 8; 2006 A Tougher Example What is the voltage drop across the 3 resistors in this circuit? V = 3.0 Volts Spring 2006 20 UCSD: Physics 8; 2006 Answer First, need to figure out the current that flows in the circuit. This depends on the total resistance in the loop. Combine the parallel resistors into an equivalent single series resistor, the parallel pair are equal to a single resistor of 10 Ohms The total resistance in the loop is 5 + 10 = 15 Ohms So the total current is I = V/R = 3/15 = 0.20 Amps Voltage across R1 is V = IR = 0.2A 5 Ohms = 1 Volt Voltage across R2, R3 is equal, V = IR = 0.2A 10 = 2 V Note that the sum of the voltage drops equals battery voltage! V = 3.0 Volts
Spring 2006 21 UCSD: Physics 8; 2006 Complex Example E Say battery is 5.5 Volts, and each bulb is 6 AB combo is 3 CDE combo is 2 total resistance is 11 current through battery is 5.5V/11 = 0.5 A A gets 0.25 A, so V = 1.5V C gets 0.1667 A, so V = 1.0 V F gets 0.5 A, so V = 3.0 V note voltage drops add to 5.5 V Use V2/R or I2R to find: PAB = 0.375 W each PCDE = 0.167 W each PF = 1.5 W
22 A B + _ C D F
Spring 2006 UCSD: Physics 8; 2006 Four Questions on Ohm's Law We now have a series of questions on the brightness's of bulbs in different circuits. Higher current makes a bulb brighter In each case assume the bulbs are identical and the wires have zero resistance You need V=IR, decreasing R will increase current I and make a bulb brighter. Spring 2006 23 ...
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