On the Brightness of Bulbs Resistance Blackbody Radiation Ohms 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 The color and intensity of the emitted radiation depend on the objects temperature Not surprisingly, our eyes are optimized for detection of light emitted by the sun, as early humans saw most things via reflected sunlight no light bulbs, TVs We now make some artificial light sources, and ideally they would have same character as sunlight better match to our visual hardware (eyes) Spring 2006 5 UCSD: Physics 8; 2006

Color Temperature Object Temperature You Heat Lamp Candle Flame Bulb Filament Suns Surface ~ 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 arent black! Blackbody??!! Black in this context just means reflected light isnt important Hot charcoal in a BBQ grill may glow bright orange when hot, even though theyre black Sure, not everything is truly black, but at thermal infrared

wavelengths (250 microns), youd 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 = AT4, where is a dimensionless number between 0 (perfectly shiny) and 1.0 (perfectly black) , recall, is 5.6710-8 in MKS units, T in Kelvin Why do we use aluminum foil? Spring 2006 9

UCSD: Physics 8; 2006 What Limits a Bulbs 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 Ohms Law: V = IR

voltage = current resistance Weve already seen voltage and current before, but whats 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 Ohms Law The best way to think about Ohms 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 Ohms 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 _ + Which bulb is brighter? WHY? A B Spring 2006 13 UCSD: Physics 8; 2006

Answer There is only one current flowing, and it goes through both bulbs. They will therefore shine with equal brightness. Imagine exchanging bulbs. Does this change anything? Spring 2006 14 + _ Bulbstravaganza Exploration of Circuits & Ohms Law UCSD: Physics 8; 2006 Reminder: Ohms 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 Ohms Law V Spring 2006 I R 16 UCSD: Physics 8; 2006 Numerical examples of Ohms Law (V = IR)

How much voltage is being supplied to a circuit that contains a 1 Ohm resistance, if the current that flows is 1.5 Amperes? If a 12 Volt car battery is powering headlights that draw 2.0 Amps of current, what is the effective resistance in the circuit? Spring 2006 17 UCSD: Physics 8; 2006 Answer #1: (How much voltage is being supplied to a circuit that contains a 1 Ohm resistance, if the current that flows is 1.5 Amperes?) Use the relationship between Voltage, Current and Resistance, V = IR. Total resistance is 1 Ohm

Current is 1.5 Amps So V = IR = (1.5 Amps)(1 Ohms) = 1.5 Volts Spring 2006 18 UCSD: Physics 8; 2006 Answer #2 (If a 12 Volt car battery is powering headlights that draw 2.0 Amps of current, what is the effective resistance in the circuit?) Again need V = IR Know I, V, need R Rearrange equation:

Spring 2006 R = V/I = (12 Volts)/(2.0 Amps) = 6 Ohms 19 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 20 UCSD: Physics 8; 2006 Multi-bulb circuits Rank the expected brightness of the bulbs in the circuits shown, e.g. A>B, C=D, etc. WHY?! _ _

+ + A B C Spring 2006 21 UCSD: Physics 8; 2006 Answer: Bulbs B and C have the same brightness, since the same current is flowing through them both.

Bulb A is brighter than B and C are, since there is less total resistance in the single-bulb loop, so A > B=C. Spring 2006 22 UCSD: Physics 8; 2006 Adding Bulbs Where should we add bulb C in order to get A to shine more brightly? _ + P

C A R B Spring 2006 Q 23 Answer UCSD: Physics 8; 2006 The only way to get bulb A to shine more brightly is to increase the current flowing through A. The only way to increase the current flowing through A is to decrease the total resistance in the circuit loop

Since bulbs in parallel produce more paths for the current to take, the best (and only) choice is to put C in parallel with B: _ + A B Spring 2006 C 24 UCSD: Physics 8; 2006 A more complex example! A

B + _ C D E F Predict the relative brightness of the bulbs Spring 2006 25 UCSD: Physics 8; 2006

Answer The entire current goes through bulb F so its going to be the brightest The current splits into 3 branches at C,D,E and they each get 1/3 of the current The current splits into 2 branches at A,B and they each get half the current, so F>A=B>C=D=E Spring 2006 26 UCSD: Physics 8; 2006 If I disconnect bulb B, does F get brighter or fainter? A

B + _ C D E F Spring 2006 27 UCSD: Physics 8; 2006 Answer By disconnecting B, the resistance of the (AB)

combination goes up, so the overall current will be reduced. If the current is reduced, then F will be less bright. Spring 2006 28 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

29 UCSD: Physics 8; 2006 How about multiple resistances? Resistances in series simply add Voltage across each one is V = IR R1=10 R2=20 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

30 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 Ohms Law to find equal current in each leg. To get twice the current of a single10 , could use 5 . Spring 2006 31

UCSD: Physics 8; 2006 A Tougher Example What is the voltage drop across the 3 resistors in this circuit? R2=20 R1=5 V = 3.0 Volts Spring 2006 R3=20 32 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! R2=20

R1=5 V = 3.0 Volts Spring 2006 R3=20 33 UCSD: Physics 8; 2006 Complex Example A B + _

C F Spring 2006 D 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 34 UCSD: Physics 8; 2006 Assignments Read pp. 224231, 332333, 407 for this lecture HW #3: Chapter 10: E.2, E.10, E.32, P.2, P.13, P.14, P.15, P.18, P.19, P.23, P.24, P.25, P.27, P.28, P.30,

P.32 Next Q/O (#2) due next Friday: only submit one this week if you missed it last week. Spring 2006 35