ANNOUNCEMENT REMINDER There will be a review session Monday night in Forney (FRNY) Room G124 (note change, many evening exams on Monday). The building just South of Civil Engr. and across the inside street on the West side of Physics. Time: from 7pm to about 8:30 pm. This will be an opportunity for you to ask questions about problem solutions, either homework or practice exam or other. 1
Electrical circuits All circuits are basically the same. There is an external source of energy which produces a voltage. In a charged battery there is chemical separation of charge. When the circuit is made positive charge flows from high to low voltage or negative charge flows from low to high voltage releasing the stored energy. Normally it is electrons which flow. As they move they collide with the atoms of the wire and lose some of their energy in the form of heat. There is resistance to the flow. 2
Current and resistance The rate of flow of charge determines how much charge is transferred per unit time I = q/t Amperes (Coulombs/sec) The direction of I is the flow of positive charge or opposite to the flow of negative charge. OHM s Law R = ΔV/I [Ohms, Ω ] or: V = IR where it is understood that V is just the voltage drop across the resistor, and we omit the Delta symbol. Every part of a circuit has resistance including an internal resistance in the battery. The higher the resistance the lower is the current for a given voltage difference 3
Electromotive force The electromotive force of a battery, ε is the voltage difference between the two terminals when no current is being drawn. When it is connected to a simple circuit I = ε/(r circuit + R battery ) A voltage difference is the energy stored or released per Coulomb of charge. So if charge +q goes from high to low voltage then the energy released is qε or qδv (20 Ω+5 Ω)*60 ma = 1.5 V = ε Part of the energy is wasted in internal resistance, the rest is delivered to the external circuit. Internal voltage drop is 0.3 V 4
Series circuit Rbattery If we add more identical light bulbs in the circuit in series the total resistance increases and the current will be reduced. The current is the same in all parts of the circuit: It s like water flowing in a pipe ε = I(R battery + R + R + R) The voltage difference across each bulb is ΔV = IR QUESTON: what is the internal voltage drop in the battery now? 5
Series circuit Rbattery ε = I(R battery + R + R + R) What is the internal voltage drop in the battery? R total = 5+20+20+20 = 65 Ω. ε= 1.5 as before I = ε /R = 23 ma so ΔV internal = 5*23 = 0.115 V 6
Series circuit R battery ε = I(R battery + R 1 + R 2 + R 3 ) [bulbs not the same] The voltage difference across a bulb is V i = IR i. The internal resistance is ALSO in series, and causes the voltage at the + terminal to be LESS THAN ε All the voltage drops add up to ε 7
Series circuit R battery ε = I(R battery + R 1 + R 2 + R 3 ) [bulbs not the same] ε -IR battery = I(R 1 + R 2 + R 3 ) = V external, the available voltage Here, ε is the chemical EMF of the battery, but if some current flows, the available voltage at the battery terminals is reduced by an amount IR battery. 8
Series circuit R battery ε = I(R battery + R 1 + R 2 + R 3 ) [bulbs not the same] In any series circuit, if one element burns out and causes an open circuit, then no currrent can flow. Some strings of Christmas tree lights have this design, which can be a big nuisance. A burned out bulb has, in effect, infinite resistance. 9
Parallel circuit R battery The current q at A divides into the three light bulbs and then recombines at B In a parallel circuit, the voltage difference across each light is the SAME. The total current is the SUM of the three currents I = I 1 + I 2 + I 3 and since a current = ΔV/R ΔV/R circuit = ΔV/R 1 + ΔV/R 2 + ΔV/R 3 and 1/R circuit = 1/R 1 + 1/R 2 + 1/R 3 The total resistance of the circuit is smaller than any of the single resistances. It is also true that if one bulb fails the other two will stay lit at the same brightness. 10
. QUIZ For circuits that obey OHM s law, ΔV = IR A 1.5 volt flashlight battery delivers 1 Amp of current to a lightbulb. There is a voltage drop of 1.25 V across the terminals of the lightbulb. What is the internal resistance of the battery? Here, the chemical EMF, ε, is 1.5 V and all of this will appear across the batterey terminals ONLY at zero current. A. 1.5 Ω B. 1.0 Ω C. 0.75 Ω D. 0.5 Ω E. 0.25Ω. 30 Mar 09 Answer: 0.25= ΔV across the internal resistance, hence since the current is 1 A, the internal resistance must be 0.25 Ω because V = IR Note: we use V and ΔV somewhat interchangeably for Voltage drops. 11
Review for Test 2, Ch. 7-12 Conversion Factors 1 inch = 2.54 cm 1 ft = 30.5 cm 1 m = 3.281 ft 1 km = 0.621 miles 1 mile = 5280 ft 1 nautical mile = 1.1508 miles 1 kg = 2.205 lbs (where g = 9.8 m/s2) Equations s = d/t speed s = d/δt v = d/δt (vector velocity) a = Δv/Δt (vector acceleration) v = v0 + at d = v0t + ½at2 v2 = v02 + 2ad d = ½(v + v0)t F = ma (vector) N2 F12 = - F21 N3 W = mg FORMULAE AND CONSTANTS P = 10 15 Peta T = 1012 Tera G = 109 Giga M = 106 Mega k = 10 3 kilo m = 10-3 milli μ = 10-6 micro n = 10-9 nano p = 10-12 pico f = 10-15 femto Pressure = Force/Area [N/m 2 = Pa = Pascal] 1 Atmosphere = 101.3 kpa = 760mmHg 33 ft H 2 O P = ρgd where d is depth below surface (or height of a liquid column Buoyant force = weight of displaced liquid Density H 2 0 is 1 Tonne/m3 or 1 gram/cm 3 12
Review for Test 2, Ch. 7-12 MOMENTUM AND IMPULSE -- ELASTIC AND INELASTIC COLLISIONS, CONSERVATION OF MOMENTUM ROTATIONAL MOTION OF SOLID OBJECTS -- TORQUE, MOMENT OF INERTIA, CONSERVATION OF ANGULAR MOMENTUM, ANGULAR ACCELERATION, KE of ROTATION FLUIDS PRESSURE, DENSITY, BUOYANCY TEMPERATURE AND HEAT -- TEMPERATURE SCALES, HEAT CAPACITY, PHASE CHANGES AND LATENT HEATS HEAT ENGINES AND 2d LAW OF THERMODYNAMICS -- CARNOT EFFICIENCY ELECTRICITY AND MAGNETISM: ELECTROSTATICS -- ELECTRIC FIELDS AND FORCES, VOLTAGE (ELECTRIC POTENTIAL), ELECTRIC CHARGE 13
Review for Test 2, Ch. 7-12 Circumference = 2πR g = 9.8 m/s 2 G = 6.67 10-11 N m 2 /kg 2 π = 3.14159 ω = θ/t = Δθ/Δt angular velocity e = 1.6x10-19 C the size of the charge on an electron or a proton m e = 9.1x10-31 kg the mass of the electron m p = 1836 m e the mass of the proton c H2O = 1 Cal/gm heat capacity of water L f = 80 Cal/gm latent heat of fusion of water L v = 540 Cal/gm latent heat of vaporization of water 14
Review for Test 2, Ch. 7-12 DEFINITIONS a = v 2 /R centripetal acceleration F = mv 2 /R centripetal force F = GM 1 M 2 /R 2 gravitational force G = 6.61x10-11 Nm 2 /kg 2 F = kq 1 q 2 /R 2 electrostatic (Coulomb) force k = 9x10 9 Nm 2 /C 2 W = Fd work P = W/t = Fv power KE = ½ mv 2 Kinetic energy PE = mgh potential energy PE = ½ kx 2 potential energy P = mv momentum FΔt = P f P i = Δp impulse F external = 0, Δp = 0 momentum conservation W = qδv work or energy, electrical P = I ΔV power, electrical E = ΔV/d for uniform electric field, where d is distance along the voltage gradient 15
Review for Test 2, Ch. 7-12 DEFINITIONS ε = (T hot T cold )/T hot Carnot efficiency of an ideal heat engine PV = NkT Ideal gas law where P,V,T are pressure, volume, and Kelvin temperature and k is Boltzmann s constant I = mr 2 Moment of rotational inertia of a point mass a distance R from the axis of rotation Torque = r x F Torque is product of Force x Lever arm distance to the axis of rotation (the component of distance at right angle to F) Torque = I α α angular acceleration, in radians/s 2 KE = 0.5 I ω 2 Kinetic energy of rotation of a solid body I = MR 2 for hoop I = 0.5 MR 2 for disk I = 0.4 MR 2 for sphere I = (1/12)ML 2 for a rod of length L KEPLER s Laws Elliptical orbit sweeps out equal areas in equal time intervals T 2 = r 3 for all satellites of a common attracting body, where T is the period and r is the average radius of the orbit 16
What is voltage? We have seen that the definition of voltage is ΔV = ΔPE/q when a charge q is moved in an electrical force field. So energy is stored as potential energy when a positive charge is moved against the direction to E (or a negative charge is moved in the same direction as E.) If we move a positive charge toward a positive charge, the potential energy and ΔV increase. Also if we move a negative charge away from a positive charge. Just as in the gravitational field, what matters is DIFFERENCES in PE. So normally we use the term ΔV. But very often for circuits we choose one point, usually the negative terminal, to be zero and then instead of ΔV we just use V. When charge is free to move, the PE will transform into KE just like dropping something. Positive charge will move to lower voltage, negative charge to higher voltage, tending to neutralize the system. In a simple circuit with resistance the KE is turned into heat and light and there is a voltage drop across every element in the circuit. 17
Voltage drop ( here we assume negligible R battery ) If we have a circuit with many different resistors then there is a voltage drop across each resistor [in the direction of the current] and there is also a voltage drop for the whole circuit. Current only flows if there is a voltage difference. In a time t, charge q passes through the resistor. I = q/t and ΔV = IR Case1 I = 6/60 = 1/10 ΔV 15 = 15/10 ΔV 20 = 20/10 ΔV 25 = 25/10 Case 2 I = 12/8 I 24 = 12/24 Case 3 I = V AB /5.5; ΔV 33 = 1.5I; ΔV 3 = 3I; ΔV 333 =1I Check: (1.5+3+1) I = 5.5 I = V AB 18
Voltage drop ( here we assume negligible R battery ) The voltage jumps from 0 to ε as we go from the to the + plate of the battery. From there on, there are a succession of voltage drops, in the direction of the current flow through the series of resistors. These bring us back to V=0 at the - plate Case1 I = 6V/60Ω = 1/10 A ΔV 15 = 15/10; ΔV 20 = 20/10; ΔV 25 = 25/10. 1.5+2+2.5 = 6V checks OK. 19
Voltage drop ( here we assume negligible R battery ) Case 3 I = V AB /5.5; ΔV 33 = 1.5I; ΔV 3 = 3I; ΔV 333 =1I Check: (1.5+3+1) I = 5.5 I = V AB NOTE: Here we lump together the two (or three) resistors in parallel with each other, before putting the lumps in series with each other and with the single 1.5 Ω 3 Ω 1 Ω 3 Ω resistor. 20
Measuring current and Voltage It is often very important to know the current in a circuit or the voltage difference between two points. A hand held meter is very useful to test batteries or a circuit. An ammeter is a device inserted into a circuit. The resistance of an ammeter is very small so as to minimize the effect on the circuit. A voltmeter is attached in parallel and V is found by measuring the current and V = I meter R meter. The resistance has to be much larger than the circuit resistance so that the current is very small and does not disturb the main circuit. Both meters actually measure current 21
Measuring current and Voltage An old fashioned voltmeter is a very sensitive coil in a magnetic field, with a pointer that swings when tiny currents flow in the coil. We put this in series with a large resistor, so that it won t disturb the current flowing in the circuit very much. The combination is what s represented by the circle-v Both meters actually measure current 22
Measuring current and Voltage Now we can use this voltmeter to make an Ammeter: The circle-a contains a very low-r shunt resistor which is put in series with the circuit, so that the current I flows though the shunt. Low R A means tiny Voltage drop, so little effect on I. Put (V) meter across the shunt to measure the tiny Voltage drop. Knowing R A, we know the current I Both meters actually measure current 23
Measuring current and Voltage Better modern meters use power amplification, so that they drain an even tinier fraction of the power from the circuit being measured. Affordable DVMs (digital voltmeters) have 10 MΩ input impedance and can measure tiny voltages (sub-mv), and tiny currents (sub-μa). Multi-meters can also measure resistance, using an known V internal battery to drive current through a R and measuring the current through the resistor. Both meters actually measure current 24
Power Electromotive force or voltage difference between two points is the difference in potential energy/unit charge. So the energy delivered if charge q is transferred is energy = Vq power = Vq/t = VI watts For any voltage difference ΔV and current I. power = ΔVI but we will generally just use V for the voltage difference across the R For circuits that obey OHM s law V = IR P = VI = I 2 R = V 2 /R watts The power used appears as heat or light Choosing the most appropriate form for the electric Power equation can simplify a solution. 25
Power P = VI = I 2 R = V 2 /R watts Just choose the most appropriate form. Example: 100 W lightbulb designed for 115 V. What is R (when filament is hot)? What do we know?: P, V What is the question? R=? Use the last form, since we aren t told OR asked what is I. R = V 2 /P = 115 2 V 2 /100 W = 132.25 Ω 26
Power P = VI = I 2 R = V 2 /R watts Just choose the most appropriate form. Example: 100 W lightbulb designed for 115 V. Now we ask what is the Power (compared to the 115V power) if the voltage is cut to 100 V? (We ll assume, not quite correctly, that the cooler filament has the same R. In fact, it will have a lower resistance than the hotter filament at 115V) Here, we still use the last form, and ALSO use ratios so that the allegedly constant R divides out: P /P = V 2 /V 2 = (100/115) 2 = 0.87 2 = 0.756 Numerical observations: divide 100 by 100 + 15%. Result is 1.00-13%. Squaring it roughly doubles the percentage difference: (1-24.4%) (1+x) 2 = 1 + 2x + x 2 and if x is small, you can more or less neglect x 2 (1-x) 2 = 1-2x + x 2 ; (1+x) -2 = 1-2x + [(-2)(-3)/2]x 2 +...x 3 +.x 4 Using this last form, 1/(1+15%) 2 1-30% and you crudely get P /P 0.700 27
Digression Numerology, but may be useful in everyday life. Let s say you lose 50% of your investment in stock in Bank of America. Then the market picks up and your BoA stock value increases by 50%. Are you happy? (1-0.5) x (1 + 0.5) = 0.5 x 1.5 = 0.75 Your BoA stock is worth 75% of your original investment. You would have needed a 100% increase to get back to where you started. Put simply, if your stock value drops to half, it needs to DOUBLE to get back where it started. So percentage decreases and increases are not exactly additive!! For SMALL percentages, increases and decreases are ALMOST additive: 1.01 x.99 = 0.9999 but you re still a tiny bit (.01 2 ) low 28
Power P = VI = I 2 R = V 2 /R watts Let s rig three 100 Watt 115 V bulbs into a circuit where two bulbs are in parallel, and the third bulb is in series with the pair. ------O------ o--------o-------- ---------o ------O------ All bulbs will run dim, if this entire arrangement is put across the same Voltage of 115V The total R T is R + R/2 where R is for one bulb and R/2 is for the parallel pair. So now I will be 2/3 as big as before, and the left bulb will have I 2 R power = (2/3) 2 = 4/9 as big as a single bulb across 115 V. 29
Power P = VI = I 2 R = V 2 /R watts Let s rig three 100 Watt bulbs into a circuit where two bulbs are in parallel, and the third bulb is in series with the pair. ------O------ o--------o-------- ---------o ------O------ More extremely: the current divides in the parallel portion, so each of those bulbs gets only 1/3 the current of a single bulb across 115V, hence only 1/9 the power. Does power add up? 4/9 + 1/9 + 1/9 = 6/9 = 2/3 for all three bulbs. Same voltage, and remember 3/2 as much resistance, hence 2/3 as much current. Now use the FIRST form of the Power: P = VI = (same x 2/3) as much as for one single bulb. Checks OK 30
Power P = VI = I 2 R = V 2 /R watts Do things add up? 4/9 + 1/9 + 1/9 = 6/9 = 2/3 for all three bulbs. Same voltage, and remember 3/2 as much resistance, hence 2/3 as much current. Now use the FIRST form of the Power: P = VI = same x 2/3 as much as for one single bulb. Checks OK *** But I put in an unnecessary extra step here, I could just have used the THIRD Power form, and not bothered to calculate the current, and found: New Power = (V) 2 /(3/2R) = 2/3 {(V) 2 /R} = 2/3 of single bulb Pwr. Everything checks out consistently. There is no wrong approach, just one approach is somewhat simpler and quicker. 31
QUIZ March/April 09 P = VI = I 2 R = V 2 /R [watts]; V = IR Ohm s Law Two 100 Watt bulbs are in series, across 115 Volts 0V o--------o-----------o---------o +115V QUESTION: Compared to a single 100 Watt bulb across 115V, how much power does ONE of these bulbs consume? Let R be the resistance of a single bulb and work out the series resistance, etc. A. same B. 2x C. 4x D. ½ x E. ¼ x. For a multi-step solution, note that the resistance is 2R, the current is 0.5 I, and EACH BULB sees HALF the115 Volts drop compared to a single 100 W bulb. At that point, you can use form 1, P =V I = 0.5 V x 0.5 I = 0.25 VI Or you can save time by going directly to form 3 with ½ 115 V across the one bulb, and you are squaring that, which gives a factor of ¼ if R is unchanged E. Is the correct answer. 32
Summary Chapter 13 Common I, voltage drops add I = q/t Amperes (Coulombs/sec) OHM s Law: R = ΔV/I [Ohms, Ω] I = ε/(r circuit + R battery ) P = ΔVI = I 2 R = (ΔV) 2 /R watts Common voltage drop, currents add 33
Current and power I = q/t Amperes (Coulombs/sec) Transferring charge q across a voltage drop ΔV involved an energy q ΔV in Joules Then when current flows across a voltage drop, there is a Power P = I ΔV in Watts or Joules/sec 34
Transmission In the distribution of electric power the goal is to deliver to the user as large a fraction as possible of the generated power. Practical cables have a specific resistance so the power losses will be I 2 R cable and we need I to be as small as possible. But we also need the delivered power P = V source I source to be as high as possible, therefore, the electrical power is distributed at very high voltage and low current. The voltage is reduced from 375,000Volts to 230Volts and/or 115 V for households by using a transformer. The current increases by the same factor since for an ideal transformer no power is lost. Transformers are the dominant reason electrical transmission is alternating current i V source R cable i V user R user V user = V source IR cable P user = iv user = iv source i 2 R cable 35
Transmission AC allows voltage step-up and step-down via magnetic induction the electric fields that arise from changing magnetic fields. With modern solid state electronics, we can step steady (DC) voltages up and down, and convert DC to AC for use in the household. DC transmission lines are better: more average voltage for a given peak voltage, and no phase-matching problems over the nation-wide grid. i V source R cable V user R user i V user = V source IR cable P user = iv user = iv source i 2 R cable 36
Transmission Above about 375 kv AC, power lines tend to leak via ionization of the surrounding air, and arcing at the insulating supports of the wires. This limits how far we can push the HV strategy. Aluminum is lighter and cheaper than copper, but has poorer conductance (higher resistance.) Choice of metal is an economic tradeoff. i V source R cable V user R user i V user = V source IR cable P user = iv user = iv source i 2 R cable 37
Household appliances Household circuits are wired in parallel so that when more than one appliance is plugged in each sees the same voltage and can get the required current. As we plug in more and more appliances the current in the circuit increases and the I 2 R losses in wires could cause a fire. This is why we have fuses and why major appliances use 230 volts and many parts of the world use 230 volts for all household use. Half the voltage means double the current and four times the dangerous waste heat. But 115 V is less likely to electrocute. As many people turn on appliances (air conditioners) the grid has to supply more power by increasing the current. P user = iv user = iv source i 2 R cable This results in a higher fraction of the power being lost in the cable In cases of very heavy load the power station reduces the transmission voltage resulting in a brown out and in extreme cases there are rolling blackouts. 38