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Transcript
Upcoming Schedule Oct. 8 19.2-19.4 Oct. 10 boardwork Quiz 4 Oct. 13 19.5-19.9 Oct. 15 boardwork Oct. 17 boardwork Quiz 5 Oct. 20 review Oct. 22 Exam 2 Ch. 18, 19 Oct. 24 20.1-20.2 Oct. 6 19.1 “As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.”—A. Einstein Examples 19-3 and 19-4 How much current flows from the battery in the circuit shown? What is the current through the 500 resistor? 500 400 a I1 c b 700 I I2 + - 12 V I What is your stragegy? Step 1—replace the 500 and 700 parallel combination by a single equivalent resistor. 500 400 a I1 I=? c b 700 I I2 I I1 = ? + - 12 V Woe is me, what to do? Always think: bite-sized chunks! Step 2—replace the 400 and Req1 series combination by a single equivalent resistor Req, net. Req1 400 a I=? c b I I I1 = ? + - 12 V Woe is me, what to do? Find another bite-sized chunk! Step 3—Solve for the current I. Req1, net c a I I + - 12 V This isn’t so complicated! Step 4—To get I1, Calculate Vbc. Use Vtotal = Vab + Vbc. Vbc 500 Vab Knowing I, Calculate I1. Woe is me! Stuck again! 400 a I1 c b 700 I I2 I + - 12 V You know Vtotal= V and I so you can get Vab and then Vbc. The voltage drop across both the 500 and 700 resistors is the same, and equal to Vbc. Use V = IR to get I1 across the 500 resistor. 500 400 a I1 c b 700 I I2 + - 12 V I I’ll post some sample solutions, worked out in detail, for similar problems. The previous slides were intended to illustrate concepts and techniques only. The next slides on the “resistor ladder” are also conceptual only. Don’t worry, there will be plenty of practice with “real” problems. Example 19-6 Resistor “ladder.” Estimate the equivalent resistance of the resistor ladder shown. All resistors have the same resistance R. A B Remember this (seriously): don’t bite off more than you can chew. Bite off little bite-sized chunks. Not a “law” of physics, but sometimes helps with circuits: look for “bite-sized” chunks sticking out at one end. Series A B The new color indicates the value of the resistance is not R. In a real problem, you would calculate the “new color” resistor’s resistance. Parallel A B Any more bite-sized chunks? Series A B Parallel A B Series A B All done! A B 19.2 EMF and Terminal Voltage We have been making calculations with voltages from batteries without asking detailed questions about the batteries. Now it’s time to ask those questions. We introduce a new term – emf – in this section. Any device which transforms a form of energy into electric energy is called a “source of emf.” “emf” is an abbreviation for “electromotive force,” but emf does not really refer to force! The emf of a source is the voltage it produces when no current is flowing. The voltage you measure across the terminals of a battery (or any source of emf) is less than the emf because of internal resistance. Here’s a battery with an emf. Also, all batteries have an emf is the potential difference between + and “internal resistance:” a + - b The “battery” is everything inside the green box. Hook up a voltmeter to measure the emf: emf a + - b The “battery” is everything inside the green box. Getting ready to connect the voltmeter (it’s not hooked up yet). Measuring the emf??? a emf + - I b The “battery” is everything inside the green box. As soon as you connect the voltmeter, current flows. You can’t measure voltage without some (however small) current flowing, so you can’t measure emf directly. You can only measure Vab. We model a battery as producing an emf, , and having an internal resistance r: r a + - r b The “battery” is everything inside the green box. Vab The terminal voltage, Vab, is the voltage you measure with current flowing. When a current I flows through the battery, Vab is related to the emf, , by Vab = ε ± I r . Why the sign? If the battery is delivering current, the V it delivers is less than the emf, so the – sign is necessary. If the battery is being charged, you have to “force” the current through the battery, and the V to “force” the current through is greater than the emf, so the + sign is necessary. This will become clear as you work (and understand) problems. Operationally, you simply include an extra resistor to represent the battery resistance, and label the battery voltage as an emf instead of V (units are still volts). Example 19-7 For the circuit below, calculate the current drawn from the battery, the terminal voltage of the battery, and the current in the 6 resistor. 10 8 6 4 5 0.5 = 9 V The following is a “conceptual” solution. Please go back and put in the numbers for yourself. In the next section, we will learn a general technique for solving circuit problems. For now, we break the circuit into manageable bits. “Bite-sized chunks.” 10 8 6 4 5 0.5 = 9 V Replace the parallel combination by its equivalent. Do you see any bite-sized chunks that are simple series or parallel? Any more “bite-sized chunks?” Pretend that everything inside the green box is a single resistor. 10 8 6 4 0.5 = 9 V Replace the series combination by its equivalent. 5 You are left with an equivalent circuit of 3 resistors in series, which you can handle. 10 8 6 4 5 0.5 = 9 V Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination by its equivalent. 19.3 Kirchoff’s Rules No, it is pronounced “KIRKOFF’s” rules. The ch sounds like “k,” not like “ch.” Analyze this circuit for me, please. Find the currents I1, I2, and I3. (I could give I1 instead of 2, and ask for I2, I3, & 2. Lots of combinations.) h 30 I1 I3 40 a 1 2 = 45 V c b d 20 I2 1 = 80 V 1 g f e I see two sets of resistors in series. This. And this. You know how to analyze those. Further analysis is difficult. For example, series1 seems to be in parallel with the 30 resistor, but what about 2? You don’t know how to analyze that combination. h 30 I1 I3 40 a 1 2 = 45 V c b series1 I2 1 = 80 V 1 g f series2 e d 20 A new technique is needed to analyze this, and far more complex circuits. Kirchoff’s Rules Kirchoff’s Junction Rule: at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction. Also called Kirchoff’s First Rule.* Kirchoff’s Loop Rule: the sum of the changes of potential around any closed path of a circuit must be zero. Also called Kirchoff’s Second Rule.** *This is just conservation of charge: charge in = charge out. **This is just conservation of energy: a charge ending up where it started out neither gains nor loses energy (Ei = Ef ). 19.4 Solving Problems with Kirchoff’s Rules Just as we had a litany for force problems in our Mechanics semester, we have a litany for circuit problems. Litany for Circuit Problems 1. Draw the circuit. 2. Label + and – for each battery (the short side is -). 3. Label the current in each branch of the circuit with a symbol and an arrow. You may choose whichever direction you wish for the arrow. 4. Apply Kirchoff’s Junction Rule at each junction. The direction of the current arrows tell you whether current is flowing in (+) or out (-). Step 4 will probably give you fewer equations than variables. Proceed to step 5 go get additional equations. 5. Apply Kirchoff’s Loop Rule for as many loops as necessary to get enough equations to solve for your unknowns. Follow each loop in one direction only—your choice. 5a. For a resistor, the sign of the potential difference is negative if your chosen loop direction is the same as the chosen current direction through that resistor; positive if opposite. 5b. For a battery, the sign of the potential difference is positive if your chosen loop direction moves from the negative terminal towards the positive; negative if opposite. 6. Collect equations, solve, and check results. We need a shortened version of the litany for quick reference. Brief litany for Circuit Problems 1. Draw the circuit. 2. Label + and – for each battery. 3. Label the current in each branch of the circuit with a symbol and an arrow. 4. Apply Kirchoff’s Junction Rule at each junction. Current in is +. 5. Apply Kirchoff’s Loop Rule for as many loops as necessary. Follow each loop in one direction only. 5a. Resistor: I V is loop 6. Solve. 5b. Battery: +- V is + loop h 30 I1 I3 40 a 1 2 = 45 V c b d 20 I2 1 = 80 V g 1 f e Back to our circuit: we have 3 unknowns (I1, I2, and I3), so we will need 3 equations. We begin with the junctions. Junction a: I3 – I1 – I 2 = 0 Junction d: -I3 + I1 + I2 = 0 --eq. 1 Junction d gave no new information, so we still need two more equations. h 30 I1 I3 40 a 1 2 = 45 V c b d 20 I2 1 = 80 V g There are three loops. 1 e f Loop 1. Loop 2. Loop 3. Any two loops will produce independent equations. Using the third loop will provide no new information. Reminders: I V is - loop +- V is + loop The “green” loop (a-h-d-c-b-a): (- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0 - 30 I1 + 45 - 41 I3 = 0 --eq. 2 The “blue” loop (a-b-c-d-e-f-g): (+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0 41 I3 -130 + 21 I2 = 0 --eq. 3 Three equations, three unknowns; the rest is “algebra.” Make sure to use voltages are in V and resistances in . Then currents will be in A. Collect our three equations: I3 – I1 – I2 = 0 - 30 I1 + 45 - 41 I3 = 0 41 I3 -130 + 21 I2 = 0 Rearrange to get variables in “right” order: – I1 – I2 + I3 = 0 - 30 I1 - 41 I3 + 45 = 0 21 I2 + 41 I3 -130 = 0 Use the middle equation to eliminate I1: I1 = (41 I3 – 45)/(-30) There are many valid sets of steps to solving a system of equations. Any that works is acceptable. Two equations left to solve: – (41 I3 – 45)/(-30) – I2 + I3 = 0 21 I2 + 41 I3 -130 = 0 Might as well work out the numbers: 1.37 I3 – 1.5 – I2 + I3 = 0 21 I2 + 41 I3 -130 = 0 – I2 + 2.37 I3 – 1.5 = 0 21 I2 + 41 I3 -130 = 0 Multiply the top equation by 21: – 21 I2 + 49.8 I3 – 31.5 = 0 21 I2 + 41 I3 -130 = 0 Add the two equations to eliminate I2: – 21 I2 + 49.8 I3 – 31.5 = 0 + ( 21 I2 + 41 I3 -130 = 0 ) 90.8 I3 – 161.5 = 0 Solve for I3: I3 = 161.5 / 90.8 I3 = 1.78 Go back to the “middle equation” two slides ago for I1: I1 = (41 I3 – 45)/(-30) I1 = - 1.37 I3 + 1.5 I1 = - (1.37) (1.78) + 1.5 I1 = - 0.94 Go back two slides to get an equation that gives I2: – I2 + 2.37 I3 – 1.5 = 0 I2 = 2.37 I3 – 1.5 I2 = (2.37) (1.78) – 1.5 I2 = 2.72 Summarize answers so your lazy prof doesn’t have to go searching for them and get irritated (don’t forget to show units in your answer): I1 = - 0.94 A I2 = 2.72 A I3 = 1.78 A These don’t look “quite like” Giancoli’s. He rounded to 2 digits. Maybe that’s why. Still, we’d better check our results. I3 – I1 – I2 = 0 - 30 I1 + 45 - 41 I3 = 0 41 I3 -130 + 21 I2 = 0 I1 = - 0.94 A I2 = 2.72 A I3 = 1.78 A 1.78 – (-0.94) – 2.72 = 0 - 30 (-0.94) + 45 - 41 (1.78) = 0.22 ? 41 (1.78) -130 + 21 (2.72) = 0.10 ? Are the last two indication of a mistake or just roundoff error? Recalculating while retaining 2 more digits gives I1=0.933, I2=2.714, I3=1.7806, and the last two results are 0.01 or less roundoff was the culprit.