PHYS 1442 – Section 004 Lecture #9 Wednesday February 12, 2014 Dr. Justin Griffiths for Dr. Andrew Brandt CH 19 “Simple” circuits Kirchoff’s Laws RC Circuits 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Announcements a) HW5 on ch 19 due Monday 17th at 5:25 pm (can get a 24 hour extension if in class) d) Test on Weds. Feb. 19th ch 16-18 plus part of ch 19th Mostly multiple guess, bring a scantron, I don’t know how many problems 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Resistors in Series and in Parallel Example : Analyzing a circuit.(a) How much current is drawn from the battery? (b) what is the current in the 10 Ω resistor a.) Overall resistance is 10.3 Ω. The current is 9.0 V/10.3 Ω = 0.87 A b.) The voltage across the 4.8 Ω is 0.87*4.8=4.2V, so the current in the 10 Ω is I=V/R=4.2/10=0.42A Solution: a. First, find the equivalent resistance. The 8-Ω and 4-Ω resistors in parallel have an equivalent resistance of 2.7 Ω; this is in series with the 6.0-Ω resistor, giving an equivalent of 8.7 Ω. This is in parallel with the 10.0-Ω resistor, giving an equivalent of 4.8 Ω; finally, this is in series with the 5.0-Ω resistor and the internal resistance of the battery, for an overall resistance of 10.3 Ω. The current is 9.0 V/10.3 Ω = 0.87 A. b. The terminal voltage of the battery is the emf less Ir = 8.6 V. c. The current across the 6.0-Ω resistor and the 2.7-Ω equivalent resistance is the same; the potential drop across the 10.0-Ω resistor and the 8.3-Ω equivalent resistor is also the same. The sum of the potential drop across the 8.3-Ω equivalent resistor plus the drops across the 5.0-Ω resistor and the internal resistance equals the emf of the battery; the current through the 8.3-Ω equivalent resistor is then 0.48 A. 7/4/2019

Kirchhoff’s Rules Some circuits are very complicated to analyze using the simple rules for combining resistors G. R. Kirchhoff devised two rules to deal with complicated circuits. Kirchhoff’s rules are based on conservation of charge and energy Kirchhoff’s 1st rule: Junction rule, charge conservation. At any junction point, the sum of all currents entering the junction must equal to the sum of all currents leaving the junction. In other words, charge is conserved At junction a in the figure, I3 goes into the junction while I1 and I2 leaves: I3 = I1+ I2 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Kirchhoff’s 2nd Rule Kirchoff’s 2nd rule: Loop rule, uses conservation of energy. The sum of the changes in potential around any closed path of a circuit must be zero. 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Kirchhoff’s 2nd Rule The current in the circuit in the figure is I=12/690 =0.017A. Point e is the highest potential point while point d is the lowest potential. When the test charge starts at e and returns to e, the total potential change is 0. Between point e and a, no potential change since there is no source of potential nor any resistance. Between a and b, there is a 400W resistance, causing IR=0.017*400 =6.8V drop. Between b and c, there is a 290W resistance, causing IR=0.017*290 =5.2V drop. Since these are voltage drops, we use negative sign for these, -6.8V and -5.2V. No change between c and d while from d to e there is +12V change. Thus the total change of the voltage through the loop is: -6.8V-5.2V+12V=0V.

Example Use Kirchhoff’s rules. Calculate the currents I1, I2 and I3. The directions of the current through the circuit is not known a priori but since the current tends to move away from the positive terminal of a battery, we arbitrarily choose the direction of the currents as shown. We have three unknowns so we need three equations. Using Kirchhoff’s junction rule at point a, we obtain This is the same for junction d as well, so it gives no additional information. Now apply K’s second rule to the top loop ahdcba The total voltage change in loop ahdcba is. 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Using Kirchhoff’s Rules Determine the flow of currents at the junctions. It does not matter which direction you assign a current. If the value of the current after completing the calculations are negative, it means you chose the wrong direction Write down the current equation based on Kirchhoff’s 1st rule at various junctions. Choose closed loops in the circuit Write down the potential in each interval of the junctions, keeping the sign properly. Write down the potential equations for each loop. Solve the equations for unknowns. 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

Example cnt’d Now the 2nd rule on the bottom loop agfedcba. The total voltage change in loop agfedcba So the three equations are We can obtain the three current by solving these equations for I1, I2 and I3. 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

EMFs in Series and Parallel: Charging a Battery When two or more sources of emfs, such as batteries, are connected in series The total voltage is the algebraic sum of their voltages, if their direction is the same Vab=1.5 + 1.5=3.0V in figure (a). If the batteries are arranged in an opposite direction, the total voltage is the difference between them Parallel arrangements (c) are used only to increase currents. Vac=20 – 12=8.0V in figure (b) Connecting batteries in opposite direction might seem wasteful. This, however, is the way a battery charger works. Since the 20V battery is at a higher voltage, it forces charges into 12V battery Some battery are rechargeable since their chemical reactions are reversible but most the batteries can not reverse their chemical reactions Why would you consider parallel arrangement

RC Circuits Circuits containing both resistors and capacitors RC circuits are used commonly in everyday life Control windshield wiper Timing of traffic light from red to green Camera flashes and heart pacemakers What does an RC circuit look like? There should be a source of emf, capacitors and resistors What happens when the switch S is closed? Current immediately starts flowing through the circuit. Electrons flow out of negative terminal of the emf source, through the resistor R and accumulate on the upper plate of the capacitor The electrons from the bottom plate of the capacitor will flow into the positive terminal of the battery, leaving only positive charge on the bottom plate As the charge accumulates on the capacitor, the potential difference across it increases The current reduces gradually to zero -- at that point the voltage across the capacitor is the same as that of the emf The charge on the capacitor increases until it reaches to its maximum C. 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt

RC Circuits What does all this look like graphically? Charge on the capacitor and current as a function of time From energy conservation (Kirchhoff’s 2nd rule), the emf  must be equal to the voltage drop across the capacitor and the resistor =IR+Q/C R includes all resistance in the circuit, including the internal resistance of the battery, I is the current in the circuit at any instant, and Q is the charge of the capacitor at that same instant 7/4/2019 PHYS 1442-004, Dr. Andrew Brandt