PHYS 1444 – Section 003 Lecture #12AnnouncementsWhat did we do at Saturday’s workshop?EMF and Terminal VoltageSlide 5Example 26 – 1Resisters in SeriesEnergy Losses in ResistersResisters in ParallelResister and Capacitor ArrangementsExample 26 – 2Example 26 – 5Kirchhoff’s Rules – 1st RuleKirchhoff’s Rules – 2nd RuleUsing Kirchhoff’s RulesExample 26 – 8Example 26 – 8, cnt’dEMFs in Series and Parallel: Charging a BatteryRC CircuitsSlide 20Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu1PHYS 1444 – Section 003Lecture #12Monday, Oct. 10, 2005Dr. Jaehoon Yu•EMF and Terminal Voltage•Resisters in series and parallel•Kirchhoff’s Rules•EMFs in series and parallel•RC CircuitsToday’s homework is homework #7, due noon, next Tuesday!!Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu2Announcements•First term exam Wednesday, Oct. 12–Time: 1 – 2:20 pm–Location: SH103–Coverage: CH. 21 – 25•Reading Assignments–CH26 – 5 –CH26 – 6Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu3What did we do at Saturday’s workshop?Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu4•What do we need to have current in an electric circuit?–A device that provides a potential difference, such as battery or generator•They normally convert some types of energy into electric energy•These devices are called source of electromotive force (emf)–This is does NOT refer to a real “force”.•Potential difference between terminals of emf source, when no current flows to an external circuit, is called the emf () of the source.•Battery itself has some internal resistance (r) due to the flow of charges in the electrolyte–Why does the headlight dim when you start the car?•The starter needs a large amount of current but the battery cannot provide charge fast enough to supply current to both the starter and the headlight EMF and Terminal VoltageMonday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu5•Since the internal resistance is inside the battery, we can never separate them out. EMF and Terminal Voltage•So the terminal voltage difference Vab=Va-Vb.•When no current is drawn from the battery, the terminal voltage equals the emf which is determined by the chemical reaction; Vab= .•However when the current I flows naturally from the battery, there is an internal drop in voltage which is equal to Ir. Thus the actual delivered terminal voltage is abV Ire= -Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu6Example 26 – 1 Battery with internal resistance. A 65.0- resistor is connected to the terminals of a battery whose emf is 12.0V and whose internal resistance is 0.5-. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistor. (a) Since abV =abV Ire= -We obtain I =Solve for I(b) The terminal voltage Vab isabV =(c) The power dissipated in R and r areP =P =IR =Ire -R re=+12.00.18365.0 0.5VA=W+ WIre - =12.0 0.183 0.5 11.9V A V- � W=2I R =( )20.183 65.0 2.18A W� W=2I r =( )20.183 0.5 0.02A W� W=What is this?A battery or a source of emf.Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu7•Resisters are in series when two or more resisters are connected end to end–These resisters represent simple resisters in circuit or electrical devices, such as light bulbs, heaters, dryers, etc Resisters in Series•What is common in a circuit connected in series?–Current is the same through all the elements in series•Potential difference across every element in the circuit is–V1=IR1, V2=IR2 and V3=IR3• Since the total potential difference is V, we obtain–V=IReq=V1+V2+V3=I(R1+R2+R3)–Thus, Req=R1+R2+R3eq iiR R=�Resisters in seriesWhen resisters are connected in series, the total resistance increases but the current decreases.Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu8•Why is it true that V=V1+V2+V3? Energy Losses in Resisters•What is the potential energy loss when charge q passes through the resister R1, R2 and R3– U1=qV1, U2=qV2, U3=qV3 •Since the total energy loss should be the same as the energy provided to the system, we obtain– U=qV=U1+U2+U3=q(V1+V2+V3)–Thus, V=V1+V2+V3Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu9•Resisters are in parallel when two or more resisters are connected in separate branches–Most the house and building wirings are arranged this way. Resisters in Parallel•What is common in a circuit connected in parallel?–The voltage is the same across all the resisters.–The total current that leaves the battery, is however, split.•The current that passes through every element is–I1=V/R1, I2=V/R2, I3=V/R3•Since the total current is I, we obtain–I=V/Req=I1+I2+I3=V(1/R1+1/R2+1/R3)–Thus, 1/Req=1/R1+1/R2+1/R31 1eq iiR R=�Resisters in parallelWhen resisters are connected in parallel, the total resistance decreases but the current increases.Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu10•Parallel Capacitor arrangements Resister and Capacitor Arrangementseq iiC C=�1 1eq iiR R=�•Parallel Resister arrangements•Series Capacitor arrangements1 1eq iiC C=�•Series Resister arrangementseq iiR R=�Monday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu11Example 26 – 2 Series or parallel? (a) The light bulbs in the figure are identical and have identical resistance R. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? (a) What are the equivalent resistances for the two cases? SeriesThe bulbs get brighter when the total power transformed is larger.seriesSP =eqR =IV =Parallel1eqR=eqR =SoIV =PP =parallel2eqVR=22VR2eqVR=22VR=4SPSo parallel circuit provides brighter lighting.(b) Car’s headlights are in parallel to provide brighter lighting and also to prevent both lights going out at the same time when one burns out. So what is bad about parallel circuits? Uses more energy in a given time.2R2R2RMonday, Oct. 10, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu12Example 26 – 5 Current in one branch. What is the current flowing through the 500- resister in the figure?What do we need to find first? Thus the total current in the circuit is1PR=I =We need to find the total current.To do that we need to compute the equivalent resistance. Req of the small parallel
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