PHYS 1444 – Section 004 Lecture #12 EMF and Terminal Voltage EMF and Terminal Voltage Resistors in Series Energy Losses in ResistorsExample 26 – 1 Resistors in Parallel Resistor and Capacitor ArrangementsExample 26 – 2 Example 26 – 5 Kirchhoff’s Rules Kirchhoff’s 2nd Rule Using Kirchhoff’s RulesExample 26 – 8Example 26 – 8, cnt’d EMFs in Series and Parallel: Charging a BatteryWednesday, Mar. 21, 2007 1PHYS 1444-004, Spring 2007Dr. Andrew BrandtPHYS 1444 – Section 004Lecture #12Wednesday, Mar. 26 2007Dr. Andrew Brandt• EMF and Terminal Voltage• Resistors in Series and Parallel• Energy loss in ResistorsHW6 due Fri 3/30 at 8 pmWednesday, Mar. 21, 2007 2PHYS 1444-004, Spring 2007Dr. Andrew Brandt• What do we need to have current in an electric circuit?– A device that provides a potential difference, such as battery or generator• typically it converts some type of energy into electric energy• These devices are called sources of electromotive force (emf)– This does NOT refer to a real “force”.• The potential difference between terminals of the source, when no current flows to an external circuit, is called the emf() of the source.• A battery itself has some internal resistance (r ) due to the flow of charges in the electrolyte– Why do headlights 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 headlightsEMF and Terminal VoltageWednesday, Mar. 21, 2007 3PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Since the internal resistance is inside the battery, we cannot separate the two.EMF and Terminal Voltage• So the terminal voltage difference is 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 from the battery, there is an internal drop in voltage which is equal to Ir. Thus the actual delivered terminal voltage is abVIrε=−Wednesday, Mar. 21, 2007 4PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Resistors are in series when two or more of them are connected end to end– These resistors represent simple electrical devices in a circuit, such as light bulbs, heaters, dryers, etc.Resistors in Series• What is common in a circuit connected in series?– the current is the same through all the elements in series• Potential difference across each element in the circuit is:V1=IR1, V2=IR2and V3=IR3• Since the total potential difference is V, we obtainV=IReq=V1+V2+V3=I(R1+R2+R3)Thus, Req=R1+R2+R3eq iiRR=∑Resistors in seriesWhen resistors are connected in series, the total resistance increases and the current decreases.Wednesday, Mar. 21, 2007 5PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Why is it true that V=V1+V2+V3?Energy Losses in Resistors• What is the potential energy loss when charge q passes through the resistor R1, R2and R3ΔU1=qV1, ΔU2=qV2, ΔU3=qV3• Since the total energy loss should be the same as the energy provided to the system, we obtainΔU=qV=ΔU1+ΔU2+ΔU3=q(V1+V2+V3)Thus, V=V1+V2+V3Wednesday, Mar. 21, 2007 6PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 26 – 1 Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0V and whose internal resistance is 0.5-Ω. 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=abVIrε=−We obtain I=Solve for I(b) The terminal voltage VabisabV=(c) The power dissipated in R and r areP=P=IR =Irε−Rrε=+12.00.18365.0 0.5VA=Ω+ΩIrε−=12.0 0.183 0.5 11.9VA V−⋅Ω=2IR=()20.183 65.0 2.18AW⋅Ω=2Ir=()20.183 0.5 0.02AW⋅Ω=What is this?A battery or a source of emf.Wednesday, Mar. 21, 2007 7PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Resisters are in parallel when two or more resistors are connected in separate branches– Most house and building wirings are arranged this way.Resistors in Parallel• What is common in a circuit connected in parallel?– The voltage is the same across all the resistors.– The total current that leaves the battery, is however, split.• The current that passes through every element isI1=V/R1, I2=V/R2, I3=V/R3• Since the total current is I, we obtainI=V/Req=I1+I2+I3=V(1/R1+1/R2+1/R3)Thus, 1/Req=1/R1+1/R2+1/R311eq iiRR=∑Resisters in parallelWhen resistors are connected in parallel, the total resistance decreases and the current increases.Wednesday, Mar. 21, 2007 8PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Parallel Capacitor arrangementsResistor and Capacitor Arrangementseq iiCC=∑11eq iiRR=∑• Series Resistor arrangements• Series Capacitor arrangements11eq iiCC=∑• Parallel Resistor arrangementseq iiRR=∑Wednesday, Mar. 21, 2007 9PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 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.2R2R2RWednesday, Mar. 21, 2007 10PHYS 1444-004, Spring 2007Dr. Andrew BrandtExample 26 – 5 Current in one branch. What is the current flowing through the 500-Ω resistor 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. Reqof the small parallel branch is: PR=Reqof the circuit is: eqR=The voltage drop across the parallel branch isbcV=The current flowing across 500-Ω resistor is thereforeWhat is the current flowing in the 700-Ω resister?700I=bcVR=34.969.92 10 9.92500mA−=×=500II−=17 9.92 7.08mA−=1112500 700 3500+=3500123500400 400 292 69212+=+=ΩeqVR=1217692mA=PIR=317 10 292 4.96 V−×⋅==500IWednesday, Mar. 21, 2007 11PHYS 1444-004, Spring 2007Dr. Andrew Brandt• Some
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