Presentation on theme: "Principles of Parallel and Series Circuits using"— Presentation transcript:
1 Principles of Parallel and Series Circuits using Development of BasicPrinciples of Paralleland Series Circuits usingconcepts of Electric Fields.
2 Potential energy is lost as charge moves through a circuit. This Is measured in Volts. As an electron moves toward the (+)Terminal it will lose potential energy which is usuallyConverted into heat. The symbol (V) actually means V.V=ED or EL means Energy lost/charge = Force/charge x LengthV = ED or EL Work = Force x DistanceVolts Potential Energy∆V = EDD
9 The currents thru the resistors are the same even when reversed. Go back and check it out!
10 Ed =∆V Series Circuit Voltage Current ∆V I < ∆VI I Ed=∆V Copper I NichromeI
11 EL= Volts VI=(volts)(3a)= HOT Compare Electric Fields in the wires. NichromeEEL= VoltsCopperLengthElectrons build up at the Cu/Nichrome junction and reduce theelectric field in copper while increasing the electric field innichrome UNTIL the two currents are the SAME.It takes less force (electric field) to push the current throughthe copper wire than through the nichrome wire.VI=(volts)(3a)= HOTVI=(volts)(3a)= cool
12 2a6vElectric field in theNichrome wire mustbe larger because ofits larger resistance.El cu +El nic.=V 6vElectric field in aSeries Circuit:Cu wireHave studentstouch the wires.Nichrome wireCool hot
13 Conclusion about Series Circuits: 1. All the currents must be the same!2. The resistor with the most resistance must have thelargest electric field in it.3. Therefore, The resistor with the most volts lost(greatest resistance) must get the hottest.∆V I = watts ∆V = E d
17 Power= VI Which wire gets Hotter? Copper WireNichrome WireEE∆V=EL∆V=EL2.8 volts2.8 voltsLengthLengthThe resistance of nichromewire is large so the currentshould be smaller than Cu.The resistance of copper wireis small so the current shouldbe larger than in the Nichrome.∆VI= hotter∆V I= coolerPower = (2.8 v)(12a)=34wattsPower=(2.8v)(3a)=8.4watts
18 ConclusionIn a parallel circuit the voltages across the resistances areequal. (Logic tells us this must be true)The currents add up to the total current.The electric fields which push the electrons around areequal in each resistor…since ∆V=Ed and the d’s are =.
22 Volts lost R Extension Cord Light bulb ∆V I + ∆V I + ∆V I = total powerL cu = L nichromenichromeVoltslostEcoppercopperVolts lostVolts lostElectric force field in the copper wires is very small; as in theprevious example the field in the nichrome must be larger toproduce the same current throughout.Therefore, negligible heat is generated in the ext. cord
24 In “Ohmic” devices…..that follow ohm’s law the current is proportional to the voltage. V α I therefore the ratio of V to Iis a constant V/I = constant This constant is called theresistance. Therefore V/I = R or V = IR
25 I V R=4 Variable resistor: Vary R and record V and I. Graph. Graphing Ohm’sLaw: A demoon the black-board.VR=4VR=28V424I
26 In “Ohmic” devices…..that follow ohm’s law the current is proportional to the voltage. V α I therefore the ratio of V to Iis a constant V/I = constant This constant is called theresistance. Therefore V/I = R or V = IRWhat happens to the total resistance of a circuit when you addresistors in series??
27 . . R=4 What happens when you add another resistance in series? Is the resistance going “up” stayingthe “same” or “going down” ??If you compare slides 4,5,and 7 youcan see that adding wires in seriesincreases the resistance and lowersthe current. Adding two 2 ohmresistors in series will make the currenthalf and therefore must double thetotal resistance. Look at the graph frompoint “a” to point “b”.Resistors placed in series have theirvalues added to find the total resistance.Variableresistor: VaryR and recordV and I. Graph.R=4V..R=2b8a424I
28 Conclusion about Series Circuits: 1. All the currents must be the same!2. The resistor with the most resistance must have thelargest electric field in it.3. Therefore, The resistor with the most volts lost(greatest resistance) must get the hottest.∆V I = watts ∆V = E d4. R total = R 1 + R 2 + R 3 etc