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Today 3/27. Circuits Current Potential (same as always) HW: 3/27 “Circuits 2” Due Monday 3/31. Rank the bulbs from brightest to dimmest. Are any equally bright?. D. 1. A. 2. B. C. 3. E. A = B = C > D = E. Play “Current Show” powerpoint presentation.
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Today 3/27 • Circuits • Current • Potential (same as always) • HW: 3/27 “Circuits 2” Due Monday 3/31
Rank the bulbs from brightest to dimmest. Are any equally bright? D 1 A 2 B C 3 E A = B = C > D = E Play “Current Show” powerpoint presentation
Rank the bulbs from brightest to dimmest. Are any equally bright? D 1 A 2 B C 3 E Current is charge in motion, = “how many coulombs per second” pass by A = B = C > D = E Rank the bulbs for current. A = B = C > D = E Rank the batteries for current. 2 > 1 > 3
Rank the bulbs from brightest to dimmest. Are any equally bright? A A gets all the current while B and C each only get part so A is brightest. At the junction the current divides 50/50 since both branches have the same resistance. B C Rank the bulbs for current. A > B = C Rank the bulbs for brightness. More flow, more glow. A > B = C
Rank the bulbs from brightest to dimmest. Are any equally bright? Since branch D-E has mor R than branch F (extra something added in series) the current does not divide 50/50. D F E Current favors the path of least R. Beware of the words “takes the path…” as it implies none goes through D-E. Rank the bulbs for current. F > D = E Current for D = current for E as all that goes through D also goes through E. (See “Current Show.”) Rank the bulbs for brightness. F > D = E
Added in... Added in... Networks A D F B C E Which circuit has the least resistance and hence the greatest current through the battery? Series Parallel Series addition increases R, less current Parallel addition decreases R, more current
Rank for resistance A B C D E Compared to A, B and C have an extra clog on an existing path. Most R Least R B C A D E Compared to A, D and E have an extra path.
Conceptual Circuits (Current) • More flow-more glow • What goes around, comes around • Current divides at junctions • Resistance inhibits current • add something in series = more R • add something in parallel = less R • Think in terms of “networks.”
Voltage Model • Voltage: how much energy each coulomb of charge gains (battery) or loses (bulb) in going through an element. (as usual) • More glow, more voltage. • The Loop Rule: • What goes up, must come down! VA,A = 0,voltage rises and drops must must cancel around any loop.
C D E Current Model Fails A Compare B to D all B gets all of the current through A. D gets half of the current through C. B lesscurrent But the current through C is bigger than the current through A. 1/2 1/2 morecurrent Can’t be sure which is brighter!!!
C VC VA D E VD VB Voltage Model Compare B to D A Compare the voltages across B and D. More volts more glow! Recall current for B and D. V B think loops C brighter than D V C has more voltage than D
Voltage Model A The voltage for series elements SPLITS because it adds up to the battery voltage. The voltage splits equally for identical Rs. Otherwise, the voltage drop is greater across the greater R. in series B
2 3 6 5 6 12 V 12 V 8 4 8 ? 12 7 12 9 Find Equivalent Resistance Req = 6
Find current through the battery I = 2 Amps 2 5 6 12 V 12 V 8 8 6 7 9 Req = 6 It is the same as the current through Req.
12 V 6 0.25 Ampeach Current through each element 0.5 Amp 0.5 Amp I = 2 Amps 2 5 6 12 V 8 8 2 Amps 1 Amp 7 1 Amp 9 What is the current through each resistor? What is the voltage across each resistor?
12 V 6 2 V each Voltage across each element 1 V I = 2 Amps 3 V 2 5 6 12 V 5 V 8 8 7 9 V 9 7 V What is the current through each resistor? What is the voltage across each resistor?
Kirchhoff’s Rules, Loop • The sum of all voltages around any closed loop is zero. (what goes up must come down) • or VA,A = 0 • !!must keep track of ups and downs!! (+/-)
Kirchhoff’s Rules, Junction • The sum of all currents at any junction is zero. (what goes in must come out) • !!must keep track of ins and outs!! (+/-)