# Today 3/27  Circuits  Current  Potential (same as always)  HW:3/27 “Circuits 2” Due Monday 3/31.

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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? A E D CB A = B = C > D = E 132 Play “Current Show” powerpoint presentation

Rank the bulbs from brightest to dimmest. Are any equally bright? A E D CB A = B = C > D = E Rank the bulbs for current. A = B = C > D = E Rank the batteries for current. 132 2 > 1 > 3 Current is charge in motion, = “how many coulombs per second” pass by

Rank the bulbs from brightest to dimmest. Are any equally bright? A CB A > B = C Rank the bulbs for current. A > B = C Rank the bulbs for brightness. 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. More flow, more glow.

Rank the bulbs from brightest to dimmest. Are any equally bright? E D F F > D = E Rank the bulbs for current. F > D = E Rank the bulbs for brightness. Since branch D-E has mor R than branch F (extra something added in series) the current does not divide 50/50. Current favors the path of least R. Beware of the words “takes the path…” as it implies none goes through D-E. Current for D = current for E as all that goes through D also goes through E. (See “Current Show.”)

Networks E D F A CB Which circuit has the least resistance and hence the greatest current through the battery? Added in... SeriesParallel Series addition increases R, less current Parallel addition decreases R, more current

Rank for resistance ABCD E Most R Least R BC E AD Compared to A, B and C have an extra clog on an existing path. 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!  V A,A = 0, voltage rises and drops must must cancel around any loop.

Current Model Fails A B Compare B to D B gets all of the current through A. D gets half of the current through C. But the current through C is bigger than the current through A. C ED Can’t be sure which is brighter!!! less current more current all 1/21/2 1/21/2

Voltage Model A B Compare B to D C ED C brighter than D C has more voltage than D VCVC VDVD Compare the voltages across B and D. More volts more glow! Recall current for B and D. VAVA VBVB think loops V V

Voltage Model 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. A B in series

Find Equivalent Resistance 12 V ?? R eq = 6  22 66 88 88 99 77 55 12 V 44 66 33 12 

Find current through the battery 22 66 88 88 99 77 55 12 V 66 I = 2 Amps 12 V It is the same as the current through R eq. R eq = 6 

Current through each element 22 66 88 88 99 77 55 12 V 66 I = 2 Amps What is the current through each resistor? 12 V 1 Amp 0.5 Amp 0.25 Amp each What is the voltage across each resistor? 2 Amps

Voltage across each element 22 66 88 88 99 77 55 12 V 66 I = 2 Amps What is the current through each resistor? 12 V 5 V 9 V 3 V 1 V 2 V each What is the voltage across each resistor? 7 V

Kirchhoff’s Rules, Loop  The sum of all voltages around any closed loop is zero. (what goes up must come down)  or  V A,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!! (+/-)

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