Capacitors Capacitance tells me how many coulombs of charge are stored in a capacitor when it has 1 Volt across it. A 25 F capacitor will have 25 C of charge on it when it has 1 volt across it. Think of capacitance as “coulombs per volt”. Units:coulombs per volt or “Farads” Equation: C = q/V
Capacitors Think of storing propane in a tank. The amount of propane depends on the pressure just as the amount of charge depends on voltage. We will use parallel plate capacitors so that all ideas from parallel plates apply. V = EdE = / o (two plates) = Q/A
Capacitors How much charge is on the capacitor? Think “6x10 -6 Coulombs per Volt and you won’t need the equation! 12 V 6F6F6F6F 72x10 -6 C or 72 C
Capacitors What is the E-field inside the capacitor if the plates have an area of 1 m 2 ? 12 V 6F6F6F6F Q = 72x10 -6 C or 72 C E = / o = 8.2x10 6 N/C What is the distance between the plates? V = Ed d = 12/8.2x10 6 = 1.5x10 -6 m Note that capacitance depends on the area and separation distance
Capacitors and Energy How many Joules of energy are stored in the capacitor? 12 V 6F6F Ask yourself, “How much work must be done to charge the plates?”
Capacitors and Energy How many Joules of energy are stored in the capacitor? 12 V 6F6F As the battery moves charge from one plate to another, the potential difference increases from zero to 12 V. The average is 6 V and the energy stored is q times 6 V (q V ave ). Energy = q V ave = 1 / 2 q V = 1 / 2 C V 2 C = q/ V
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.”)
Conceptual Circuits 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.”
Networks E D F A CB Which circuit has the least resistance and hence the greatest current through the battery? Added in... SeriesParallel More R, less current through battery Less R, more current through battery
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.