Chapter 28Direct Current Circuits1. R connections in series and in parallel2. Define DC (direct current), AC (alternating current)3. Model of a battery4. Circuits with 2+ batteries – Kirchhoff’s Rules5. RC circuit子曰子曰子曰子曰:"温故而知新温故而知新温故而知新温故而知新"Confucius says, reviewing helps one learn new knowledge.Concepts:Charge: positive, negative, conserve, induction.potentialElectric: fieldfluxElectrostatic equilibrium: no moving charge.Current: moving charge Capacitance: charge over potentialResistance: potential over currentResistance and Resistivity (conductivity) and temperature: testqFE≡EFq=Φ = ⋅∫ surfaceEdE ABAV d∆− = ⋅∫E sVxxxV )( kjiE∂∂+∂∂+∂∂≡−∇=VQC∆≡dtdQI =VRI∆≡[1 ( )]o oρ ρ α T T= + −= + −[1 ( )]o oR Rα T TLaws:Coulomb's: force and charge.Gauss’s: electric flux and charge.Ohm’s: electric potential and current.121212rrF221022141rqqrqqkeπε==0AEεΦqdE=⋅=∫VRI∆≡Circuits and components:Symbols: wire, battery, C, Switch…Power: current times potential difference:Capacitor: in parallel in seriesResistor: discuss today= ⋅ ∆p I V...CCCCeq+++=321...CCCCeq+++=3211111Resistor connectionsIn series. Condition: In parallel.Condition:= =∆ = ∆ + ∆1 21 2I I IV V V∆ + ∆ ∆ ∆∆≡ = = + ≡ +1 2 1 21 21 2eqV V V VVR R RI I I I= +∆ = ∆ = ∆1 21 2I I IV V V+≡ = = + ≡ +∆ ∆ ∆ ∆1 2 1 21 2 1 21 1 1eqI I I IIR V V V V R R∆1V∆2V∆1V∆2VResistor connectionsIn series, :voltage sharingpower sharing∆=∆1 12 2V RV R=1 12 2P RP RIn parallel, : current sharingpower sharing= =∵1 2I I I∆1V∆2V∆ = ∆ = ∆∵1 2V V V= =1 21 1 2 22 1, or I RI R I RI R= =1 21 1 2 22 1, or P RPR P RP R∆1V∆2VResistors connections, summaryIn seriesIn parallel...= + + +1 2 3eqR R R R...= + + +1 2 31 1 1 1eqR R R R∆ ∆ ∆ =1 2 3 1 2 3: : :... : : :...V V V R R R...= = =1 1 2 2 3 3I R I R I R...= = =1 1 2 2 3 3PR P R P R...= = = =1 2 3I I I I...∆ = ∆ = ∆ = ∆ =1 2 3V V V V=1 2 3 1 2 3: : :... : : :...P P P R R RResistors in Series – Example Use the active figure to vary the battery voltage and the resistor valuesObserve the effect on the currents and voltages of the individual resistorsPLAYACTIVE FIGUREResistors in Parallel – Example Use the active figure to vary the battery voltage and the resistor valuesObserve the effect on the currents and voltages of the individual resistorsPLAYACTIVE FIGURECombinations ofResistorsThe 8.0-Ω and 4.0-Ωresistors are in series and can be replaced with their equivalent, 12.0 ΩThe 6.0-Ω and 3.0-Ωresistors are in parallel and can be replaced with their equivalent, 2.0 ΩThese equivalent resistances are in series and can be replaced with their equivalent resistance, 14.0 ΩMore examplesR1R2R3R4R5Direct Current and Alternating CurrentWhen the current direction (not magnitude) in a circuit does not change with time, the current is called direct currentMost of the circuits analyzed will be assumed to be in steady state: with constant magnitude and direction, like the one powered through a battery.When the current direction (often also the magnitude) in a circuit changes with time, the current is call alternating current.The current from your car’s alternator is AC.Model of a batteryTwo parameters, electromotive force (emf), ε, and the internal resistance r, are used to model a battery.When a battery is connected in a circuit, the electric potential measured at its + and –terminals are called The terminal voltage ∆V, with ∆V = ε– IrIf the internal resistance is zero, the terminal voltage equals the emfε.The internal resistance, r, does not change with external load resistance R, and this provides the way to measure the internal resistance.∆Vbatteryload∆VPLAYACTIVE FIGUREBattery power figurebatteryload∆VThe power the battery generates (through chemical reactions):= ⋅ = + ⋅2( )pε I R r IThe power the battery delivers to the load, hence efficiency:= ∆ ⋅ = ⋅2loadp V I R I=+efficiency = loadpRp R rThe maximum power the battery can deliver to a load= ⋅2loadp R IFromand= + ⋅( )ε R r IWe have=+22( )loadRpεR rWhere the emf is a constant once the battery is given.ε = − = + + 22 31 20( ) ( )loaddpRεdR R r R rFrom We get to be the condition for maximum , or power delivered to the load.=R rloadpBattery power figurebatteryload∆VOne can also obtain this result from the plot of reaches the maximum value=+22( )loadRpεR rWhere whenThe efficiency of the battery at this point is 50% because =R rloadp=+efficiency = loadpRp R rMore complicated circuits, circuits with 2+ batteries: Kirchhoff’s RulesA typical circuit that goes beyond simplifications with the parallel and series formulas: the current in the diagram. Kirchhoff’s rules can be used to solve problems like this.Rule 1:Kirchhoff’s Junction Rule Junction Rule, from charge conservation: The sum of the currents at any junction must equal zero Mathematically: The example on the left figure:I1- I2 - I3 = 00junctionI=∑Rule 2:Kirchhoff’s Loop RuleChoose your loopLoop Rule, from energy conservation:The sum of the potential differences across all elements around any closed circuit loop must be zeroMathematically: One needs to pay attention the sign (+ or -) of these potential changes, following the chosen loop direction.closedloop0V∆ =∑∆1V∆2VLoop directionRemember two things:1. A battery supplies power. Potential rises from the “–”terminal to “+” terminal.2. Current follows the direction of electric field, hence the decrease of potential.Kirchhoff’s rulesStrict steps in solving a problemStep 1: choose and markthe loop.L1L2Step 2: choose and markcurrent directions. Mark the potential change on resistors.I1I3I2Step 3: apply junction rule:Step 4: apply loop rule:+ − =1 2 30I I I. . .. . .− + =− − − =3 23 1L1: +2 00 12 0 4 00 0L2: 8 00 2 00 6 00 0I II IStep 5: solve the three equations for the three variables.+–––++One more exampleL1L2Step 1: choose and markthe loop.Step 2: choose and markcurrent directions. Mark the potential change on resistors.Step 3: apply junction rule:Step 4: apply loop rule:+ − =1 2 30I I I. . . .. . .− − − =− + − =1 23 1L1: +6 0 10 0 4 0 14 0 0L2: 2 0 10 0 6 0 0I II IStep
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