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SMU PHYS 1304 - Direct Current Circuits

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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≡EFq=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.121212rrF221022141rqqrqqkeπε==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, summaryIn seriesIn 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 valuesObserve 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 valuesObserve the effect on the currents and voltages of the individual resistorsPLAYACTIVE FIGURECombinations ofResistorsThe 8.0-Ω and 4.0-Ωresistors are in series and can be replaced with their equivalent, 12.0 ΩThe 6.0-Ω and 3.0-Ωresistors are in parallel and can be replaced with their equivalent, 2.0 ΩThese equivalent resistances are in series and can be replaced with their equivalent resistance, 14.0 ΩMore examplesR1R2R3R4R5Direct Current and Alternating CurrentWhen the current direction (not magnitude) in a circuit does not change with time, the current is called direct currentMost 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 batteryTwo 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– IrIf 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 RulesA 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 RuleChoose your loopLoop Rule, from energy conservation:The sum of the potential differences across all elements around any closed circuit loop must be zeroMathematically: 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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SMU PHYS 1304 - Direct Current Circuits

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