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UCF PHY 2049C - DC CIRCUITS

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DC CircuitsPower Source in a CircuitSlide 3A REAL Power Source is NOT an ideal batteryA Physical BatteryBack to PotentialConsider a “circuit”.To rememberNEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.Take a trip around this circuit.Circuit ReductionMultiple BatteriesReductionAnother Reduction ExampleNOTICE ASSUMED DIRECTION OF TRAVELSlide 16The Unthinkable ….RC CircuitClose the SwitchReally Close the SwitchThis is a differential equation.Slide 22Time ConstantResult q=CE(1-e-t/RC)q=CE(1-e-t/RC) and i=(CE/RC) e-t/RCDischarging a CapacitorSlide 27If the resistance is doubled what is the power dissipated by the circuit?DC CircuitsMarch 1, 2006Power Source in a CircuitVThe ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.9. Consider the circuit shown in Figure P28.9. Find (a) the current in the 20.0-Ω resistor and (b) the potential difference between points a and b.A REAL Power Sourceis NOT an ideal batteryVE or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another.By the way …. this is called a circuit!Internal ResistanceA Physical BatteryInternal ResistanceRrEmfiBack to PotentialRepresents a charge in spaceChange in potential as one circuitsthis complete circuit is ZERO!Consider a “circuit”.This trip around the circuit is the same as a path through space.THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!To rememberIn a real circuit, we can neglect the resistance of the wires compared to the resistors.We can therefore consider a wire in a circuit to be an equipotential – the change in potential over its length is slight compared to that in a resistorA resistor allows current to flow from a high potential to a lower potential.The energy needed to do this is supplied by the battery. VqW NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.LOOP EQUATIONThe sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero.Sometimes known as Kirchoff’s loop equation.NODE EQUATIONThe sum of the currents entering (or leaving) a node in a circuit is ZEROTake a trip around this circuit.Consider voltage DROPS:-E +ir +iR = 0orE=ir + iRriseCircuit Reductioni=E/ReqMultiple BatteriesReductionComputes iAnother Reduction ExamplePARALLEL12121600503012011RRNOTICE ASSUMED DIRECTION OF TRAVELVoltage Drops:-E1 –i1R1 + i2R2 + E2 -i1R1 = 0Second Loop -From “a”Let’s do it!NODEI3 +i2 = i1In the figure, all the resistors have a resistance of 4.0  and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?The Unthinkable ….RC CircuitInitially, no current through the circuitClose switch at (a) and current begins to flow until the capacitor is fully charged.If capacitor is charged and switch is switched to (b) discharge will follow.Close the SwitchI need to use E for ENote RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)Really Close the SwitchI need to use E for ERERCqdtdqorECqdtdqRCqiREdtdqi since0Equation LoopNote RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)This is a differential equation.To solve we need what is called a particular solution as well as a general solution.We often do this by creative “guessing” and then matching the guess to reality.You may or may not have studied this topic … but you WILL!RCREaeCERERCqdtdqCEqRERCqdtdqKeqqatpatp1RCEEaE/R0CEa0for t/)e-CE(1)()e-CE(1q and -CEKKCE0solution from and 0q 0,When tand 0dq/dt charged,fully is device When the:solution particularat Look Solution Generalat-at-Time ConstantRCResult q=CE(1-e-t/RC)q=CE(1-e-t/RC) and i=(CE/RC) e-t/RCRCteREi/Discharging a Capacitorqinitial=CE BIG SURPRISE! (Q=CV)iiR+q/C=0RCtRCteRCqdtdqieqqsolutionCqdtdqR/0/00In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy Eth in the resistor as a function of time t. (a)What is the electric potential across the battery? (60)(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)(c) Did you put your name on your paper? (1) Looking at the graph, we see that theresistor dissipates 0.5 mJ in one second.Therefore, the POWER =i2R=0.5 mWma 88.41088.41038.2Ω 21mW 0.53252ampiampRPimVamp 10221104.88iRVor iR reisitor theacross drop Voltage3-If the resistance is doubled what is the power dissipated by the circuit?mJRiPmaR248.043.24210102RVimV 102V


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UCF PHY 2049C - DC CIRCUITS

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