4 slides per page(.pdf)

20.7 RL CircuitsRL Circuit (continued)Recall Ohm’s s Law to find the voltage drop on RΔ V = IRWe have something similar with inductorsΔ IEL= − LΔ tSimilar to the case of the capacitor, we get an equationfor the current as a function of time (series circuit).IE= −R−Rt/L( 1 e )(voltage across a resistor)Lτ =R(voltage across an inductor)IV= −R−Rt / L( 1 e )6/12/200796/12/20071020.8 Energy stored in a magnetic fieldThe battery in any circuit that contains a coil has to dowork to produce a currentSimilar to the capacitor, any coil (or inductor) would storepotential energyPEL1= LI2Summary of the properties of circuit elements.Resistor2CapacitorInductorExample: stored energyA 24V battery is connected in series with a resistor and an inductor,where R = 8.0Ω and L = 4.0H. Find the energy stored in the inductorwhen the current reaches its maximum value.6/12/2007unitssymbolrelationpower dissipatedenergy storedohm, Ω = V / ARV = I RP = I V = I² R= V² / R0farad, F = C / VCQ = C V0PE C= C V² / 2henry, H = V s / ALemf = -L (ΔI / Δt)0PE L= L I² / 2116/12/2007123

A 24V battery is connected in series with a resistor and an inductor, where R =8.0Ω and L = 4.0H. Find the energy stored in the inductor when the current creaches its maximum value.Given:V = 24 VR = 8.0 ΩL = 4.0 HRecall that the energy stored in theinductor isPEL1= LI22Chapter 21Find:PE L=?The only thing that is unknown inthe equation above is current. Themaximum value for the current isImaxV 24V= = = 3.0AR 8.0ΩInserting this into the above expression for the energy givesAlternating Current Circuitsand Electromagnetic Waves6/12/2007PE 1 ( 4.0 )( 3.0 ) 2L= H A = 18J213AC Circuit• An AC circuit consists of a combination ofcircuit elements and an AC generator orsource• The output of an AC generator is sinusoidaland varies with time according to thefollowing equation• Δv v = ΔV max sin 2πƒt2• Δv v is the instantaneous voltage• ΔV max is the maximum voltage of the generator• ƒ is the frequency at which the voltage changes, in Hz6/12/2007 15Resistor in an AC Circuit• Consider a circuitconsisting of an AC sourceand a resistor• The graph shows thecurrent through and thevoltage across the resistor• The current and thevoltage reach theirmaximum values at thesame time• The current and thevoltage are said to be inphase6/12/2007 164

More About Resistors in an ACCircuit• The direction of the current has no effect onthe behavior of the resistor• The rate at which electrical energy isdissipated in the circuit is given by• P = i 2 R= (I max sin 2πƒt) 22 R• where i is the instantaneous current• the heating effect produced by an AC current with amaximum value of I max is not the same as that of a DCcurrent of the same value• The maximum current occurs for a small amount of time6/12/2007 17rms Current and Voltage• The rms current is the direct currentthat would dissipate the same amountof energy in a resistor as is actuallydissipated by the AC currentImaxIrms= =20.707 Imax• Alternating voltages can also bediscussed in terms of rms valuesΔVΔV=2= 0.707 ΔVmaxrmsmax6/12/2007 18Ohm’s s Law in an AC CircuitExample: an AC circuit• rms values will be used when discussingAC currents and voltages• AC ammeters and voltmeters are designedto read rms values• Many of the equations will be in the sameform as in DC circuits• Ohm’s s Law for a resistor, R, in an ACcircuit• ΔV rms = I rms R• Also applies to the maximum values of v and i 6/12/2007 206/12/2007 19An ac voltage source has an output of ΔV V = 150 sin (377 t). Find(a) the rms voltage output,(b) the frequency of the source, and(c) the voltage at t = (1/120)s.(d) Find the rms current in the circuit when the generator isconnected to a 50.0Ω resistor.ΔVmaxΔ Vrms= = 0.707 Δ Vmax= 0.707 x 150V = 106 V2ω = 377 rad/sec, ω = 2 π f, f = ω/ 2 π = 377/ 2π= 60 HzΔV = 150 sin (377 x 1/120) = 0 VΔV rms= I rmsR thus, I rms= ΔV rms/R = 2.12 A5

Capacitors in an AC Circuit• Consider a circuit containing a capacitor andan AC source• The current starts out at a large value andcharges the plates of the capacitor• There is initially no resistance to hinder the flow ofthe current while the plates are not charged• As the charge on the plates increases, thevoltage across the plates increases and thecurrent flowing in the circuit decreases6/12/2007 21More About Capacitors in anAC Circuit• The current reversesdirection• The voltage acrossthe plates decreasesas the plates losethe charge they hadaccumulated• The voltage acrossthe capacitor lagsbehind the currentby 90°6/12/2007 22Capacitive Reactance andOhm’s s Law• The impeding effect of a capacitor on thecurrent in an AC circuit is called the capacitivereactance and is given byX C1=2πƒC• When ƒ is in Hz and C is in F, X C will be in ohms• Ohm’s s Law for a capacitor in an AC circuit• ΔV rms = I rms X CInductors in an AC Circuit• Consider an AC circuitwith a source and aninductor• The current in thecircuit is impeded bythe back emf of theinductor• The voltage across theinductor always leadsthe current by 90°6/12/2007 236/12/2007 246

Inductive Reactance andOhm’s s Law• The effective resistance of a coil in anAC circuit is called its inductivereactance and is given by• X L = 2πƒL2• When ƒ is in Hz and L is in H, X L will be inohms• Ohm’s s Law for the inductor• ΔV rms = I rms X L6/12/2007 25Example: AC circuit withcapacitors and inductorsA 2.40mF capacitor is connected across an alternating voltage with anrms value of 9.00V. The rms current in the capacitor is 25.0mA. (a) Whatis the source frequency? (b) If the capacitor is replaced by an ideal coilwith an inductance of 0.160H, what is the rms current in the coil?ΔV rms = I rms X C , first we find X C :ΔV rms / I rmsNow, solve for ƒ: ƒ = 1/ 2π X C C = 0.184 Hz= 9.00V/25.0 x 10 - 3 A = 360 ohms1 ⎞= ⎟2πƒC⎠For and inductor X L = 2πƒL, 2try solving for I rms = ΔV rms / X L6/12/2007 26⎛⎜⎝X CThe RLC Series Circuit• The resistor,inductor, andcapacitor can becombined in a circuit• The current in thecircuit is the same atany time and variessinusoidally withtime6/12/2007 27Current and VoltageRelationships in an RLC Circuit• The instantaneousvoltage across theresistor is in phase withthe current• The instantaneousvoltage across theinductor leads thecurrent by 90°• The instantaneousvoltage across thecapacitor lags thecurrent by 90°6/12/2007 287

Phasor Diagrams• To account for thedifferent phases of thevoltage drops, vectortechniques are used• Represent the voltageacross each element asa rotating vector, calleda phasor• The diagram is called aphasor diagram6/12/2007 29Phasor Diagram for RLCSeries Circuit• The voltage across theresistor is on the +xaxis since it is in phasewith the current• The voltage across theinductor is on the +ysince it leads thecurrent by 90°• The voltage across thecapacitor is on the –yaxis since it lags behindthe current by 90°6/12/2007 30Phasor Diagram, cont• The phasors areadded as vectors toaccount for thephase differences inthe voltages• ΔV L and ΔV C are onthe same line and sothe net y componentis ΔV L - ΔV C6/12/2007 31ΔV max From the PhasorDiagram• The voltages are not in phase, so they cannotsimply be added to get the voltage across thecombination of the elements or the voltagesource22ΔV= ΔV+ ( ΔV− ΔV)maxRΔVL− ΔVCtanφ =ΔVR• φ is the phase angle between the current andthe maximum voltage6/12/2007 32LC8

Impedance of a Circuit• The impedance, Z,can also berepresented in aphasor diagramImpedance and Ohm’s s Law• Ohm’s s Law can be applied to theimpedance• ΔV max = I max ZZ =R2+ (XXL− Xtanφ =RL− XCC)26/12/2007 336/12/2007 34Summary of Circuit Elements,Impedance and Phase AnglesProblem Solving for ACCircuits• Calculate as many unknown quantitiesas possible• For example, find X L and X C• Be careful of units -- use F, H, Ω• Apply Ohm’s s Law to the portion of thecircuit that is of interest• Determine all the unknowns asked forin the problem6/12/2007 356/12/2007 369

Power in an AC Circuit• No power losses are associated withcapacitors and pure inductors in an AC circuit• In a capacitor, during one-half of a cycle energy isstored and during the other half the energy isreturned to the circuit• In an inductor, the source does work against theback emf of the inductor and energy is stored inthe inductor, but when the current begins todecrease in the circuit, the energy is returned tothe circuit6/12/2007 37Power in an AC Circuit, cont• The average power delivered by thegenerator is converted to internalenergy in the resistor• P av = I rms ΔV R = I rms ΔV rms cos φ• cos φ is called the power factor of thecircuit• Phase shifts can be used to maximizepower outputs6/12/2007 38Resonance in an AC Circuit• Resonance occurs atthe frequency, ƒ o ,where the current hasits maximum value• To achieve maximumcurrent, the impedancemust have a minimumvalue• This occurs when X L = X Cƒ o1=2πLC6/12/2007 3910