R-C-L Preparation for the Final Exam
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Lecture 23. Impedance, Resonance in R-C-L Circuits Preparation for the Final Exam (a) Start earlier! (b) Review the concepts (lectures + textbook) and prepare your equation sheet. Think how you can use every equation on your sheet, what types of problems can be solved with these equations. (c) Work on practice exams. (d) Review all HW and Iclicker questions. (e) Go over the end-of-chapter problems (you don’t need to solve them, just check that you know how to approach them). At the Exam (a) Make sure you understand the problem, read the problem formulation carefully. Make a drawing!!! If you remain uncertain raise your hand and ask the proctors. (b) Get the units right. It is easy to eliminate the answers with wrong units. This applies to formulas too. 1 Reactance (recap) Resistor ππ = π π = πΌπ = πΌππ Capacitor 1 ππΆ = ππ π π‘ =πΌ π‘ Inductor −π ππΆ ππΏ = ππΏ π π‘ = πΌ π‘ πππΏ AC (cos ππ + π ) driven circuits! 2 Impedance π π‘ πΌ π‘ Impedance π is a measure of how much the circuit impedes the flow of current. The impedance is a complex number (time-independent phasor), it relates timedependent phasors V(t) and I(t). RLC π= π π ππ π0 ≡ π π π‘ = πΌ π‘ π π‘ =πΌ π‘ π ππππ π πππ‘ = πΌπππ π πππ−ππ π π ππ ππππ = πΌπππ π −ππ π π ππ Reactances: Impedances: ππππ = πΌπππ π πΌπΌ πΌ −π π π π π π π is the reference phasor all terms are real ππ = π ππ = π 1 ππΏ = ππΏ ππ 1 ππΏ = πππΏ ππΆ = πππ ππΆ = 3 C circuit π π‘ πΌ π‘ ππππ = πΌπππ ππΆ πΌπππ Can we plug a 1-µF capacitor into a wall outlet πππ (π = 2π β 60 , ππππ = 120π) if the circuit breakers π can take 15A ? πΌπππ = ππ ππΆ ππππ 120π = = = 45ππ ππ 2650Ω 1 1 = ππΆ = = Ω = 2650Ω ππ 2π β 60 β 1 β 10−6 This current is sufficiently small. The primary concern is the voltage rating of the capacitor, which should be around 200V. Current (reference phasor) Voltage −π π π‘ = πΌ π‘ ππΆ = πΌ π‘ ππΆ 4 L circuit π π‘ πΌ π‘ What happens when we plug a 1-H inductor into a wall outlet? ππππ = πΌπππ ππΏ = ππ πΌπππ πΌπππ ππππ 120π = = = 0.32π΄ ππΏ 377Ω ππΏ = ππΏ = ππ = 2π β 60 β 1Ω = 377Ω Again, the current won’t blow a circuit breakers. The inductor must be designed to carry 0.32A without overheating or saturating the iron core. Current (reference phasor) Voltage π π‘ = πΌ π‘ ππΏ = πΌ π‘ πππΏ 5 Series R-C circuit π = ππ + ππΆ = π − πππ πΌπΌ − 1 ππ π = π π π π π π 2 + ππ 2 π = π π ππ πΌπΌπΌ −ππΆ π = ππππππ = ππππππ π π π π ππππ = πΌπππ π 2 + ππ 2 = πΌπππ −π π 1 2 π + ππΆ 2 π π‘ = πΌ π‘ π = πΌ(π‘) π − πππ π0 π πππ = πΌ0 π π ππ−π π π ππ 6 R-C circuits: Example πΌπΌ − π 1 ππ πΌπΌ −π π π π πΌ π π π πΌπΌπΌ −ππΆ 1 π = ππππππ = ππππππ π =π −π π π π π ππ 1 1 ππ tan π = − =− = −50 3 −6 π 1 β 10 β 100 β 0.2 β 10 π ππ Note that π is negative (as it should be for the RC circuits). πππ π π π(π‘) = πΌ(π‘)π π0 π πππ = πΌ0 π π ππ−π π π ππ 7 Low-Pass Filter Goal: to suppress high-frequency (π > π0 ) components in the spectrum of a signal. πππ = πΌ π − πππ π = π − πππ π = πππ = πΌ π 2 + ππ 2 π 2 + ππ 2 ππππ = πππ Output power: ππππ πππ 2 = 1 πππ π 2 +1 ππ π 2 + ππ 2 1 1 πππ π Cutoff frequency: π0 = 2ππ0 = = π 2 πβͺ 2 1 π π πβ« We want to suppress the high-frequency (π > 10πππ) components in the output of an audio amplifier with the output resistance 100 β¦. What capacitance do you need? π° ππππ = πΌ −πππ 1 ππΆ 1 + ππ 1 ππ π 1 ππ π πΆ= 2 1 ππ πΆ πππ π½πͺ = π½πππ ππππ = πΌππ = π½πΉ 2 +1 = 1 πππ π 2 π +1 = two times 1 1 = πΉ = 160ππ 2ππ0 π 2π104 β 100 8 Series R-L-C Circuits For R, C, and L in series: π = π + πππΏ − πππ = π + π ππ − π = π β π∗ = πΌ 1 π 2 + ππ − ππ ππππ = πΌπππ π = πΌπππ ππΌπΌπΏ −ππΌπΌπΆ 2 1 2 π + ππ − ππ π(π‘) = πΌ(π‘) π + π ππ − 1 ππ ? ππ 1 ππ 2 1 ππ 9 Series R-L-C circuits: Example 1 ππΏ = πππΏπΏ ππππ πΏ = 2.2 β 80 = 176π ππ ππππ π = 2.2 β 40 = 88π ππ ππππ πΆ = 2.2 β 110 = 242π ππΆ = −π πΌ ππ ππππ = πΌπππ π = πΌπππ 1 π 2 + ππ − ππ ππππ = 2.2 402 + 80 − 110 2 = 110π 2 10 Series R-L-C circuits: Example 2 An R-L-C series circuit with an inductance of 0.119H , a resistance of 244 β¦, and a capacitance of 7.27 µF carries an rms current of 0.446A with a frequency of 391Hz . 1. What is the impedance of the circuit? 1 π = π + πππΏ − πππ = π + π ππ − ππ 2. What is the phase angle? tan π = π0 = π = 2455 πππ/π 1 π 2 + ππ − ππ 1 −6 ππΆ = 2455 β 0.119 − 2455 β 7.27 β 10 π 244 ππΏ − 3. What is the rms voltage of the source? −1 ≈ 0.97 2 = 339Ω πΌπΌ π −π arctan 0.97 ≈ 0.77 πππ π πΌ π π π ππππ = πΌπππ β π0 = 0.446π΄ β 339Ω = 151π 4. What average power is delivered by the source? cos 0.77 = 0.72 - power factor for this circuit πππ = ππππ β πΌπππ cos π = 151 β 0.446 β 0.72 = 48.6π - average rate at which electrical energy is converted to thermal energy in the resistor 11 Parallel R-L-C Circuit: Example πππ = ππππ β πΌπππ cos π 1 1 1 = + π ππ − π π ππ 1 = π πΌπΌ π −π π πΌ π π π tan π = πΌπππ 1 ππ = 3 1/π 4 cos π = 1 1 + π‘π‘π‘2 π 1 ππππ 1 = = ππππ 2 + ππ − π π ππ πππ = ππππ β πΌπππ cos π = 2 1 1 − π ππ − ππ π= = π 2 1 1 1 1 + π ππ − + ππ − ππ π ππ π 2 1 ππ − 1 1 + ππ − π 2 ππ 2 = = 4 5 1 + 6−3 2 0.252 10 10 50 4 β β = 200π 2 2 5 2 = 50 2 π΄ 12 Series Resonance in the R-L-C circuits πΌ For R, C, and L in series: 1 π 2 + ππ − ππ π = π β π∗ = πΌ = Resonance condition: ππ = 1 ππ π = π π0 = 1 1 π = π + πππΏ − πππ = π + π ππ − ππ π 2 π 1 + ππ − ππ 2 2 - resonance πΏπΏ frequency At π =π0 minimum (real) impedance, max current. Note that at π =π0 , ππΆ and ππΏ can be greater than π. π > π0 π < π0 π = π0 13 Parallel Resonance in the R-L-C circuits πΌ 1 1 1 ππΆ 1 1 1 1 1 = + + = −π +π = + π ππ − π π πππ −π π ππΏ ππΆ π ππ 1 = π At the resonance frequency π0 = 1 πΏπΏ 1 π 1 πΌ = = π 2 + ππ − π ππ π min at ππ = 2 1 1 + ππΆ − π 2 ππΏ 1 π 2 is at its minimum π → 0 πΏ is a “short” π → ∞ πΆ is a “short” 1 ππ Note that at π =π0 , πΌπΆ and πΌπΏ can be greater than πΌ. R = 1β¦, C = 1F, L = 1H, and V = 1V 16 Transformer Φπ΅ - the flux per turn πΦπ΅ ππ πΦπ΅ βπ = −ππ ππ βπ = −ππ For an ideal transformer (π π = π π = 0): Energy conservation: πΌπ ππ = πΌπ ππ ππ ππ = ππ ππ 2 πΌπ 2 1 ππ πΌπ π≡ ππ ππ = πΌπ ππ 2 π Φπ ππ Φπ΅ = πΌπ πΌπ π ππΌπ πΦπ΅ = ππ ππ ππ πΏπ πΏπ : βπ = π ππ ππ = ππ ππ ππ πΌπ = ππ πΌπ 2 ππ π ππ “impedance transformation” - as if the source had been connected directly to a resistance Using mutual inductance π = βπ ππ = βπ ππ ππΌπ ππ 17 Example Sloppy formulation βπ π= ππΌπ =π = πΌπ π π ππ πΌπ π π 0.4 β cos 377π‘ 12 = = 2.55ππ ππΌπ 377 β 5 β cos 377π‘ ππ βπ = π ππΌπ = π 6π‘ = 10 β 10−3 β 6 β 3 ππ = 0.18π 18 Next time: Lecture 24. Electromagnetic Waves, §§ 32.1 - 4 19
