# LC Circuit Analysis

###### LC Circuit Analysis
AC Circuits III Physics 2415 Lecture 24 Michael Fowler, UVa Today’s Topics • LC circuits: analogy with mass on spring • LCR circuits: damped oscillations • LCR circuits with ac source: driven pendulum, resonance. LC Circuit Analysis • The current . I dQ / dt • . • With no resistance, the voltage across the capacitor is exactly balanced by Q Q the emf from the inductance: C I Q dI S  L C dt • From the two equations above, L 2 d Q Q S in the diagram is  2 the closed switch dt LCQuick review of simple harmonic motion from Physics 1425… Force of a Stretched Spring • If a spring is pulled to • A extend beyond its Natural length natural length by a distance x, it will pull back with a force F kx Spring’s force where k is called the F kx “spring constant”. Extension x The same linear force is also generated when the spring is compressed. Quick review of simple harmonic motion from Physics 1425… Mass on a Spring • Suppose we attach a • A Natural length mass m to the spring, free to slide backwards m and forwards on the frictionless frictionless surface, then Spring’s force pull it out to x and let go. F kx • F = ma is: m 22 md x / dt kx Extension x Quick review of simple harmonic motion from Physics 1425… Solving the Equation of Motion • For a mass oscillating on the end of a spring, 22 md x / dt kx • The most general solution is x Acos t  • Here A is the amplitude,  is the phase, and by 2 putting this x in the equation, mω = k, or  km / • Just as for circular motion, the time for a complete cycle T1/ f 2 / 2 m / k ( f in Hz.)Back to the LC Circuit… • The variation of charge with time is 2 Q Q • . d Q Q  2 dt LC C I S • We’ve just seen that 22 md x / dt kx L has solution x Acost , k / m  from which Q Q cos t, 1/ LC. 0Where’s the Energy in the LC Circuit • The variation of charge with time is Q Q cos t, 1/ LC 0 Q Q • . so the energy stored in the capacitor is C 2 2 2 I S U Q / 2C Q / 2C cost  E 0 • The current is the charge flowing out L IdQ / dt Q sin t 0 so the energy stored in the inductor is 2 2 2 2 2 2 2 11 U LI LQ sint Q / 2C sint1/ LC  B2200 Compare this with the energy stored in the capacitor Energy in the LC Circuit • We’ve found the energy in the capacitor is 2 2 2 Q Q U Q / 2C Q / 2C cost  E 0 • . • The energy stored in the inductor is C I S 2 2 2 1 U LI Q / 2C sint  B 0 2 • So the total energy is L 2 2 2 2 U Q / 2C cos tsin t Q / 2C.  B00 • Total energy is of course constant: it is cyclically sloshed back and forth between the electric field and the magnetic field. Energy in the LC Circuit • . • Energy in the capacitor: electric field energy • Energy in the inductor: magnetic field energy The LRC Circuit • Adding a resistance R to the LC circuit, adds a voltage drop IR, so • . Q Q Q dI  L IR C dt C I R • Remembering I  dQ / d , t we find 2 L d Q dQ Q LR 0. 2 dt dt C • A differential equation we’ve seen before… Damped Harmonic Motion • In the real world, oscillators • C Drag force Spring’s force experience damping forces: F kx F bv friction, air resistance, etc. m • These forces always oppose the motion: as an example, Extension x we consider a force F = −bv proportional to velocity. The direction of drag force shown is on the assumption that • Then F = ma becomes: the mass is moving to the left. ma = −kx −bv 22 md x / dt bdx / dt kx 0 • That is, LRC is just a Damped Oscillator • Compare our charge equation with the displacement equation for a damped harmonic oscillator: 2 d Q dQ Q LR 0. 2 dt dt C 22 md x / dt bdx / dt kx 0 • They are the same: Q x, L m, R b, 1/ C k.Equation Solution From Physics 1425: • Therefore • The equation of motion 2 d Q dQ Q 22 LR 0 2 md x / dt bdx / dt kx 0 dt dt C has solution has solution t t  Q Q e cos t  x Ae cos t 0 where where RL / 2 , bm / 2 , 22 22   1/ LC R / 4L     k / m b / 4m   Q x, L m, R b, 1/ C k. Spreadsheet AC Source and Resistor • . • For an AC source (denoted by a wavy line in a circle) VV sint 0 the current is: I I sin t V / R sin t.  00 • The current and voltage peak at the same time. • Power: the ac source is working at a rate 2 1 P IV I V sint I V 0 0 0 0 2AC Source and Inductor • For a purely inductive circuit, for V  V s in t , the current is 0 given by V sint LdI / dt 0 so I  I co s t where 0 IV /L 00 ωL is the inductive reactance. Power: P IV I V sin t cos t 0 00AC Source and Inductor… IV /L 00 ωL is inductive reactance. • Notice that this increases with frequency: faster oscillations mean more back emf. • Note also that the peak in current occurs after the peak in voltage in the cycle. AC Source and Capacitor • For pure capacitance, V sin t Q / C Q sin t / C  00 so I I cost dQ / dt Q cost 00 and from this we see that ICV 00 and the capacitive reactance is: 1 X C  CComparing Pure L and Pure C • . • For L, peak emf is before peak current, for C peak current is first. • Mnemonic: ELI the ICE man. • No power is dissipated in inductors nor in capacitors, since emf and current are 90 out of phase: 1 sint cost sin 2t 0 2L and C in Series • The same current is passing through both: the red curve is the emf drop over L and C respectively—notice they’re in opposite directions • (We show here a special case ω = L = C = 1 where no external emf is needed to keep current going— this is resonance.) Clicker Question • This shows ac emf and • . current for ω = C = 1. • What happens to the current if ω is increased to 2, but emf kept constant A. Current doubles B. Current halved C. Current same maximum value, but phase changes. Clicker Answer • This shows ac emf and • . current for ω = C = 1. • What happens to the current if ω is increased to 2, but emf kept constant A. Current doubles • Notice the axis is rescaled • Capacitances pass higher frequency ac more easily— opposite to inductances Circuit with L, R, C in Series • For a current of amplitude I passing through 0 all three elements, the emf drop across R is I R, in phase with the current. 0 • Remember the emf drops across L, C have opposite sign—the total emf drop is I (ωL1/ωC), but this emf is 90 out of phase. 0 • The current will therefore be ahead of the total emf by a phase angle  given by:  LC1/ tan RMaximum emf and Total Impedance Z • For a given ac current, we find the emf driving it through an LCR circuit has two components which are 90 out of phase. • To find the maximum total emf V , these two 0 amplitudes must be added like vectors. • The amplitudes are: I R, I (ωL1/ωC). 0 0 • So 2 1  2 V I RL I Z 0 0  0 C Geometry of Z and  2 1  2 V I RL I Z 0 0  0 C  • . The emf across the resistor is in phase with the current. The total emf is represented by Z, Z 1  L and if ωL 1/ωC, the  C  emf is ahead of the current by phase .  R 22 P I R I Z cos Power dissipation only in R: rms rms LCR Impedance Z as a Function of ω 2 1  2 V I RL I Z 0 0  0 C  • Notice that if ωL = 1/ωC, V = I R, the 0 0 minimum possible impedance. The capacitor and inductor generate emf’s that exactly cancel. This is resonance. • At very high frequencies, Z approaches ωL. • At very low frequencies, Z approaches 1/ωC. Spreadsheet link Clicker Question • Is it possible in principle to construct an LCR series circuit, with nonzero resistance, such that the current and applied ac voltage are exactly 90 out of phase A. Yes B. No Clicker Answer • Is it possible in principle to construct an LCR series circuit, with nonzero resistance, such that the current and applied ac voltage are exactly 90 out of phase A. Yes B. No Because there is always energy dissipated, hence power used, in a resistor, and 90 out PVIV I sin t cos t 0 of phase means . 00 Clicker Question • This is for my information: all answers will score 2. i ei  cos sin • Do you know the equation A. Yes, I’ve covered it in a math (or other) course, and think I can probably work with it. B. I’ve seen it before, but haven’t really used it. C. I have no idea what this equation is about. Matching Impedances • A power supply (red box), say • . an amplifier, has internal resistance R , and neglibible V 0 1 inductance and capacitance. It R 1 generates an emf V . 0 • What speaker resistance R 2 takes maximum power from the amplifier R V 2 2 0 I R,. I • Power = 2 RR 12Matching Impedances V • . 2 0 • Power P I R,. I 2 RR 12 V 0 2 R  V 1 0 • So power PR .  2 RR  12 • Notice this is small for R small, 2 and small for R large. 2 R 2 dP / dR 0. • The maximum power is at 2 • You can check this is at R = R 2 1.Matching Impedances in Transmission • Typical coax cable is labeled 75, this • . means that the ratio V /I for an ac rms rms signal, the impedance Z = 75. • For the ribbon conductor shown, the corresponding impedance is 300. • Transmission from one to the other is done via a transformer such that the 22 powers are matched I Z I Z . 1 1 2 2 • Therefore the ratio of the number of Balun transformer turns in the transformer coils is: N / N Z / Z . 1 2 1 2
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