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LC Circuit Analysis

LC Circuit Analysis
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Published Date:23-07-2017
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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.)