Chapter 31b

# Chapter 31b - Last time we considered a fully charged...

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Last time we considered a fully charged capacitor to which we connected an inductor. We found that the instantaneous voltage across the elements and the current flowing in the circuit both oscillate harmonically in time (forever). o 1 LC ω= To more realistically model the fact that the wires have some resistance we added a small resistor to the circuit. Specific elements of capacitance C and inductance L will have current and voltage oscillations at an angular frequency given by, So this is called the natural angular frequency of the LC oscillator. C L

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Giving the RLC oscillator. C R L Here the instantaneous voltages across the elements and current through them still oscillate harmonically but, because of the energy dissipated in the resistor, the amplitudes of these oscillations decay with time. The angular frequency of oscillation in the RLC circuit, ω ', was given by, 2 2 2 o 2 1R 1 R LC 4L 4 L ⎛⎞ ω= = ω − ⎜⎟ ⎝⎠ Hence, so long as ω o >>R/L the frequency remains approximately the natural frequency of the LC oscillator, o 1 ω≈ω =
We now want to consider the driven RLC oscillator in which the circuit is continuously powered by a sinusoidally oscillating emf: md sin t ξ=ξ ω The angular frequency of the AC power supply, , forces the voltages across the components and the current throughout the circuit to oscillate at this same frequency (rather than the natural frequency of the LC) but as we’ll see the natural frequency does have important significance . d ω Before we consider the combined elements we first consider each of the elements driven individually by the AC supply. Do not assume that the same frequency implies that the voltages across elements have the same phase (as we’ll see, they don’t).

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HITT–10/2/09 t 0.0040 s i v 0.020 s 0.040 s 0.060 s The phasor diagram that could describe the harmonic functions shown here is: ω = 314 rad/s v(t), i(t) φ = 1.26 rad v(t), i(t) φ = 1.88 rad ω = 414 rad/s A B
Driving AC potential : The purely resistive load The circuit is shown to the right. The loop rule holds for any instant in time so, R v0 ξ− = md (t) sin t ξ ω Rm d vs i n t =ξ=ξ ω Ohm’s law still holds for any instant in time. So, with i the current, Since V R is the voltage amplitude across the resistor, which equals ξ m , we can write this as, RR d vV s i n t R vi R = d is i n t =

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Our general expression for current was, R d V is i n t R d iI s i n (t ) φ So for a pure resistive load the current amplitude , and the phase are, R R V I R = & φ = 0 The current through the resistor is in phase with the voltage across it.
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## This note was uploaded on 03/12/2010 for the course PHY PHY taught by Professor Mueller during the Spring '09 term at University of Florida.

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Chapter 31b - Last time we considered a fully charged...

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