L L, R, C. Kirchhoff s rules applied to AC circuits. C Examples: Resonant circuits: series and parallel LRC. Filters: narrowband,

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1 Today in Physics 1: A circuits Solving circuit problems one frequency at a time. omplex impedance of,,. Kirchhoff s rules applied to A circuits. it V in Examples: esonant circuits: i series and parallel. Filters: narrowband, lowpass, highpass V out November 1 Physics 1, Fall 1 1

2 By the book, not. Note that we are not following the techniques introduced in the book, in A circuits. it In the book, A circuit problems are solved by the phasor approach, in which the currents in circuit components are determined by a form of vector addition. In the judgment of your professors, this is a poor choice for introducing an important topic. It is much easier to solve A circuit problems by using the algebra of complex numbers. All but a few have seen complex numbers before, and we can bring the others up to speed quickly. eal engineers and scientists solve A circuit problems with complex numbers, not phasors. November 1 Physics 1, Fall 1

3 One frequency at a time The different time-dependent behavior of,, and makes for circuit problems that are superficially difficult. For example: What is the current, I(t), in the circuit at right, for a voltage V(t) which is a general function of time? Vt We can use the second Kirchhoff rule to generate an equation to solve for I = dq/dt: Q di Vt I dt I t dq dq 1 Q V t dt dt November 1 Physics 1, Fall 1 3

4 One frequency at a time (continued) That looks hard to solve, and indeed it is. an t be integrated directly, like all the other differential equations we ve met. But it turns out, as you will find in MTH 81 or ME 1, that any function V(t) can be represented td as a sum of sines and cosines; for example, 1 V t V costd Fourier s theorem where is an angular frequency. Suppose we deal with one frequency at a time. Vt I t November 1 Physics 1, Fall 1 4

5 One frequency at a time (continued) For one frequency component, dq dq 1 Q V cost dt dt and we solve for I dq dt. Q() and I() will still depend on t.? V cost Virtue: Q, and I, will also vary sinusoidally how could they not? apart from a delay resulting from a compromise between the delays between V and I for and : I Q Q cos t sin I I sin t November 1 Physics 1, Fall 1 5

6 One frequency at a time (continued) Even easier, in the algebra and calculus sense, is to use imaginary numbers: it V V e V cos t iv sin t under the agreement that we always take the real part at the end of the problem, as there are no imaginary voltages and currents in nature. In these terms, 1 i t Fourier s V t V e d theorem (again) and we use i only to simplify the math. i t? eve e I November 1 Physics 1, Fall 1 6

7 eminders about imaginary and complex numbers i i 1 i 1 e i e Im A i i i A A e A cos ia sin A e A Im A arctan Im A e 1 Phasor Im A representations of A e A i 1 i complex numbers e A A Electrical engineers usually refer to i as j, for perverse reasons best known to themselves. November 1 Physics 1, Fall 1 7

8 One frequency at a time (continued) et s prove that Q Q e is the solution to the differential equation above: i t dq dq Q V e dt dt Q e i 1 it iq e i t i t 1 Q e i t i t V e i t? eve Qe 1 i V e I November 1 Physics 1, Fall 1 8

9 One frequency at a time (continued) The term in brackets can be written as u1 i u e u euim u 1 i arctan 1 i V i so Qe e, whence u 1 Q V 1, and 1 i t? eve e I November 1 Physics 1, Fall 1 9

10 One frequency at a time (continued) arctan 1 arctan, 1 q.e.d. Furthermore, i t? eve dq it it I iq e Qe dt e I Q cos t Q sin t e I with Q and as given above. November 1 Physics 1, Fall 1 1

11 eactance and impedance eturn for a moment to the differential equation we solved: dq dq Q V e dt dt 1 i t and insert the solution. In the first and third terms we get i t? eve d i t d i t Q e i Q e i I dt dt 1 it 1 d i t Qe Qe 1 I i dt i e I November 1 Physics 1, Fall 1 11

12 eactance and impedance (continued) Thus we can rewrite the differential equation for Q a loop equation, remember as i I I 1 I i t V e i Apparently, at a given frequency, and have a property rather like resistance which relates their current and voltage. We call it reactance, X: Im X 1 e X i X i e V = IX: thus V lags I Phasor X 1 i e g by / for, and leads by / for. representation of the reactances November 1 Physics 1, Fall 1 1

13 eactance and impedance (continued) And thus the series A circuit looks just like a series of resistors in D circuits, the only change being a complex version of Ohm s law: it 1 V V e Ii IZ i Here Z is the impedance of the series combination. So what have we done? 1. By taking one frequency at a time, we have swapped a complicated calculus problem for a simple algebra problem. That s always a good trade. Im X 1 Z e X In phasor-speak, speak Z is the vector sum of the resistances and reactances. November 1 Physics 1, Fall 1 13

14 eactance and impedance (continued). To find currents and voltages in A circuits, inductors and capacitors can be treated with the Kirchhoff rules precisely as resistors are, using the reactances X and X. For example, the current in the series circuit again: 1 1 i V it Z i i Z e I e i Z 1 where Z Z 1 and arctan Im Z e Z arctan Three lines, instead of three pages. November 1 Physics 1, Fall 1 14

15 Parallel circuit Problem 3-93 in the book, except for a 9 phase shift: Determine the current through each component in this parallel circuit, and the total current leaving the voltage source. V cost November 1 Physics 1, Fall 1 15

16 Parallel circuit (continued) V cost I Individual currents, I I I it I V e X: V i t V i t i t I e I e I i V e i i V V e i t e or, as they would be measured, V V I sin t I cos t I V sin t t November 1 Physics 1, Fall 1 16

17 Parallel circuit (continued) V cost I I I In the second part of the problem we could just add the currents, but it will be useful to compare to the series if we work out the impedance of the parallel combination: i i Z i I Z November 1 Physics 1, Fall 1 17

18 Parallel circuit (continued) V cost I I I 1 arctan Im 1 Z e 1 Z arctan i So the impedance is Z Z e, with Z and as given above, and 1 V V 1 it it I 1 e I e Z I of which the real part gets measured: November 1 Physics 1, Fall 1 18 e I I cos t.

19 Parallel circuit (continued) V cost I I I I 11 4 The circuit draws a lot of current except for angular I frequencies near 1 ; 1 V the resonance in this case is a maximum of impedance. (See 1 below.) November 1 Physics 1, Fall 1 19

20 Parallel circuit (continued) ompare the impedance of the series and parallel circuits at the resonance frequency, 1 : 1 1 Z,series Z,parallel 1 For the series the impedance reaches its minimum at this frequency; for the parallel one, its maximum. November 1 Physics 1, Fall 1

21 Narrowband filter (continued) V in V cos t Vout This suggests the use of circuits as narrowband or notch th filters. For example, series as narrowband: V it V out I e Z Amplitude: V Z,series 1 V 1 V 1 Zseries,series out November 1 Physics 1, Fall 1 1

22 Narrowband filter (continued) g Vout V From this we define the voltage gain, and plot: g ) 1.5 For this plot, component values are chosen such that / = 3., This is a useful building block for signal-processing circuits: only voltages with frequencies near the resonance frequency are transmitted from the input to the output. November 1 Physics 1, Fall 1

23 ow-pass filter Many other useful signal-processing gizmos make good A circuit examples, like this one. Problem 3-13 in the book: Find the voltage gain, V out V in, in this circuit, and discuss the behavior of the gain when f and f. V in V cos t Vout Note: f = / is the frequency of oscillation, in cycles per unit time. November 1 Physics 1, Fall 1 3

24 ow-pass filter (continued) V in V cos t Vout As before, we compute the impedance and current: 1 i Z X X i i t V in V e i V i e i t I Z i i 1 V V IX it 1 i e out November 1 Physics 1, Fall 1 4 1

25 ow-pass filter (continued) V in V cos t Vout i Express in the form Ae : it Ve V out 1 e 1 V 1 e iarctan i tarctan Again, recall that t only the real part is measured. November 1 Physics 1, Fall 1 5

26 ow-pass filter (continued) V in V cos t Vout The voltage gain is the ratio of the input and output voltages: V g e out 1 cos arctan i t Ve 1 If tan ab, then cos b a b. November 1 Physics 1, Fall 1 6

27 ow-pass filter (continued) Note the limits of low and high frequency: 1 g 1 g g 1 g 1 1 lim g The circuit efficiently transmits angular frequencies below 1/, and attenuates those above this value; hence its name November 1 Physics 1, Fall 1 7

28 Single-pole lowpass and highpass filters owpass Highpass V in V out V in V out Vin V out V in Vout November 1 Physics 1, Fall 1 8