INDUCTOS AND AC CICUITS We begin by introducing the new circuit element provided by Faraday s aw the inductor. Consider a current loop carrying a current I. The current will produce a magnetic flux in the loop proportional to I. Hence by Faraday s aw a change in I will produce an EMF around the loop. We define the self-inductance of the loop by: d EMF M dt EMF di dt where = inductance. There are two equivalent definitions which are often useful. Since EMF dm di dt dt we have I M (We are ignoring signs just as we did for capacitance). Finally consider the energy involved in setting up the field: di P I I dt Hence: t t t di d U P dt I dt I dt I dt dt
U M I where U M is the magnetic field energy. Now consider two current loops: oop () will produce a flux in loop () and vice-versa. Hence a change in the current in () will produce an EMF around (). We define mutual inductance as: di EMF dt Just as with capacitance, inductance is purely geometric depends only on the shape of the circuit or circuits. Consider the solenoid shown: INDUCTANCE OF SOENOID As we have seen there will be a field in the solenoid given by: B N Izˆ Hence each loop has a flux: Since there are N loops, the total flux is: N loop I
N I I N AC CICUITS We now turn to the use of the elements resistors, capacitors, and inductors in circuits with an AC generator instead of a battery to provide the EMF. We begin by considering each in turn. ESISTO As usual the voltages around a closed loop must add to zero. Thus: cos ti cost I Now consider the power provided by the generator. Hence the average power is: PI cos t P A cos t dt where τ is the period = π/ω. Thus
cos t PA dt cos zdz Thus the average power is that of a battery of voltage: MS This root mean square voltage is what is specified for AC systems. The term comes from the fact that we are taking the average value of the square and then taking the square root. CAPACITO q cost C I sin t C I Csin t We would like to be able to treat this like we did resistors Ohm s aw. This would require: I eff xc But this won t work since ~cos(t) while I~sin(t). In other words, they are out of phase. We can solve this problem by using complex numbers. COMPEX NUMBES Complex numbers can be thought of as points in a plane as opposed to real numbers which are points on a line.
We denote by i. Points in the plane are then of the form: a + ib in Cartesian coordinates. For our purpose polar coordinates are more useful. We have: We now remember that: Thus where a r cos b rsin 3 4 3 5 ix ix ix x x x x e ix ixi i cos xisin x! 3!! 4! 3! 5! Now consider the product: Hence the angles add. i a ib re / b tan, r a b a i i i re r e rr e
CONENTIONS We adopt the convention that physical quantities are to be the real part of complex numbers. Hence: i t cos t e Then in the capacitor circuit we considered above: et it e I Csin t X C XC i C Then it e it i Ce i Ccos t isin t i C i C cos t C sintc sint Hence if we take: XC i C we can use Ohm s aw in the form: = IX C Why is X C what it is? The voltage arises from a charge build up on the capacitor. The bigger the less time for charge to build up before the voltage reverses and charge is removed. The bigger C the more charge is required to produce a given voltage. Hence: i X C ~ C The i comes from the fact that voltage lags current by 9 o (I first, then ). But:
e i i Hence XC i C INDUCTO di cost dt di costi I sint dt In our case there is no constant I. Thus I = and: Try Then Hence we can take: I sint X i it it e e i sint I costisin t X i X = i
Why this form? The larger the bigger (di/dt) and the larger voltage produced. The bigger the larger voltage for a given (di/dt). The i comes from the fact that leads I (first, then I). SEIES CICUIT Now let s see how this works in a series circuit. We have q di costi C dt di I d I sin t dt C dt We solve this in the usual way: I = I H + I P where I H is the general solution of the homogeneous equation and I P is one solution of the complete equation. et IH mt Ae Then mt mt mt Am e me Ae C m m C
4 C m / Hence I M will have a factor t e and thus will go to zero at long times. Then Next I P. As usual we try a solution of the form: IP AcostBsin t A costb sin t Asin tbcost A cos t B sin t sin t C A A B C B B A C (Set coefficients of sin and cos each to zero so that the solution will work at all times.) Then: B A C C A A C
A C C B C C C C / A B P / / I A cos t B sin t A B cos t sin t A B A B But cosθ cosφ + sinθ sinφ = cos(θ φ). Hence We have A B / P / I A B cos t C C / / C C /
B tan C A et i z X XC X i i C C / i z e C where tan C Then I cos t C exactly as above. Hence our complex solution gives the values at long times once the transients have died out. We now have a general procedure for any AC circuit: solve it exactly as you would a DC circuit, but use complex X instead of, and take physical quantities to be the real part of the complex numbers. / Then POWE We will always end up with: it it e e i I e cos t z i ze z z t
P o cos t cos t cos tcos t cos sin t sin z z This is exactly like the DC result: P MS A cos cos z z P A ff except for the cos factor. This factor is called the power factor and means that it is possible to have both a current and a voltage and yet no average power. This has important consequences as we will now see. Consider the circuit shown: PAAE CICUITS We solve it exactly as we would if it were DC. We first note that and C are in parallel. They have an effective impedance (term for effective resistance in AC circuits) given by: Then and z are in series with i ic z z XC X i i C C i z z z e C i
/ z C C tan Then A P cos z / cos C cos z A P z z Hence the average power will be a max or min when Z is. This occurs when:
C C / At that point Z becomes infinite and no power is provided by the source. At this resonance frequency the energy in the parallel section is just passed back and forth between the inductor (magnetic field) and the capacitor (electric field). Now recall our result for the series circuit: / z C tan C cos / z C PA z z Again the resonance frequency is: / (C) but this times it leads to maximum power from the source because it makes Z a minimum instead of a maximum. A curious fact about AC circuits is that the sum of the MS voltages around a circuit need not be zero. The sum of the instantaneous voltages must by zero, but the sum of the averages (MS voltages) need not be. You should check this numerically for a couple of circuits. This method can be used to solve any AC problem even those where Kirchoff s aws must be used. The only restriction is that we get the long term steady state solution, not the transients. If the transients are desired we have to use the entire solution and boundary conditions. This is not usually of interest.