Advanced rcuts Topcs - Part by Dr. olton (Fall 07) Part : Some thngs you should already know from Physcs 0 and 45 These are all thngs that you should have learned n Physcs 0 and/or 45. Ths secton s organzed lke ths so you can see the parallels between resstor problems and complex mpedance problems: Outlne:. Voltage and urrent. esstors and Ohm s aw 3. Seres and Parallel esstors 4. Solvng rcut Problems Wth Equvalent esstances 5. Interlude: A Quck Summary of omplex Numbers 6. omplex Voltage and urrent 7. omplex Impedances and Ohm s aw 8. Seres and Parallel Impedances 9. Solvng rcut Problems Wth Equvalent Impedances. Voltage and urrent The basc water analogy works well for me: f electrcty s lke water flowng n ppes, then the current s the rate at whch the water s flowng n the ppe (kg/s) and the voltage s the heght of the ppe (potental energy of the water). Batteres whch rase the potental of the charges are lke pumps pumpng the water up to a hgher elevaton. The potental dfference of a crcut element s the dfference n heght from one sde of the element to the other.. esstors and Ohm s aw esstors and conductors are the ppes themselves. A conductor s lke a very large dameter ppe that allows for a lot of current flow. A resstor s a lke a small dameter ppe whch lmts the flow: the larger the resstor, the smaller the dameter and the smaller the flow. The resstor ppe s angled downwards, so that the end at whch the water comes out s lower than the end at whch the water goes n. The potental dfference of a resstor s gven by Ohm s law: Δ 3. Seres and Parallel esstors A combnaton of resstors can be labeled ether as seres or parallel, dependng on how they are connected: ppes can ether be one after the other (seres) or sde-by-sde (parallel). Seres resstors share the same current; parallel resstors share the same potental dfference. SEIES PAAE / / Shortcut notaton: // / /
4. Solvng rcut Problems Wth Equvalent esstances Networks of resstors can often be decomposed nto combnatons of resstors that are n seres and parallel wth each other, called equvalent resstances of the resstor combnatons. Example Problem : What s the total resstance of the 4 resstors? 3 4 Soluton: // // Suppose the four resstors were hooked up to a battery wth voltage V B. The total current flowng from the battery would be gven by Ohm s law for the crcut as a whole: / /. The current and voltage of any of the resstors n ths problem can be obtaned by applcaton of seres/parallel resstor formulas along wth Ohm s law. The same s true for nearly any crcut problem nvolvng a sngle voltage source. For more complcated stuatons, Krchhoff s two crcut laws can be used to obtan smultaneous equatons, but that s beyond the scope of ths document. 5. Interlude: A Quck Summary of omplex Numbers We wll be usng complex numbers n Physcs 44 and 44 as a tool for descrbng oscllatng electromagnetc phenomena: A crcuts n Physcs 44, and electromagnetc waves n Physcs 44. A complex number can be wrtten ether n rectangular form, or polar form,. The polar form can equvalently be expressed as a complex exponental. The complex exponental form follows drectly from Euler s formula: cossn, and by lookng at the x- and y-components of the polar coordnates, cos, sn. o s called the ampltude, magntude, or modulus of the complex number. In Mathematca t s obtaned va the command Abs. o s called the phase angle, phase factor, phase, angle, argument, or ampltude of the complex number. In Mathematca t s obtaned va the command Arg. For example, consder the complex number 3 + 4: = (3, 4) n rectangular form, = (5, 53.3º) n polar form, and 53.3 0.973 = 5 e or 5e n complex exponental form, snce 53.3º = 0.973 rad. I often wrte the polar form usng ths shortcut notaton: A. The symbol can be read as, at an angle of. By the rules of exponents, when you multply two complex numbers n polar form, the ampltudes multply and the angles add. Smlarly, when you dvde n polar form, the ampltudes dvde and the angles subtract. Thus you can wrte: (3 + 4) (5 + ) = 553.3 367.38 = 650.5 (3 + 4) / (5 + ) = 553.3 / 367.38 = 0.3846 4.5 Yes, unfortunately the word ampltude can at tmes refer to ether A or ϕ. I myself wll reserve ampltude for A, though.
6. omplex Voltage and urrent Suppose you have an oscllaton such as an A voltage or current that you want to represent va complex numbers. The equaton for a generc oscllaton would be ths, wth arbtrary ampltude V 0 and phase ϕ: cos One represents the oscllatory functon va complex numbers lke ths: cos e e e Startng e e wth ths step, there s an mpled take the real part of everythng on the rght hand sde There s often an mpled e as well, so ths can become: ϕ Thus a voltage oscllatng n tme can be represented as a sngle complex number. Ths trck wll make the math much easer for many calculatons. Therefore, f for example you see 3.V30, what t means s that a voltage s oscllatng n tme as 3.V cos. Presumably the oscllaton frequency ω would be gven elsewhere n the problem, or maybe you d be lookng for thngs as a functon of ω. Smlarly, f you see.5a0 t means that a current s oscllatng as.5a cos. In summary: cos cos 7. omplex Impedances and Ohm s aw If you have a crcut wth a snusodal drvng functon (e.g. voltage) and f you are only concerned about the steady-state (long term) response, then the concept of complex mpedances smplfes matters consderably. Actually, any perodc drvng functons can be handled ths way snce wth Fourer analyss you can decompose any perodc sgnal nto a sum of snusodal ones. If you are supplyng a snusodal voltage at frequency ω, after a long tme all currents and potental dfferences of ndvdual crcut elements wll also be snusodal and oscllatng at the same frequency ω. As dscussed n the last secton, we wll use complex numbers to represent these voltages and currents, and we wll restrct ourselves to crcuts made of resstors, capactors, and nductors. As you wll see, complex numbers wll allow us to easly handle phase shfts between crcut elements. To solve such systems we want to defne an Ohm s aw -lke quantty for capactors and nductors; then we can use the seres and parallel resstance formulas whch were derved va Ohm s aw to descrbe complcated crcuts. That quantty s called the mpedance, and wll be a complex number Z: Δ s lke Δ for a resstor. 3
We wll set the zero of tme such that the drvng battery voltage s a cosne functon wth no phase shft,.e. t reaches ts maxmum at t = 0. et s call the ampltude V 0 : cos 0 The total current suppled by the battery wll also be snusodal but t may have a dfferent phase than the voltage. et s call the current s ampltude, and say that t lags the voltage by a phase (f the current s actually leadng the voltage, then wll be negatve): cos Keep n mnd that the voltages and currents are real quanttes, not complex; the pont of the complex exponentals s just to smplfy the mathematcs of snusodal sgnals wth phase shfts. At the end of a problem, f you want to know the actual voltages or currents just convert back from the complex numbers usng the ϕ cos types of correspondences. Impedance of esstor. esstors are easy: Z =. Ohm s aw apples drectly wth no werd modfcatons. Impedance of apactor.δ, and f, then when you ntegrate I you get a factor of / from the exponental, tmes the current I tself. Thus the capactor equaton becomes: Δ, and the resstance-lke quantty s (also often wrtten as ). Impedance of Inductor. As s later n hapter 7 of Grffths, and you learned n Phys 0, Δ. When you take the dervatve of I, you get a factor of from the exponental, tmes the current I tself. Thus the nductor equaton becomes: Δ, and the resstance-lke quantty s. In summary: If you use these complex mpedances (and have snusodal responses), then Ohm s law works for capactors and nductors and for the crcut as a whole just lke t does for resstors: Δ 8. Seres and Parallel Impedances Because the seres and parallel resstor formulas were derved va Ohm s law, they carry over completely wth complex mpedances. Seres: Parallel: / / // These equatons are very powerful because they can be used wth combnatons of any types of mpedances. 4
9. Solvng rcut Problems Wth Equvalent Impedances Usng these mpedances to determne the ampltude and phase of the current from the power supply s straghtforward: Step. Fnd the total mpedance of the crcut,, usng seres/parallel rules. It s a complex number. Step. Wrte n polar form: ϕ Step 3. Use Ohm s law to get the total current, also a complex number: Step 4. onvert back to a cosne oscllaton to get your real answer f desred: cos, usng the same phase conventon as wrtten n secton 7 above (I lags V when s postve). Note that f the phase angle of the complex mpedance s negatve, then the phase angle of the complex current wll be postve, and vce versa. Once the total current s known n complex form, other quanttes of nterest can generally also be found usng Phys 0 technques nvolvng Ohm s law and seres/parallel mpedances. Example Problem What current s suppled here? Soluton: Step : Zeq Z Z 3 3 4 0.5 3 Step : Now wrte 3 4 n polar form: 5 0.97 Step 3: 0.0.97 4 length=5, angle = tan (4/3) = 0.97 rad Step 4: If desred, take the real part to get the actual formula for the current: 0.A cos 0.97 Example Problem 3 For ths crcut, what s (a) V and (b) V? 5
6 Step : Z eq Z eq Step : Put n polar form, ϕ: Z eq tan (I assumed that ω > /ω so the vector shown above s ndeed n the frst quadrant. auton: f not n the frst or fourth quadrants the tan functon wll gve wrong answers. As mentoned above Mathematca s Arg command s helpful, and much better than the regular tan functon.) Steps 3 & 4: I = V/Z, and convert to a real oscllaton: t V I tan cos 0 (a) Everythng s n seres, so ths same current goes through the resstor. Use V = I by Ohm s aw, and the answer s: t V V tan cos 0 (b) Use the complex Ohm s law on the nductor: Δ Δ tan Δ tan That last step was done by wrtng tself n polar form. From there t s just complex number multplcaton and convertng to real numbers f you want: Z /
Δ Δ tan cos tan Example Problem 4 What s the current comng from the power supply, n terms of V 0, ω, t,,,, and? Frst fnd the total mpedance of the crcut: Z Z Z // Z Z eq I ll use Mathematca to help wth the algebra: It needs to be put nto polar form n order to fnd ts magntude and phase. The polar coordnates can easly be found n Mathematca wth these commands: and Abs Arg Then the current s gven by cos for that Z and ϕ. It s probably not worth the space to wrte out the full-blown answer. 7