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Review Part 3 of Course Passive Circuit Elements + + i v R - v iR 1 i v R v i + i + L - di vL dt 1 i vdt L C v - - 1 v idt C dv iC dt Energy stored in the capacitor The instantaneous power delivered to the capacitor is dv p(t ) vi Cv dt The energy stored in the capacitor is thus t dv w p(t )dt C v dt C vdv dt t 1 2 w Cv (t ) joules 2 Energy stored in the capacitor Assuming the capacitor was uncharged at t = -, and knowing that q Cv 2 1 2 q (t ) w Cv (t ) 2 2C represents the energy stored in the electric field established between the two plates of the capacitor. This energy can be retrieved. And, in fact, the word capacitor is derived from this element’s ability (or capacity) to store energy. Parallel Capacitors + i v i1 C1 i2 iN + i CN C2 - v i Ceq - dv i1 C1 dt dv i2 C2 dt dv iN C N dt dv dv i i1 i2 iN C1 C2 C N Ceq dt dt N Ceq Ck k 1 Thus, the equivalent capacitance of N capacitors in parallel is the sum of the individual capacitances. Capacitors in parallel act like resistors in series. Series Capacitors C1 C2 CN + + DC v 1- + v2 - + vN - DC v Ceq - i 1 v1 idt C1 v i 1 v2 idt C2 vN 1 idt CN 1 1 1 1 v v1 v2 vN idt idt CN Ceq C1 C2 N 1 1 Ceq k 1 Ck The equivalent capacitance of N series connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitors. Capacitors in series act like resistors in parallel. Energy stored in an inductor The instantaneous power delivered to an inductor is di p (t ) vi Li dt The energy stored in the magnetic field is thus t di wL (t ) p(t )dt L i dt L idi dt t 1 2 wL (t ) Li (t ) joules 2 Series Inductors L1 L2 LN + + DC v 1- + v2 - + vN - DC v v i Leq - i di v1 L1 dt di v2 L2 dt di vN LN dt di di v v1 v2 vN L1 L2 LN Leq dt dt N Leq Lk k 1 The equivalent inductance of series connected inductors is the sum of the individual inductances. Thus, inductances in series combine in the same way as resistors in series. Parallel Inductors + i v i2 i1 L1 iN L2 LN i v i Leq - - 1 i1 vdt L1 + 1 i2 vdt L2 1 iN vdt LN 1 1 1 1 i i1 i2 iN vdt vdt LN Leq L1 L2 N 1 1 Leq k 1 Lk The equivalent inductance of parallel connected inductors is the reciprocal of the sum of the reciprocals of the individual inductances. Thus inductances in parallel combine like resistors in parallel. Complex Numbers A x jy A A cos j jA sin j j tan 1 A y x j e jj cos j j sin j e j j j measured positive counter-clockwise A Ae jj Note: real x A cos j Euler's equation: A y A sin j A x2 y 2 Complex Plane imag e e cos j j sin j jj j Comparing Sinusoids -sin t cos t cos t Im -cos t Re cos t 2 Note: positive angles are counter-clockwise Im 45 135 45 cos t sin t 2 sin t cos t 45 sin t sin t sin t cos t sin t Re cos t cos t leads sin t by 90 cos t lags - sin t by 90 sin t 45 cos t 135 leads cost by 45 and leads sint by 135 AC Circuits and Phasors A phasor is a complex number that represents the amplitude and phase of a sinusoid. X M e jj X M j X XM j j t Recall that when the assumed form of the current i(t ) I M e was substituted into the differential equations, the e j t cancelled out. We were then left with just the phasors Impedance Impedance Z i(t) VM cos t + AC - V phasor voltage I phasor current R + VR I - + VL - L Units = ohms V R j L V Z R j L Z z I Z R jX R resistance Z R2 2 L2 X reactance 1 L z tan R Note that impedance is a complex number containing a real, or resistive component, and an imaginary, or reactive, component. Admittance 1 I phasor current Admittance Y = = Z V phasor voltage i(t) VM cos t + Units = siemens R + VR AC - conductance G V I R j L - + VL - R R 2 2 L2 L I 1 R j L Y= G jB 2 V R j L R 2 L2 susceptance B L R 2 2 L2 Im V V I I Re I in phase with V V Im I V Re I lags V V I Im I V I I leads V Re I Z3Z 4 Zin Z1 Z 2 Z3 Z 4 DC Z1 V Z2 I1 Z3 I2 Z4 Zin We see that if we replace Z by R the impedances add like resistances. Impedances in series add like resistors in series Impedances in parallel add like resistors in parallel Voltage Division Z1 + V1 DC V I Z2 + V2 - V I Z1 Z 2 But V1 Z1I V2 Z2I Therefore Z1 V1 V Z1 Z 2 Z2 V2 V Z1 Z 2 Instantaneous Power v(t ) VM cos t v i(t ) I M cos t i p(t ) v(t )i(t ) VM I M cos t v cos t i VM I M cos v i cos 2t v i p(t ) 2 Note twice the frequency Average Power T 2 1 P T t0 T t0 1 P T 1 p(t )dt T t0 T t0 t0 T t0 VM I M cos t v cos t i dt VM I M cos v i cos 2t v i dt 2 P 1 VM I M cos v i 2 Purely resistive circuit Purely reactive circuit v i 0 v i 90 P 1 VM I M 2 P 1 VM I M cos 90 0 2 Effective or RMS Values We define the effective or rms value of a periodic current (voltage) source to be the dc current (voltage) that delivers the same average power to a resistor. 1 PI R T 2 eff I eff t0 T t0 i 2 (t )Rdt 1 t0 T 2 i (t )dt T t0 I eff I rms Veff2 1 P R T t0 T t0 v 2 (t ) dt R 1 t0 T 2 Veff v (t )dt T t0 root-mean-square Veff Vrms Effective or RMS Values Vrms Using Vrms Vrms 1 t0 T 2 v (t )dt T t0 v(t ) VM cos t v cos 1 1 cos 2 2 2 2 VM 2 VM 2 2 0 2 0 T 2 and 1 1 cos 2t 2v dt 2 2 1 dt 2 1 2 VM 2 2 t 20 1 2 1 2 VM 2 Ideal Transformer - Voltage AC i1 i2 + + v1 N1 N2 - d v1 (t ) N1 dt v2 Load - This changing flux through coil 2 induces a voltage, v2 across coil 2 d v1 N1 dt N1 v2 N 2 d N 2 dt The input AC voltage, v1, produces a flux 1 v1 (t )dt N1 d v2 (t ) N 2 dt N2 v2 v1 N1 Ideal Transformer - Current AC i1 i2 + + v1 N1 N2 - Magnetomotive force, mmf v2 Load F Ni - The total mmf applied to core is F N1i1 N 2i2 R For ideal transformer, the reluctance R is zero. N1i1 N 2i2 N1 i2 i1 N2 Ideal Transformer - Impedance AC i1 i2 + + v1 N1 v2 N2 - - Load V2 ZL I2 N1 V1 V2 N2 Input impedance V1 Zi I1 Load impedance 2 N1 Zi ZL N2 ZL Zi 2 n Turns ratio N2 I1 I2 N1 N2 n N1 Ideal Transformer - Power AC i1 i2 + + v1 N1 N2 - Load P vi - Power delivered to primary Power delivered to load P2 v2i2 P1 v1i1 N2 v2 v1 N1 v2 N1 i2 i1 N2 P2 v2i2 v1i1 P1 Power delivered to an ideal transformer by the source is transferred to the load. Force on current in a magnetic field Force on moving charge q -- Lorentz force F q (v B ) Current density, j, is the amount of charge passing per unit area per unit time. N = number of charges, q, per unit volume moving with mean velocity, v. L S j F N V q(v B) F ( j B)V F (i B)L vt V S L dQ NqSvt i j S dt t j Nqv Force per unit length on a wire is iB Rotating Machine B Force out commutator i + i brushes X Force in Back emf B B d dt Force out +- i a b r l i X X Force in area A lw l 2r cos w 2r cos flux BA Β2rl cos emf E ds flux B2rl cos t t emf eab B 2rl sin 2Brl sin t B Back emf Force out commutator i + - eab 2Brl sin i brushes X Force in eab Armature with four coil loops eab S X X X X N Motor Circuit Ia Ra Vt Ea Vt Ea I a Ra I a Vt Ea / Ra I a Vt Ka / Ra Ea K a Power and Torque Pd Ea I a d d Ka I a Vt K a d Ra Ka Ra 2 AC Nodal Analysis For steady-state AC circuits we can use same the method of writing nodal equations by inspection that we learned for resistive circuits. To do so, we must replace resistances with impedances. We solved Problem 4.31 in Class: V1 j2 V2 40 A 20 A -j1 Change impedances to admittances V1 j2 V2 -j/2 S S 20 A 40 A -j1 j1 S Nodal Analysis for Circuits Containing Voltage Sources That Can’t be Transformed to Current Sources 1. 2. 3. Assume temporarily that the current across each voltage source is known and write the nodal equations in the same way we did for circuits with only independent voltage sources. Express the voltage across each independent voltage source in terms of the node voltages and replace known node voltages in the equations. Rewrite the equations with all unknown node voltages and source currents on the l.h.s. of the equality and all known currents and voltages on the r.h.s of the equality. We solved #4.33 in text in Class: V1 -j4 Note: V2 = 10 V2 j/4 S 645 A 1/2 S I0 + j2 -j/2 S 100 V AC I2 assume I2 0.5 j 0.25 0 V1 4.243 j 6.743 I j 0.25 1 j 2.5 2 AC Mesh Analysis For steady-state AC circuits we can use same the method of writing mesh equations by inspection that we learned for resistive circuits. To do so, we must replace conductances with admittances. We solved Problem 4.38 in text in Class: Find I1 and I2: j1 + I1 AC 120 V - -j1 + I2 60 A AC - 2 j1 1 j1 I1 12 1 j1 1 j 0 I 2 6 What happens if we have independent current sources that can’t be transformed in the circuit? 1. Assume temporarily that the voltage across each current source is known and write the mesh equations in the same way we did for circuits with only independent voltage sources. 2. Express the current of each independent current source in terms of the mesh currents and replace known mesh currents in the equations. 3. Rewrite the equations with all unknown mesh currents and voltages on the left hand side of the equality and all known voltages and currents on the r.h.s of the equality. We solved Problem 4.40 in text in Class: Find I0: Assume you know V2 -j1 + j2 + AC 60 V - I2 I1 I0 3 j1 2 j 2 20 A V2 - Note I2 = -2 0 I1 2 j 4 1 V2 4 j 4 Matlab Solution: Know how to program I1 1.414 81.87 V 2 7.37612.53 AC Thevenin's Theorem AC Thevenin's Theorem Thevenin’s theorem states that the two circuits given below are equivalent as seen from the load ZL that is the same in both cases. I a Zth Linear Circuit ZL b AC I a ZL VTh b Zin Zin VTh = Thevenin’s voltage = Vab with ZL disconnected (= ) = the open-circuit voltage = VOC AC Thevenin's Theorem I a Zth Linear Circuit ZL b AC I a ZL VTh b Zin Zin ZTh = Thevenin’s impedance = the input impedance with all independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). This is the impedance seen at the terminals ab when all independent sources are turned off. We solved Problem 4.40 using Thevenin's Thm. in Class: -j1 + j2 + AC 60 V -j1 20 A - + VOC - AC 60 V - 20 A I0 + 8 j2 VOC 6 2(1 j ) 8 j 2 -j1 j2 AC -j1 I0 I0 8 j2 8 j 2 8.246 14.04 1 j1 j 2 2 3 j1 3.16218.43 I 0 2.608 32.47 ZTH 1 j1 AC Superposition Superposition Principle The superposition principle states that the voltage across (or the current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. Steps in Applying the Superposition Principle 1. Turn off all independent sources except one. Find the output (voltage or current) due to the active source. 2. Repeat step 1 for each of the other independent sources. 3. Find the total output by adding algebraically all of the results found in steps 1 & 2 above. Example Done in Class: 30sin 5t + + v0(t) - - AC 0.2F 2 cos10t 1H Note that the voltage source and the current source have two different frequencies. Thus, if we want to use phasors, the only way we've solved sinusoidal steady-state problems, we MUST use superposition to solve this problem. We considered each source acting alone, and then found v0(t) by superposition. Remember that sin t cos t 90 Example 30sin 5t + + v0(t) - - AC 0.2F 2 cos10t 1H By superposition: v0(t) = Response due to voltage source + Response due to voltage source v0 (t ) v01 (t ) v02 (t ) v0 (t ) 4.631sin 5t 81.12 1.05cos 10t 86.24