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R, C, and L Elements and Their V and I Relationships We Deal with Three Essential Elements in Circuit Analysis

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R, C, and L Elements and Their V and I Relationships We Deal with Three Essential Elements in Circuit Analysis R, C, and L Elements and their v and i relationships We deal with three essential elements in circuit analysis: • Resistance R • Capacitance C • Inductance L Their v and i relationships are summarized below. Resistance: a a Time domain: v(t)= Ri(t) i + I + v(t) R v R V i(t)= R − − b b Time Phasor Domain Domain Resistance is a static element in the sense v(t) versus i(t) relationship is instan- taneous. For example, v(t) at time t = 2 seconds simply depends only on i(t) at t = 2 seconds and nothing else. This implies that the resistance does not know what happened in the past, in other words it does not store any energy unlike other elements C and L as we see soon. Capacitance: a a Time domain: q(t)= Cv(t) i I + + dq dv C v C V i(t)= = C − − dt dt dv 1 = i(t) b b dt C t dv 1 t Time Phasor = v(t) − v(to)= i(τ)dτ. Domain Domain Zto dt C Zto 1 t v(t)= v(to)+ i(τ)dτ. C Zto Unlike in the case of resistance, for a capacitance the v(t) versus i(t) relationship and vice versa at any time t depends on the past as they involve differentials and integrals. This implies that the capacitance is a dynamic element. What happened in the past influences the present behavior. As we shall see soon, capacitance stores energy. 2 Inductance: a a di i I Time domain: v(t)= L + + dt L v L V di 1 = v(t) − − dt L t di 1 t b b = i(t) − i(to)= v(τ)dτ. Zto dt L Zto Time Phasor 1 t Domain Domain i(t)= i(to)+ v(τ)dτ. L Zto Once again, unlike in the case of resistance, for an inductance the v(t) versus i(t) relationship and vice versa at any time t depends on the past as they involve differentials and integrals. This implies that the inductance is a dynamic ele- ment. What happened in the past influences the present behavior. As we shall see soon, inductance stores energy. 332:221 Principles of Electrical Engineering I Basic properties of R, L, and C Both voltage v and current i are considered as functions of time t Property R L C v - i relationship v = Ri v = L di v = 1 R t i(τ)dτ + v(t ) dt C t0 0 i - v relationship i = v i = 1 R t v(τ)dτ + i(t ) i = C dv R L t0 0 dt 2 di dv p power consumed p = vi = i R p = vi = Li dt p = vi = Cv dt 1 2 1 2 w energy stored w = 0 w = 2 Li w = 2 Cv C1C2 Series connection Req = R1 + R2 Leq = L1 + L2 Ceq = C1+C2 R1R2 L1L2 Parallel connection Req = Leq = Ceq = C1 + C2 R1+R2 L1+L2 DC Steady State Behavior v = Ri, No change v = 0 Short Circuit i = 0 Open Circuit Can v change instantaneously Yes Yes No without i being an impulse? Can i change instantaneously Yes No Yes without v being an impulse? AC Steady State behavior To be discussed To be discussed To be discussed Resistance is a static element and has no memory. On the other hand, both capacitance and inductance are dynamic elements and have memory. Principles of Electrical Engineering I Integrating and Differentiating Circuits Integrating Circuits Differenciating Circuits Passive Integrating Circuit: Passive Differenciating Circuit: v + C− R dvo dvC i = C dt C i = C dt + + v + C v v + R v in − − o in − o − Figure 1 Figure 3 Consider the RC circuit of Figure 1. Consider the RC circuit of Figure 3. Its equation is Its equations are dvo dvC vo v = RC + v . C = , v = vC + v . in dt o dt R in o Suppose RC >> 1. This renders vo small in com- By differentiating the second equation, and then parison with vin. Then, we have substituting the first, we get dvo dv v dv v = RC . in = o + o . in dt dt RC dt This implies that 1 dvo Suppose RC >> 1. This renders dt small and negligible in comparison with vo . Then, we have 1 t RC v (t)= v (t0)+ v (τ)dτ. o o RC Z in dv t0 v = RC in . o dt Active Integrating Circuit: Active Differenciating Circuit: v dvo − o if = C + dt i vo − vo + f = R vN vin C v is = vin − i N R R + N N dvin v − is = C in− i dt + N N − R iP P + + i + C P P + vin − v + P vo v + in − vP v − o G − G Figure 2 Figure 4 For an ideal Op-Amp, we know that vP − vN =0 v − v and iP = −iN = 0. Since vP = 0, this imples that For an ideal Op-Amp, we know that P N =0 i −i v vN = 0. Then, the circuit equation is and P = N = 0. Since P = 0, this imples that vN = 0. Then, the circuit equation is dvo vin C + =0. dv v dt R C in + o =0. dt R This implies that This implies that − 1 vo(t)= vin(τ)dτ. dv RC Z v (t)= −RC in . o dt Principles of Electrical Engineering I Current, Voltage, Power, and Energy associated with an Inductance Example: Consider an inductance L in Henries. Let the current i(t) through it be Im sin(ωt) where Im is the amplitude and ω is the angular frequency in radians per second. Write down explicit expressions for the voltage v(t) across it at time t, the power p(t) consumed by it at time t, and the energy e(t) consumed by it during the time interval from 0 to t. The voltage v(t) across it at time t is given by a di i + v(t)= L = ωLIm cos(ωt). di dt L v = L dt − The power p(t) consumed by it at time t is given by b 2 1 2 p(t)= v(t)i(t)= ωLIm sin(ωt) cos(ωt)= ωLIm sin(2ωt). Time 2 Domain The energy e(t) consumed by it during the time interval from 0 to t is given by t t 1 2 e(t) = p(t)dt = ωLIm sin(2ωt)dt Z0 2 Z0 1 2 1 2 2 1 2 = LIm [1 − cos(2ωt)] = LIm sin (ωt)= Li (t). 4 2 2 Current, Voltage, Power, and Energy associated with an Inductance 2 Energy Current 1 Power 0 −1 Voltage −2 0 2 4 6 8 10 12 Time t Thw above figure shows the graphs of current i(t)= Im sin(ωt) (solid line graph), voltage v(t)= ωLIm cos(ωt) (dashed line graph), 1 2 the power p(t)= 2 ωLIm sin(2ωt) (dotted line graph), and 1 2 the energy e(t)= 2 Li (t) (dash dotted line graph). For numerical computations, we used Im = 2 A, ω = 0.25π radians per second, and L = 1 Henry. Principles of Electrical Engineering I Voltage, Current, Power, and Energy associated with a Capacitance Example: Consider a Capacitance C in Farads. Let the voltage v(t) across it be Vm sin(ωt) where Vm is the amplitude and ω is the angular frequency in radians per second. Write down explicit expressions for the current i(t) through it at time t, the power p(t) consumed by it at time t, and the energy e(t) consumed by it during the time interval from 0 to t. The current i(t) through it at time t is given by a dv dv i = C dt i(t)= C = ωCVm cos(ωt). + dt v C − The power p(t) consumed by it at time t is given by b 2 1 2 p(t)= v(t)i(t)= ωCVm sin(ωt) cos(ωt)= ωCVm sin(2ωt). Time 2 Domain The energy e(t) consumed by it during the time interval from 0 to t is given by t t 1 2 e(t) = p(t)dt = ωCVm sin(2ωt)dt Z0 2 Z0 1 2 1 2 2 1 2 = CVm [1 − cos(2ωt)] = CVm sin (ωt)= Cv (t). 4 2 2 Voltage, Current, Power, and Energy associated with a Capacitance 2 Energy Voltage 1 Power 0 −1 Current −2 0 2 4 6 8 10 12 Time t Thw above figure shows the graphs of voltage v(t)= Vm sin(ωt) (solid line graph), current i(t)= ωCVm cos(ωt) (dashed line graph), 1 2 the power p(t)= 2 ωCVm sin(2ωt) (dotted line graph), and 1 2 the energy e(t)= 2 Cv (t) (dash dotted line graph). For numerical computations, we used Vm = 2 V, ω = 0.25π radians per second, and C = 1 Farad. 332:221 Principles of Electrical Engineering I Suggested HW Problems in the text book by Ulaby and Maharbiz Partial answers are in Appendix E These problems are not collected and graded. Problem 5.2 Exercise to get familiarity with Step function Problem 5.7 Time Constant of an exponential Problem 5.15 RC Circuit in DC steady state Problem 5.19 Capacitors Series-Parallel Problem 5.25 Voltage signal across an inductance Problem 5.29 RLC Circuit in DC steady state 332:221 Principles of Electrical Engineering I i Voltage, Current, Power, Energy The voltage across an element is given by v(t) = 100sin(2π 100t) while + current through it is i(t) = 2cos(2π 100t).
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