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: Time domain: v(t) = Ri(t) i(t) = v(t) R R i a v b Time Domain R I a V b Phasor Domain Resistance is a static element in the sense v(t) versus i(t) relationship is instantaneous. 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: Time domain: q(t) = Cv(t) i(t) = dq = Cdv dv = 1 C i(t) t t o dv = v(t) v(t o) = 1 C v(t) = v(t o ) 1 C t t o i(τ)dτ. t t o i(τ)dτ. a i C v b Time Domain a I C V b Phasor Domain 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: Time domain: v(t) = L di di = 1 L v(t) t di t o = i(t) i(t o) = 1 L i(t) = i(t o ) 1 L t t t o v(τ)dτ. t o v(τ)dτ. i L a v b Time Domain L I a V b Phasor Domain 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 element. 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 C t t i(τ)dτ v(t ) i - v relationship i = v R i = 1 L t t v(τ)dτ i(t ) i = C dv p power consumed p = vi = i 2 R p = vi = Li di p = vi = Cv dv w energy stored w = w = 1 2 Li2 w = 1 2 Cv2 Series connection R eq = R 1 R 2 L eq = L 1 L 2 C eq = C 1C 2 C 1 C 2 Parallel connection R eq = R 1R 2 R 1 R 2 L eq = L 1L 2 L 1 L 2 C eq = C 1 C 2 DC Steady State Behavior v = Ri, No change v = Short Circuit i = 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.
Integrating Circuits Passive Integrating Circuit: v in R C v o Principles of Electrical Engineering I Integrating and Differentiating Circuits i = C dvo Differenciating Circuits Passive Differenciating Circuit: v in vc C R i = C dvc v o Figure 1 Consider the RC circuit of Figure 1. Its equation is v in = RC dv o v o. Suppose RC >> 1. This renders v o small in comparison with v in. Then, we have This implies that v in = RC dv o. v o (t) = v o (t ) 1 t v in (τ)dτ. RC t Active Integrating Circuit: v in i s = vin R G v in R v N v P N P i N i P Figure 2 vo C i f = C dvo For an ideal Op-Amp, we know that v P v N = and i P = i N =. Since v P =, this imples that v N =. Then, the circuit equation is This implies that C dv o v in R =. v o (t) = 1 RC v in (τ)dτ. v o Figure 3 Consider the RC circuit of Figure 3. Its equations are C dv C = v o R, v in = v C v o. By differentiating the second equation, and then substituting the first, we get dv in = v o RC dv o. 1 dvo Suppose RC >> 1. This renders small and negligible in comparison with vo RC. Then, we have v o = RC dv in. Active Differenciating Circuit: i s = C dvin v in G v in C v N v P N P v o i N i P Figure 4 R i f = vo R For an ideal Op-Amp, we know that v P v N = and i P = i N =. Since v P =, this imples that v N =. Then, the circuit equation is This implies that C dv in v o R =. v o (t) = RC dv in. v o
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 I m sin(ωt) where I m 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 to t. The voltage v(t) across it at time t is given by v(t) = L di = ωli m cos(ωt). The power p(t) consumed by it at time t is given by p(t) = v(t)i(t) = ωli 2 m sin(ωt) cos(ωt) = 1 2 ωli2 m sin(2ωt). The energy e(t) consumed by it during the time interval from to t is given by e(t) = t p(t) = 1 2 ωli2 m t sin(2ωt) = 1 4 LI2 m [1 cos(2ωt)] = 1 2 LI2 m sin 2 (ωt) = 1 2 Li2 (t). a i L v = L di b Time Domain Current, Voltage, Power, and Energy associated with an Inductance 2 1 Power Current Energy 1 Voltage 2 2 4 6 8 1 12 Time t Thw above figure shows the graphs of current i(t) = I m sin(ωt) (solid line graph), voltage v(t) = ωli m cos(ωt) (dashed line graph), the power p(t) = 1 2 ωli2 m sin(2ωt) (dotted line graph), and the energy e(t) = 1 2 Li2 (t) (dash dotted line graph). For numerical computations, we used I m = 2 A, ω =.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 V m sin(ωt) where V m 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 to t. The current i(t) through it at time t is given by i(t) = C dv = ωcv m cos(ωt). The power p(t) consumed by it at time t is given by p(t) = v(t)i(t) = ωcv 2 m sin(ωt) cos(ωt) = 1 2 ωcv 2 m sin(2ωt). The energy e(t) consumed by it during the time interval from to t is given by e(t) = t p(t) = 1 2 ωcv 2 m t sin(2ωt) = 1 4 CV 2 m [1 cos(2ωt)] = 1 2 CV 2 m sin 2 (ωt) = 1 2 Cv2 (t). v a i = C dv C b Time Domain Voltage, Current, Power, and Energy associated with a Capacitance 2 1 Power Voltage Energy 1 Current 2 2 4 6 8 1 12 Time t Thw above figure shows the graphs of voltage v(t) = V m sin(ωt) (solid line graph), current i(t) = ωcv m cos(ωt) (dashed line graph), the power p(t) = 1ωCV 2 2 m sin(2ωt) (dotted line graph), and the energy e(t) = 1 2 Cv2 (t) (dash dotted line graph). For numerical computations, we used V m = 2 V, ω =.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 v G i v Voltage, Current, Power, Energy The voltage across an element is given by v(t) = 1 sin(2π 1t) while current through it is i(t) = 2 cos(2π 1t). Sketch the voltage v(t), current i(t), power consumed p(t), and energy consumed w(t) with respect to time. All sketches should start at time t = and should show at least one cycle of the expected wave form. Indicate the scales on both the axes of your graphs properly. (a) The voltage v in Volts: t (b) The current i in Amperes: t (c) The power consumed p in Watts: t (d) The energy consumed W in Joules: t
v g G i C Symmetrical triangular waveform v g (t) A 2 1 1 C A 2 t in milli Secs A voltage signal v g (t) having a symmetrical triangular wave form (with a period of 2 milli secs) is shown above. It is across a capacitor as shown on the left. Let A = 1V and C = 1µF (micro Farad). Draw to scale (indicate the scale on the vertical axis) the following items for the time period shown: 2 (a) The current i C in Amperes: 1 2 t in milli Secs (b) The power p consumed by the capacitor in watts: 1 2 t in milli Secs (c) The energy stored in the capacitor at t = : (d) The energy stored in the capacitor at t =.5 milli Secs: (e) The energy stored in the capacitor at t = 1 milli Secs:
332:221 Principles of Electrical Engineering I Determine the equivalent capacitance between the terminals a and b of Figure A. c c Determine the equivalent inductance between the terminals a and b of Figure B. c c 24µf 3µf 18µf 36mH 18mH 6mH 36µf a b 9µf 72mH a b 15mH c Figure A c c Figure B c