Inductors and Capacitors: Equivalent Circuits and Energy Storage - Prof. Jeffrey A. Miller, Study notes of Engineering

Download Inductors and Capacitors: Equivalent Circuits and Energy Storage - Prof. Jeffrey A. Miller and more Study notes Engineering in PDF only on Docsity! 1 ES309 Elements of Electrical Engineering Lecture #14 Jeffrey Miller, Ph.D. Outline • Chapter 6.3 Inductor Review • Looking at the equation v = L di/dt • If the current is constant . . . – The voltage across the inductor is 0 – So the inductor behaves as a short-circuit in the presence of a constant (DC) current • Current cannot change instantaneously in an inductor because that would require an infinite voltage Current Across Inductor We know v = L di/dt, but how can we express the current as a function of the voltage? v = L di/dt v dt = L di Integrate both sides from t0 to t ∫ v dt = ∫ L di ∫ v dt = L ∫ di (1/L) ∫ v dt = ∫ di (1/L) ∫ v dt = i(t) – i(t0) i(t) = (1/L) ∫ v dt + i(t0) NOTE: If the initial current i(t0) is in the same direction as the reference direction for i, then i(t0) is a positive quantity. Otherwise, it’s negative. 2 Power and Energy in the Inductor Power p = vi v = L di/dt p = Li di/dt Energy p = dw/dt = Li di/dt dw = Li di Integrate both sides w = ½ Li2 Series-Parallel Inductors and Capacitors • Just as series-parallel combinations of resistors can be reduced to a single equivalent resistor, series-parallel combinations of inductors or capacitors can be reduced to a single inductor or capacitor Inductors in Series • The inductors all carry the same current, but the voltage drops across each individual inductor v1 = L1 di/dt v2 = L2 di/dt v3 = L3 di/dt v = v1 + v2 + v3 = (L1 + L2 + L3) di/dt Inductors in Series • From the previous equation, it can be seen that the equivalent inductance for n series- connected inductors is Leq = L1 + L2 + … + Ln 5 Capacitors in Parallel • The capacitors all carry the same voltage, but the current may be different across each individual capacitor i1 = C1 dv/dt i2 = C2 dv/dt i3 = C3 dv/dt i = i1 + i2 + i3 = (C1 + C2 + C3) dv/dt Capacitors in Parallel • From the previous equation, it can be seen that the equivalent capacitance for n parallel-connected capacitors is Ceq = C1 + C2 + … + Cn Parallel Inductors Sample Problem • In the following diagram, v = -30e- 5tmV for t ≥ 0. – What is the equivalent inductance? – What is the initial current and its reference direction in the equivalent inductor? – What is i(t) in the equivalent inductor? Parallel Inductors Sample Problem • To find the equivalent inductance for inductors connected in parallel, we use the following formula 1/Leq = 1/L1 + 1/L2 = 1/60mH + 1/240mH = 4/240mH + 1/240mH = 5/240mH Leq = 48mH 6 Parallel Inductors Sample Problem • To find the initial current through the equivalent inductor, we use the following formula i(t0) = i1(t0) + i2(t0) = 3A + (-5A) = -2A Parallel Inductors Sample Problem • To find the current through the equivalent inductor, we use the following formula i = (1/Leq) ∫ v dt + i(t0) from 0 to t i = (1/48mH) ∫ -30e-5t dt + (-2A) i = (1/48mH)(-30)(-1/5)(e-5t) – 1/48mH(-30)(-1/5) – 2A i = (1/48mH)(6)(e-5t) – (1/48mH)(6) – 2A i = 0.125e-5t – 2.125A for t ≥ 0 Series Capacitors Sample Problem • The current in the following diagram is i = 240e- 10t µA for t ≥ 0 Calculate the total energy trapped in the capacitors as t → ∞. The initial voltages are -10V and -5V, as shown. Series Capacitors Sample Problem • To find the total energy as t → ∞, we need the voltage of each capacitor as t → ∞. v1(t) = (1/C1) ∫ i dt + v1(t0) from 0 to ∞ = (1/2µF) ∫ 240x10-6 e-10t dt + (- 10V) = (1/2µF)( (240x10-6)(- 1/10)(e-10(∞)) – 240x10-6)(- 1/10)(e-10(0)) ) - 10V = - 12e-10(∞) + 12e0 – 10V = 12V – 10V = 2V As t → ∞, e- 10t → 0, so we have 2V for our first capacitor 7 Series Capacitors Sample Problem • To find the total energy as t → ∞, we need the voltage of each capacitor as t → ∞. v2(t) = (1/C2) ∫ i dt + v2(t0) from 0 to ∞ = (1/8µF) ∫ 240x10-6 e-10t dt + (- 5V) = (1/8µF)( (240x10-6)(- 1/10)(e-10∞) – (240x10-6)(- 1/10)(e-10(0)) ) - 5V = - 3e-10∞ + 3e-10(0)– 5V = 3V – 5V = - 2V As t → ∞, e- 10t → 0, so we have -2V for our second capacitor Series Capacitors Sample Problem • To find the total energy, we use the following formula w1 = ½ Cv12 = ½ 2µF (2V)2 = 4µJ w2 = ½ Cv12 = ½ 8µF (-2V)2 = 16µJ w = w1 + w2 = 4µJ + 16µJ = 20µJ Homework No chapter 6 homework yet!