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

University of Alaska Anchorage Engineering

Prof. Jeffrey A. Miller

An in-depth analysis of the behavior of inductors and capacitors in electrical circuits. It covers the equations for calculating the voltage and current in inductors and capacitors, as well as the concepts of series and parallel combinations of these components. The document also includes sample problems to illustrate the application of these concepts.

Typology: Study notes

2009/2010

Uploaded on 03/28/2010

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ES309

Elements of Electrical Engineering

Lecture #14

Jeffrey Miller, Ph.D.

Outline

• Chapter 6.3

Inductor Review

• Look ing at the equation

v = L di/dt

• If the curr ent 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 can not 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 t0to 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.

Partial preview of the text

Download Inductors and Capacitors: Equivalent Circuits and Energy Storage - Prof. Jeffrey A. Miller and more Study notes Engineering in PDF only on Docsity!

ES

Elements of Electrical Engineering

Lecture

Jeffrey Miller, Ph.D.

Outline

• Chapter 6.

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 t 0 to t

∫ v dt = ∫ L di

∫ v dt = L ∫ di

(1/L) ∫ v dt = ∫ di

(1/L) ∫ v dt = i(t) – i(t 0 )

i(t) = (1/L) ∫ v dt + i(t 0 )

NOTE: If the initial current i(t 0 ) is in the same direction as the

reference direction for i, then i(t 0 ) is a positive quantity. Otherwise,

it’s negative.

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 Li^2

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

v 1 = L 1 di/dt

v 2 = L 2 di/dt

v 3 = L 3 di/dt

v = v 1 + v 2 + v 3 = (L 1 + L 2 + L 3 ) di/dt

Inductors in Series

• From the previous equation, it can be seen

that the equivalent inductance for n series-

connected inductors is

Leq = L 1 + L 2 + … + Ln

Voltage Across Capacitor

We know i = C dv/dt, but how can we express the

voltage as a function of the current?

i = C dv/dt

i dt = C dv

Integrate both sides from t 0 to t

∫ i dt = ∫ C dv

∫ i dt = C ∫ dv

(1/C) ∫ i dt = v(t) – v(t 0 )

v(t) = (1/C) ∫ i dt + v(t 0 )

Power and Energy in the Capacitor

Power

p = vi

i = C dv/dt

p = Cv dv/dt

Energy

p = dw/dt = Cv dv/dt

dw = Cv dv

Integrate both sides

w = ½ Cv^2

Capacitors in Series

• Capacitors in series have the same current, but the

voltage across each capacitor may be different

v 1 (t) = (1/C 1 ) ∫ i dt + v 1 (t 0 )

v 2 (t) = (1/C 2 ) ∫ i dt + v 2 (t 0 )

v 3 (t) = (1/C 3 ) ∫ i dt + v 3 (t 0 )

Capacitors in Series

The voltage v of the three series capacitors is the sum of the capacitor

voltages

v = v 1 + v 2 + v 3

i = (1/C 1 + 1/C 2 + 1/C 3 ) ∫ i dt + v 1 (t 0 ) + v 2 (t 0 ) + v 3 (t 0 )

Then, assuming

1/Ceq = 1/C 1 + 1/C 2 + 1/C 3

v(t 0 ) = v 1 (t 0 ) + v 2 (t 0 ) + v 3 (t 0 )

Then

v = (1/Ceq) ∫ i dt + v(t 0 )

Capacitors in Parallel

The capacitors all carry the same voltage, but the current

may be different across each individual capacitor

i 1 = C 1 dv/dt

i 2 = C 2 dv/dt

i 3 = C 3 dv/dt

i = i 1 + i 2 + i 3 = (C 1 + C 2 + C 3 ) dv/dt

Capacitors in Parallel

• From the previous equation, it can be seen

that the equivalent capacitance for n

parallel-connected capacitors is

Ceq = C 1 + C 2 + … + Cn

Parallel Inductors Sample Problem

• In the following diagram, v = -30e-^5 t^ mV 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/L 1 + 1/L 2

= 1/60mH + 1/240mH

= 4/240mH + 1/240mH

= 5/240mH

Leq = 48mH

Series Capacitors Sample Problem

• To find the total energy as t → ∞, we need the voltage of

each capacitor as t → ∞.

v 2 (t) = (1/C 2 ) ∫ i dt + v 2 (t 0 ) from 0 to ∞ = (1/8μF) ∫ 240x10-6^ e-10t^ dt + (- 5 V) = (1/8μF)( (240x10-6)(- 1/10)(e-10∞) – (240x10-6)(- 1 /10)(e-10(0)) )- 5V =- 3 e-10∞^ + 3e-10(0)– 5V = 3V – 5V =- 2 V

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

Related documents

Partial preview of the text

Download Inductors and Capacitors: Equivalent Circuits and Energy Storage - Prof. Jeffrey A. Miller and more Study notes Engineering in PDF only on Docsity!

ES

Elements of Electrical Engineering

Lecture

Jeffrey Miller, Ph.D.

Outline

• Chapter 6.

Inductor Review

v = L di/dt

constant (DC) current

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 t 0 to t

∫ v dt = ∫ L di

∫ v dt = L ∫ di

(1/L) ∫ v dt = ∫ di

(1/L) ∫ v dt = i(t) – i(t 0 )

i(t) = (1/L) ∫ v dt + i(t 0 )

NOTE: If the initial current i(t 0 ) is in the same direction as the

reference direction for i, then i(t 0 ) is a positive quantity. Otherwise,

it’s negative.

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 Li^2

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

v 1 = L 1 di/dt

v 2 = L 2 di/dt

v 3 = L 3 di/dt

v = v 1 + v 2 + v 3 = (L 1 + L 2 + L 3 ) di/dt

Inductors in Series

• From the previous equation, it can be seen

that the equivalent inductance for n series-

connected inductors is

Leq = L 1 + L 2 + … + Ln

Voltage Across Capacitor

We know i = C dv/dt, but how can we express the

voltage as a function of the current?

i = C dv/dt

i dt = C dv

Integrate both sides from t 0 to t

∫ i dt = ∫ C dv

∫ i dt = C ∫ dv

(1/C) ∫ i dt = v(t) – v(t 0 )

v(t) = (1/C) ∫ i dt + v(t 0 )

Power and Energy in the Capacitor

Power

p = vi

i = C dv/dt

p = Cv dv/dt

Energy

p = dw/dt = Cv dv/dt

dw = Cv dv

Integrate both sides

w = ½ Cv^2

Capacitors in Series

• Capacitors in series have the same current, but the

voltage across each capacitor may be different

v 1 (t) = (1/C 1 ) ∫ i dt + v 1 (t 0 )