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Lecture #13: Inductors and Capacitors - Prof. Jeffrey A. Miller, Study notes of Engineering

A lecture note from elements of electrical engineering (es309) course, focusing on inductors and capacitors. Inductors are electrical components that oppose changes in electrical current and consist of a coil of wire wound around a core. Capacitors, on the other hand, are electrical components that store electrical charge and consist of two conductors separated by an insulator or dielectric material. The properties, equations, and sample problems related to inductors and capacitors.

Typology: Study notes

2009/2010

Uploaded on 03/28/2010

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ES309
Elements of Electrical Engineering
Lecture #13
Jeffrey Miller, Ph.D.
Outline
Chapter 6.1-6.2
Inductors
An electrical component that opposes any change
in electrical current
Composed of a coil of wire wound around a
supporting core whose material may be magnetic
or non-magnetic
A time-varying magnetic field induces a voltage in
any conductor linked by the field
Inductance
Relates the induced voltage to the current
Inductors
Inductance is the circuit parameter used to
describe an inductor
The letter L is used for inductance, measured in
henrys (H)
henry => Wb/A => V * s / A
pf3
pf4
pf5

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ES

Elements of Electrical Engineering

Lecture

Jeffrey Miller, Ph.D.

Outline

• Chapter 6.1-6.

Inductors

  • An electrical component that opposes any change in electrical current
  • Composed of a coil of wire wound around a supporting core whose material may be magnetic or non-magnetic
  • A time-varying magnetic field induces a voltage in any conductor linked by the field
  • Inductance
    • Relates the induced voltage to the current

Inductors

  • Inductance is the circuit parameter used to describe an inductor
  • The letter L is used for inductance, measured in henrys (H) - henry => Wb/A => V * s / A

Inductors

v = L di/dt where v is in volts, L in henrys, i in amps, and t in seconds

If the current is in the opposite direction of the voltage drop across the inductor, the equation will be negated

Inductors

  • 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

Inductor Sample Problem

  • The independent current source in the circuit below generates no current for t < 0 and a pulse 10te-^5 t^ A for t > 0 - At what instant of time is the current maximum? - Express the voltage across the inductor as a function of time.

Inductor Sample Problem

To find when the current is maximum, find the derivative of the current with respect to time i = 10te-5t di/dt = 10 (1 * e-5t^ + t * (-5e-5t) ) = 10 e-5t^ (1 – 5t) The current is maximum when di/dt = 0 10 e-5t^ (1 – 5t) = 0 10 e-5t^ can never be 0, so 1 - 5t = 0 t = 1/5 seconds

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

Capacitors

  • An electrical component that consists of two conductors separated by an insulator or dielectric material, which stores electrical charge
  • A time-varying electric field produces a displacement current in the space occupied by the electric field
  • Capacitance
    • The displacement current is equal to the conduction current at the terminals of the capacitor

Capacitors

  • Capacitance is the circuit parameter used to describe a capacitor
  • The letter C is used for capacitance, measured in farads (F) - farad => C/V => A * s / V

Capacitors

  • Applying a voltage to the terminals of a capacitor cannot move a charge through the dielectric, but it can displace a charge within the dielectric
  • As the voltage varies with time, the displacement of charge also varies with time, causing displacement current
  • The current is proportional to the rate at which the voltage across the capacitor varies with time i = C dv/dt

Capacitors

  • The current i = C dv/dt abides by the passive sign convention, so it would be negated if the current is not in the direction of the voltage drop
  • If the voltage is constant...
    • The current across the capacitor is 0
    • So the capacitor behaves as an open-circuit in the presence of a constant voltage
  • Voltage cannot change instantaneously in a capacitor because that would produce infinite current

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

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

Capacitor Sample Problem

  • The voltage pulse described by the following equations is impressed across the terminals of a 0.5μF capacitor 0 t ≤ 0s v(t) = 4t V 0s ≤ t ≤ 1s 4e-^ (t-^1 )V t ≥ 1s Derive the expressions for the capacitor current, power, and energy. Specify the intervals of time when energy is being stored in and delivered by the capacitor.

Homework

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