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Power Dilemma and Resistor Circuits in Electrical Engineering - Prof. Jeffrey A. Miller, Study notes of Engineering

University of Alaska AnchorageEngineering
Prof. Jeffrey A. Miller

Lecture #3 of elements of electrical engineering by jeffrey miller, ph.d. The lecture discusses the power dilemma, where power is negative for power sources and positive for components that dissipate power. The document also explains kirchhoff's laws for series and parallel resistors, series-parallel simplification, and voltage-divider and current-divider circuits.

Typology: Study notes

2009/2010

Uploaded on 03/28/2010

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ES309
Elements of Electrical Engineering
Lecture #3
Jeffrey Miller, Ph.D.
Outline
• Clear up power dilemma
• Chapter 3.1-3.4
Power Dilemma
• Power will be negative for sources that generate
power and positive for components that dissipate
power
• So if there is a positive current leaving a positive
terminal from a voltage source, the power will be
negative
• If there is a negative current leaving a positive
terminal from a voltage source, the power will be
negative
• A positive current flowing through a current
source will have a negative power
Power Dilemma
• I0is -3A, so the P120V = -iv = 360W
• But the curr ent from the 6A current source is positive,
though flowing toward the positive v1terminal, so
P6A = -iv = -6A * v1= -6A * 150V = -900W
pf3
pf4
pf5
pf8

Partial preview of the text

Download Power Dilemma and Resistor Circuits in Electrical Engineering - 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

  • Clear up power dilemma
  • Chapter 3.1-3.

Power Dilemma

  • Power will be negative for sources that generate power and positive for components that dissipate power
  • So if there is a positive current leaving a positive terminal from a voltage source, the power will be negative
  • If there is a negative current leaving a positive terminal from a voltage source, the power will be negative
  • A positive current flowing through a current source will have a negative power

Power Dilemma

  • I 0 is -3A, so the P120V = -iv = 360W
  • But the current from the 6A current source is positive,though flowing toward the positive v P^1 terminal, so 6A = -iv = -6A * v 1 = -6A * 150V = -900W

Resistors in Series

  • Series-connected circuit elements carry the same current, which can be shown by using Kirchhoff’s current law

Kirchhoff’s Laws on Series Resistors

is = i 0 = -i 1 = i 2 = -i 3 = i 4 So if we know any of the currents, we know them all

Kirchhoff’s Laws on Series Resistors

-vs + isR 0 + isR 1 + isR 2 + isR 3 + isR 4 = 0 vs = is (R 0 + R 1 + R 2 + R 3 + R 4 )

Combining Resistors in Series

Req = R 0 + R 1 + R 2 + R 3 + R 4

Req = ∑ Ri

Then, vs = is Req

i=

k

Series-Parallel Simplification Series-Parallel Simplification

Series-Parallel Simplification

is = 120V / 10Ω = 12A v 1 = 12A * 6Ω = 72V

i 1 = v 1 / 18Ω = 72V / 18Ω = 4A i 2 = v 1 / 9Ω = 72V / 9Ω = 8A

Voltage-Divider Circuits

  • Using resistors in series, a voltage-divider circuit allows more than one voltage level from a single voltage source

Voltage-Divider Circuits

vs = iR 1 + iR 2 i = vs / (R 1 + R 2 ) v 1 = iR 1 = vs (R 1 / (R 1 + R 2 ) ) v 2 = iR 2 = vs (R 2 / (R 1 + R 2 ) )

Current-Divider Circuits

  • Using resistors in parallel, a current-divider circuit allows different currents from a single current source

Current-Divider Circuits

v = i 1 R 1 = i 2 R 2 = is (R 1 R 2 / (R 1 + R 2 ) ) i 1 = is (R 2 / (R 1 + R 2 )) i 2 = is (R 1 / (R 1 + R 2 ))

Current-Divider Example

Find the current across the 6Ω resistor

Voltage Division

  • Assume there are n resistors connected in series from a voltage source
  • Find the voltage drop vj across an arbitrary resistor Rj in terms of the voltage v from the voltage source

Voltage Division Equation

i = v / (R 1 + R 2 + + Rn) = v / Req

vj = i Rj = v Rj / Req

…

Current Division

  • Assume there are n resistors connected in parallel from a current source
  • Find the current ij across an arbitrary resistor Rj in terms of the current i from the current source

Current Division Equation

v = i Req

ij = v / Rj = i Req / Rj

Voltage and Current Division

Find i 0 and v 0

Voltage and Current Division

Req = 1 / (1/80 + 1/10 + 1/80 + 1/24) = 6Ω iv = 24Ω * 2A = 48V 0 = 6Ω/24Ω * 8A = 2A which is the voltage drop across the 3 resistors parallel to the 24 Ω resistor v 0 = 30Ω/80Ω * 48V = 18V

Homework

  • No homework for Chapter 3 yet
  • Assigned on Tuesday