Chapter 2 Basic Laws

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Chapter 2 (Basic Laws)
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Contents

• Kirchhoff's Law
• Series, parallel and series-parallel circuits
• Voltage and current division
• Introduction to Wye-Delta transformation

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Nodes, Branches and Loops

• Branch represents a single element such as a
  voltage source or a resistor.

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• Node is the point of connection between two or
  more branches.
• The above circuit have 4 node.

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• Loop is any closed path in a circuit.
• The above circuit have 3 loops.

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• Relationship between branches, nodes and loops.
• b l n 1
• b branches
• n - nodes
• l independent loops

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Kirchhoffs Law

• Kirchhoff current law

The algebraic sum of all the currents at any node
in a circuit equals zero
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• Kirchhoff current law (KCL) states that the sum
  of currents entering a node is zero. Or,
• The sum of current entering a node the sum of
  current leaving the node.
• Based on KCL we can write the current eq. for all
  node in a circuit.
• Ex.

Ia
Ib
Ic
Id
Ie
Try to write down the current equation base on
KCL
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• Remember!!!

Sum current in sum current out.(KCL).
Therefore Ia Ic Ib Id Ie or
Ia Ic Ib Id Ie 0
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Kirchhoffs Law

• Kirchhoff voltage law

The algebraic sum of all the voltages around any
closed path, in a circuit equals zero
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Applying KVL around the loop
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Find the currents and voltages in the circuit

• Applying Ohms Law and
• Kirchhoffs Law
• V18i1, V23i2, V36i3 ----(1)
• At node a KCL gives,
• i1-i2-i3 0 ----(2)
• Applying KVL gives,
• -30 V1 V2 0
• -30 8i1 3i2 0
• i1 (30-3i2)/8 ----(3)
• -V2 V3 0 ---(4)
• V2 V3

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• We express V1 and V2 in terms of i1 and i2 as equ
  (1), therefore equ (4) becomes
• 6i3 3i2, i3 i2/2 ----(5)
• Substitute equ 3 and 5 into 2 gives
• (30-3i2)/8 i2 i2/2 0
• i2 2 A
• So i13A, i31A, V124V, V26V, V36V

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Resistors in Series

• A series circuit provides only one path for
  current between two points so that the current is
  the same through each series resistor.

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Current in a Series Circuit

• The current is the same through all points in a
  series circuit. The current through each
  resistor in a series circuit is the same as the
  current through all the other resistors that are
  in series with it.

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Total Series Resistance

• The total resistance of a series circuit is
  equal to the sum of the resistances of each
  individual series resistor.

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Series Resistance Formula

• For any number of individual resistors connected
  in series, the total resistance is the sum of
  each of the individual values.
• RT R1 R2 R3 . . . Rn

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Voltage Sources in Series

• When two or more voltage sources are in series,
  the total voltage is equal to the the algebraic
  sum (including polarities of the sources) of the
  individual source voltages.

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Power in a Series Circuit

• The total amount of power in a series resistive
  circuit is equal to the sum of the powers in each
  resistor in series.
• PT P1 P2 P3 . . . Pn

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Resistors in Parallel

• Each current path is called a branch.
• A parallel circuit is one that has more than one
  branch.

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Voltage in Parallel Circuits

• The voltage across any given branch of a
  parallel circuit is equal to the voltage across
  each of the other branches in parallel.

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Total Parallel Resistance

• When resistors are connected in parallel, the
  total resistance of the circuit decreases.

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Formula for Total Parallel Resistance

• 1/RT 1/R1 1/R2 1/R3 . . . 1/Rn

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Notation for Parallel Resistors

• To indicate 5 resistors, all in parallel, we
  would write
• R1R2R3R4R5

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Two Resistors in Parallel

• The total resistance for two resistors in
  parallel is equal to the product of the two
  resistors divided by the sum of the two
  resistors.
• RT R1R2/(R1 R2)

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Current Sources in Parallel

• The total current produced by all current
  sources is equal to the algebraic sum of the
  individual current sources.

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Power in Parallel Circuits

• Total power in a parallel circuit is found by
  adding up the powers of all the individual
  resistors.
• PT P1 P2 P3 . . . Pn

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• 6O3 O (6 x 3)/(63) 2O
• 1O5O6O
• 2O 2O 4 O
• 4O 6O 2.4O
• Req 4O 2.4O 8O
• 14.4O

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Determine Req
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Find the equivalent resistance at terminal a-b
for each networks
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Find the equivalent resistance at terminal a-b
for each networks
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Find the equivalent resistance at terminal a-b
for each networks
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Voltage-divider circuit

• Vs V1 V2
• iR1 iR2
• i Vs / (R1 R2)

V1 iR1 Vs R1/ (R1 R2)
V2 iR2 Vs R2/ (R1 R2)
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Find Va

• Va 18/(189)15
• 10V

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Find Va
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Current-divider circuit

• V i1R1 i2R2
• (R1R2)/ (R1R2) is

i1 V/R1 R2/(R1 R2) is
i2 V/R2 R1/(R1 R2) is
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Find i

• i 6 / (63) 2
• 4/3 A

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Find i
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• Find the no-load value of V0 in circuit shown
• Find V0 when RL is 450KO
• How much power is dissipated in the 30KO resistor
  is the load terminals are accidentally
  short-circuited?
• What is the maximum power dissipated in the 50KO
  resistor.

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• Find the power
• dissipated in the 6O
• resistor

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Determine V0 and i in the circuit shown
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Delta to Wye Conversion
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Wye to Delta Conversion
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Obtain the equivalent resistance Rab for the
circuit shown and use it to find i