Electric circuits-chapter-2 Basic Laws

Chapter 2Basic Laws

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DKS1113 Electric Circuits

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Introduction

Fundament laws that govern electric circuits: Ohm’s Law. Kirchoff’s Law.

These laws form the foundation upon which electric circuit analysis is built.

Common techniques in circuit analysis and design: Combining resistors in series and parallel. Voltage and current divisions. Wye to delta and delta to wye transformations.

These techniques are restricted to resistive circuits.


Ohm’s Law

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Ohm’s Law

Relationship between current and voltage within a circuit element.

The voltage across an element is directly proportional to the current flowing through it v α i

Thus::v=iR and R=v/i Where:

R is called resistor. Has the ability to resist the flow of electric current. Measured in Ohms (Ω)


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Ohm’s Law

v=iR

*pay careful attention to current direction


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Ohm’s Law

Value of R :: varies from 0 to infinity

Extreme values == 0 & infinity.

Only linear resistors obey Ohm’s Law.

Short circuitShort circuit Open Circuit Open Circuit


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Ohm’s Law

Conductance (G) Unit mho or Siemens (S).

Reciprocal of resistance R

G = 1 / R

Has the ability to conduct electric current


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Ohm’s Law

Power:

P = iv i ( i R ) = i2R watts (v/R) v = v2/R watts

R and G are positive quantities, thus power is always positive.

R absorbs power from the circuit Passive element.


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Ohm’s Law

Example 1: Determine voltage (v), conductance (G) and power

(p) from the figure below.


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Ohm’s Law

Example 2: Calculate current i in figure below when the switch

is in position 1. Find the current when the switch is in position 2.


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Nodes, Branches & Loops

Elements of electric circuits can be interconnected in several way.

Need to understand some basic concepts of network topology.

Branch: Represents a single element (i.e. voltage, resistor & etc)

Node: The meeting point between two or more branches.

Loop:Any closed path in a circuit.DKS1113 Electric Circuits

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Example 3: Determine how many branches and nodes for the

following circuit.



5 Branches 1 Voltage Source 1 Current Source 3 Resistors

3 Nodes a b c

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Example 4: Determine how many branches and nodes for the

following circuit.


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Kirchoff’s Laws


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Kirchoff’s Laws

Kirchoff’s Current Law (KCL)

The algebraic sum of current entering / leaving a node (or closed boundary) is zero.

Current enters = +ve

Current leaves = -ve

∑ current entering = ∑ current leaving


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Kirchoff’s Laws

Example 5: Given the following circuit, write the equation for

currents.


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Kirchoff’s Laws

Example 6: Current in a closed boundary


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Kirchoff’s Laws

Example 9: Use KCL to obtain currents i1, i2, and i3 in the circuit.


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Kirchoff’s Laws

Kirchoff’s Voltage Law (KVL)

Applied to a loop in a circuit.

According to KVL The algebraic sum of voltage (rises and drops) in a loop is zero.

+

-

+ v1 -

- v3 +

+

V2

-

vs


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Kirchoff’s Laws

Example 10: Use KVL to obtain v1, v2 and v3.


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Kirchoff’s Laws

Example 11: Use KVL to obtain v1, and v2.


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Kirchoff’s Laws

Example 12: Calculate power dissipated in 5Ω resistor.

10


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Series Resistors & Voltage Division

Series resistors same current flowing through them.

v1= iR1 & v2 = iR2

KVL: v-v1-v2=0 v= i(R1+R2) i = v/(R1+R2 ) =v/Req

or v= i(R1+R2 ) =iReq iReq = R1+R2


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Series Resistors & Voltage Division

Voltage Division:

Previously: v1 = iR1 & v2 = iR2 i = v/(R1+R2 )

Thus: v1=vR1/(R1+R2) v2=vR2/(R1+R2)


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Parallel Resistors & Current Division

Parallel resistors Common voltage across it.

v = i1R1 = i2R2

i = i1+ i2

= v/R1+ v/R2

= v(1/R1+1/R2) =v/Req

v =iReq

1/Req = 1/R1+1/R2

Req = R1R2 / (R1+R2 )


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Parallel Resistors & Current Division

Current Division:

Previously: v = i1R1 = i2R2

v=iReq = iR1R2 / (R1+R2 ) and i1 = v /R1 & i2 =v/ R2

Thus: i1= iR2/(R1+R2) i2= iR1/(R1+R2 )


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Conductance (G)

Series conductance: 1/Geq = 1/G1 +1/G2+…

Parallel conductance: Geq = G1 +G2+…


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Voltage and Current Division

Example 13: Calculate v1, i1, v2 and i2.


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Example 14: Determine i1 through i4.


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Example 15: Determine v and i.

Answer v = 3v, I = 6 A.


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Example 16: Determine I1 and Vs if the current through 3Ω

resistor = 2A.


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Example 17: Determine Rab.


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Example 18: Determine vx and power absorbed by the 12Ω

resistor. Answer v = 2v, p = 1.92w.


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Wye-Delta Transformations

Given the circuit, how to combine R1 through R6? Resistors are neither in series nor parallel…

Use wye-delta transformationsDKS1113 Electric Circuits

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Y network T network


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Δ network π network


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Delta (Δ) to wye (y) conversion.


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Thus Δ to y conversion ::

R1 = RbRc/(Ra+Rb+Rc)

R2 = RaRc/(Ra+Rb+Rc)

R3 = RaRb/(Ra+Rb+Rc)

# Each resistors in y network is the product of two adjacent branches divide by the 3 Δ resistors


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Y to Δ conversions:

Ra = (R1R2 +R2 R3 +R1R3)/R1

Rb = (R1R2 +R2 R3 +R1R3)/R2

Rc= (R1R2 +R2 R3 +R1R3)/R3


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Example 19: Transform the circuit from Δ to y. Answer R1=18, R2=6, R3=3.


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Example 20: Determine Rab. Answer Rab=142.32.


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Example 21: Determine Io.


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Electric circuits-chapter-2 Basic Laws