Chapter 2 Basic Laws - Cracksmodqu.edu.iq/eng/wp-content/uploads/2015/12/electrical-engineering-Lecture3.pdf · 12/27/2015 Basic Laws Series Resistors Voltage Divider rule (VDR)

Chapter 2 Basic Laws

28/12/2015

ME117 Electrical Engineering(1)

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Basic Laws

Series Resistors

Voltage Divider rule (VDR).

Parallel Resistors

Current Divider rule (CDR).

Mesh analysis


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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/RT

or v= i(R1+R2 ) =iRT

RT= R1+R2



EXAMPLE a. Find the total resistance for the series circuit b. Calculate the source current I c. Determine the voltages V1, V2, and V3. d. Calculate the power dissipated by R1, R2, and R3. e. Determine the power delivered by the source, and compare it to the sum of the power levels of part (d). Solutions:

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To find the total resistance of N resistors of the same value in series, simply multiply the value of one of the resistors by the number in series; that is,

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EXAMPLE: Using the voltage divider rule (VDRa), determine the voltages V1 and V3 for the series circuit of Fig

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Parallel Resistors & Current Division

Parallel resistors Common voltage across it.

v = i1R1 = i2R2

KCL

i = i1+ i2

= v/R1+ v/R2

= v(1/R1+1/R2)

=v/RT

v =iRT

1/RT = 1/R1+1/R2

RT = R1R2 / (R1+R2 )



EXAMPLE For the parallel network of Fig a. Calculate RT. b. Determine Is. c. Calculate I1 and I2, and demonstrate that Is =I1 + I2. d. Determine the power to each resistive load. e. Determine the power delivered by the source, and compare it to the total power dissipated by the resistive elements

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Current Division:

Previously:

v = i1R1 = i2R2

v=iRT = iR1R2 / (R1+R2 )

and i1 = v /R1 & i2 =v/ R2

Thus:

i1= iR2/(R1+R2)

i2= iR1/(R1+R2 )


CDR

EXAMPLE: Determine the current I2 for the network of using the current divider rule (CDR)

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CDR

EXAMPLE Find the current I1 for the network of Fig.

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Conductance (G)

Series conductance:

1/GT = 1/G1 +1/G2+…

Parallel conductance:

GT = G1 +G2+…


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Voltage and Current Division

Example :

Determine i1 through i4.


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Example :

Determine v and i.

Answer v = 3v, I = 6 A.


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Example :

Determine I1 and Vs if the current through 3Ω resistor = 2A.


Mesh Analysis: Basic Concepts:

R1

Rx

R2

+

_ I1 I2

+

_VA VB

+ +

+

_

_

_V1

VL1

V2

XL

AL

RIIVRIVwhere

VVV

211111

11

;

mesh 1:

Mesh Analysis


Mesh Analysis: Basic Concepts: R1

Rx

R2

+

_ I1 I2

+

_VA VB

+ +

+

_

_

_V1

VL1

V2

222121

21

;)(;

2

RIVRIIVwith

VVV

havewemesh

XL

BL

Mesh Analysis


Steps to determine the mesh currents:

1. Assign mesh currents i1, i2, …, in to the n meshes.

2. Apply KCL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.

3. Solve the resulting n simultaneous equations to get the mesh currents.

Mesh Analysis


Mesh Analysis: Example.

Write the mesh equations and solve for the currents I1, and I2.

+

_10V

4 2

6 7

2V20V

I1 I2+

+_

_

Mesh 1 4I1 + 6(I1 – I2) = 10 - 2

Mesh 2 6(I2 – I1) + 2I2 + 7I2 = 2 + 20

Mesh Analysis


Mesh Analysis: Example , continued.

Simplifying Eq. gives,

10I1 – 6I2 = 8 -6I1 + 15I2 = 22

» % A MATLAB Solution » » R = [10 -6;-6 15]; » » V = [8;22]; » » I = inv(R)*V I = 2.2105 2.3509

I1 = 2.2105 I2 = 2.3509

Mesh Analysis


Documents

Chapter 2 Basic Laws - Cracksmodqu.edu.iq/eng/wp-content/uploads/2015/12/electrical-engineering-Lecture3.pdf · 12/27/2015 Basic Laws Series Resistors Voltage Divider rule (VDR)