chapter 2 basic laws -...
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Chapter 2 Basic Laws
28/12/2015
ME117 Electrical Engineering(1)
12/27/2015
Basic Laws
Series Resistors
Voltage Divider rule (VDR).
Parallel Resistors
Current Divider rule (CDR).
Mesh analysis
ME117 Electrical Engineering(1)
12/27/2015
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
ME117 Electrical Engineering(1)
Series Resistors & Voltage Division
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:
12/27/2015
ME117 Electrical Engineering(1)
12/27/2015
ME117 Electrical Engineering(1)
Series Resistors & Voltage Division
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,
12/27/2015
ME117 Electrical Engineering(1)
12/27/2015
Series Resistors & Voltage Division
ME117 Electrical Engineering(1)
Series Resistors & Voltage Division
EXAMPLE: Using the voltage divider rule (VDRa), determine the voltages V1 and V3 for the series circuit of Fig
12/27/2015
ME117 Electrical Engineering(1)
12/27/2015
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 )
ME117 Electrical Engineering(1)
Parallel Resistors & Current Division
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
12/27/2015
ME117 Electrical Engineering(1)
12/27/2015
Parallel Resistors & Current Division
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 )
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CDR
EXAMPLE: Determine the current I2 for the network of using the current divider rule (CDR)
12/27/2015
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CDR
EXAMPLE Find the current I1 for the network of Fig.
12/27/2015
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12/27/2015
Conductance (G)
Series conductance:
1/GT = 1/G1 +1/G2+…
Parallel conductance:
GT = G1 +G2+…
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12/27/2015
Voltage and Current Division
Example :
Determine i1 through i4.
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12/27/2015
Voltage and Current Division
Example :
Determine v and i.
Answer v = 3v, I = 6 A.
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12/27/2015
Voltage and Current Division
Example :
Determine I1 and Vs if the current through 3Ω resistor = 2A.
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Mesh Analysis: Basic Concepts:
R1
Rx
R2
+
_ I1 I2
+
_VA VB
+ +
+
_
_
_V1
VL1
V2
XL
AL
RIIVRIVwhere
VVV
211111
11
;
mesh 1:
Mesh Analysis
ME117 Electrical Engineering(1)
Mesh Analysis: Basic Concepts: R1
Rx
R2
+
_ I1 I2
+
_VA VB
+ +
+
_
_
_V1
VL1
V2
222121
21
;)(;
2
RIVRIIVwith
VVV
havewemesh
XL
BL
Mesh Analysis
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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
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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
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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
ME117 Electrical Engineering(1)