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Transcript of Ee204 01 DC Review
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EE 204 Basic Electronics and Electrical Power
Chapter 1: DC Review
EE 204 Chapter 1: DC Review 1
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Some Useful Analogies
• E(V) Voltage ~ Pressure Æ causes ‘something’ to flow • I(A) Current ~ ‘something’ Æ water, charge, flux • R(Ω) Resistance ~ ‘restriction’ Æ prevents the flow
EE 204 Chapter 1: DC Review 2
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Resistor in Direct Current (DC) Circuits
• Ohm’s Law:
• Voltage in Volts (V) • Current in Amperes (A) • Resistance in Ohms (Ω)
• Will see “E” used instead of “V” when referring to the voltage
of a source (e.g. battery)
EE 204 Chapter 1: DC Review 3
V I RVoltage Current Resistance u
u
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Resistors in Parallel
EE 204 Chapter 1: DC Review 4
1 1 1
1
1
6 16
E I RV II A
u
u :
1 2 2
2
2
6 32
E I RV II A
u
u :
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Resistors in Series
EE 204 Chapter 1: DC Review 5
1 1 2
1
6 1 31.5
E I R I RV I II A
u u
u : u :
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Kirchhoff’s Voltage Law
• The total sum of the voltage rises and drops around a closed loop must be zero:
• In a passive component like a resistor (a sink for the energy), the voltage is higher on the input end (current), and lower” on the output
• For a source of energy like a battery however, the current flows out of the positive terminal and back into the negative terminal
EE 204 Chapter 1: DC Review 6
0V ¦
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Example 1
• What is the voltage across Resistor R5?
EE 204 Chapter 1: DC Review 7
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Example 1 Solution
• Consider two loop currents I2 and I3:
• Using KVL on the middle loop:
EE 204 Chapter 1: DC Review 8
I1 I2 I3
2 2 5 2 3 1
2 2 3
2 3
0
6 2 12 0
8 2 12 0 Eq.1
R I R I I E
I I I V
I I V
: :
: : m
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Example 1 Solution (cont.)
• Using KVL on the right loop:
EE 204 Chapter 1: DC Review 9
I1 I2 I3
3 3 6 3 2 5 3 2
3 3 3 2
2 3
0
3 3 4 2 0
2 8 4 0 Eq.2
R I R I E R I I
I I V I I
I I V
: : :
: : m
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Example 1 Solution (cont.)
• Determine I3:
EE 204 Chapter 1: DC Review 10
2 3
2 3
3
3
3
8 2 12 0
8 32 16 0
30 4 0
30 40.133
I I V
I I V
I V
I VI A
: :
: :
:
:
Using Eq.1Using Eq.2×4Eq.1 + Eq.2 4
m
m
m u
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Example 1 Solution (cont.)
• Determine I2:
• Determine VR5:
EE 204 Chapter 1: DC Review 11
2 3
2
2
2
8 2 12 0
8 2 0.133 12 0
8 11.7341.466
I I V
I A V
I VI A
: :
: :
:
5
5
5
5 2 3
2 1.466 0.133
3.2 positive voltageon top
R
R
R
V R I I
V A A
V V
:
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Thévenin’s Theorem
• Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source Vth in series with an equivalent resistance Rth
EE 204 Chapter 1: DC Review 12
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Thévenin’s Theorem (cont.)
• The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited
• The equivalent resistance Rth is the resistance obtained at terminals A-B of the network will all its independent current sources open circuited and all its independent voltage sources short circuited
EE 204 Chapter 1: DC Review 13
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Example 2
• Determine the voltage across Resistor R5 using Thévenin’s Theorem
EE 204 Chapter 1: DC Review 14
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Example 2 Solution
EE 204 Chapter 1: DC Review 15
3 62 1 2
2 3 6
3 34 12 46 3 3
8
th ab
th
th
R RV V E E ER R R
V V V V
V V
§ · ¨ ¸ © ¹
: :§ · ¨ ¸: : :© ¹
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Example 2 Solution (Cont.)
EE 204 Chapter 1: DC Review 16
2 3 62 3 6
2 3 6
6 3 33
6 3 3
th
th
R R RR R R R
R R R
R
: : : :
: : :
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Example 2 Solution (Cont.)
EE 204 Chapter 1: DC Review 17
5
5
5
5
5
2 83 2
3.2 positive voltageon top
R thth
R
R
RV VR R
V V
V V
u
: u
: :
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Norton’s Theorem
• Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent current source Ino in parallel with an equivalent resistance Rno
EE 204 Chapter 1: DC Review 18
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Norton’s Theorem (cont.)
• The equivalent current Ino is the current obtained at terminals A-B of the network with terminals A-B short circuited
• The equivalent resistance Rno is the resistance obtained at terminals A-B of the network will all its independent current sources open circuited and all its independent voltage sources short circuited (note that this is the same as calculating the Thévenin equivalent resistance)
EE 204 Chapter 1: DC Review 19
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Example 3
• Determine the voltage across Resistor R5 using Norton’s Theorem
EE 204 Chapter 1: DC Review 20
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Example 3 Solution
EE 204 Chapter 1: DC Review 21
1 2
2 3 6
12 46 3 32.667
no ab
no
no
E EI IR R R
V VI
I A
: : :
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Example 3 Solution (Cont.)
EE 204 Chapter 1: DC Review 22
2 3 62 3 6
2 3 6
6 3 33
6 3 3
no
no
R R RR R R R
R R R
R
: : : :
: : :
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Example 3 Solution (Cont.)
EE 204 Chapter 1: DC Review 23
5
5
5
5
5 5
55
3 2.667 23 2
3.2 positive voltageon top
R
noR no
no
R
R
V I R
RV I RR R
V A
V V
u
§ · u u¨ ¸© ¹§ ·:
u u :¨ ¸: :© ¹
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Power Dissipated in a Resistor
• The power (P) dissipated in a resistor is given by:
𝑃 = 𝑉 × 𝐼
• Making use of Ohm’s law (𝑉 = 𝐼 × 𝑅):
𝑃 = 𝐼 𝑅
𝑃 =𝑉𝑅
EE 204 Chapter 1: DC Review 24
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KVL Review:
• What is the voltage across R2? • 3V 4V 6V 20V
• What is the current through R3? • 1.2A 10A .30A 3.33A
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KCL Review
• What is the current through R2? • .5A 1A 2A 2.5A
• What is the total current coming from the source, E? • .5A 1.5A 2.5A 3.5A
• What is the voltage across R3? • 2V 10V 12V 5V
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Thévenin Review
• What is the Thévenin Voltage? • 3.33V 6.67V 10V 20V
• What is the Thévenin resistance? • 6.67 Ohms 10 Ohms 20 Ohms 26.67 Ohms
EE 204 Chapter 1: DC Review 27