Introduction to Circuits Analysis
28
1 AGBell – EECT 111 1 AGBell – EECT 111 by Andrew G. Bell [email protected] (260) 481-2288 Lecture 3
description
Introduction to Circuits Analysis. by Andrew G. Bell [email protected] (260) 481-2288 Lecture 3. PART II. Basic Circuit Analysis. CHAPTER 3. Ohm’s Law. Relationship of Current to Voltage*. A Direct Relationship An increase in voltage will result in an increase in current. - PowerPoint PPT Presentation
Transcript of Introduction to Circuits Analysis
2AGBell – EECT 111
2AGBell – EECT 111
PART II
Basic Circuit Analysis
3AGBell – EECT 111
3AGBell – EECT 111
CHAPTER 3
Ohm’s Law
4AGBell – EECT 111
4
Relationship of Current to Voltage*
• A Direct Relationship– An increase in voltage will result in an
increase in current.– A decrease in voltage will result in a
decrease in current.
*Resistance held constant
5AGBell – EECT 111
5
Relationship of Current to Resistance*
• Indirect Relationship– An increase in resistance will result in a
decrease in current.– A decrease in resistance will result in an
increase in current.
* Voltage held constant
6AGBell – EECT 111
6
Mathematical Relationship ofV, I, and R
• Formulated with three variables: – V, I, and R
• Relationship called Ohm’s law
• Three forms exist:
R
VI IRV
I
VR
7AGBell – EECT 111
7
The Circuit
8AGBell – EECT 111
8
Ohm’s Law Examples
Voltage = 25 V
Resistance = 25 I = _____
Amp125
V25
R
VI
9AGBell – EECT 111
9
Ohm’s Law Examples (cont.)
Voltage = 15 V
Resistance = 1 kI = _______
mA15Amps015.k1
V15
R
VI
10AGBell – EECT 111
10
Direct and Inverse Relationships
• Inverse relationship
11AGBell – EECT 111
11
Direct and Inverse Relationships (cont.)
• Direct relationship
12AGBell – EECT 111
12
Ohm’s Law Examples
Resistance = 50 Current = 2.5 A
Voltage = _______
Volts12550A5.2 IRV
13AGBell – EECT 111
13
Ohm’s Law Examples (cont.)
Resistance = 10 kCurrent = 4 mA
Voltage = ______
Volts40k10mA4 IRV
14AGBell – EECT 111
14
Ohm’s Law Examples (cont.)
Current = 15 mA
Voltage = 15 V
Resistance = _______
k1mA15
V15
I
VR
15AGBell – EECT 111
15
Ohm’s Law Examples (cont.)
Current = 4 mA
Voltage = 60 V
Resistance = _______
k15mA4
V60
I
VR
16AGBell – EECT 111
16
Current Flow Notations
• Electron Flow: Flow is from (–) to (+)• Conventional Flow*: Flow is from (+) to (–)
• Polarity: One point in a component or circuit being considered more negative or positive than another
*Older notation
17AGBell – EECT 111
17
Circuit Representation ofCurrent Flow
18AGBell – EECT 111
18
DC Characteristics
19AGBell – EECT 111
19
AC Characteristics
20AGBell – EECT 111
20
AC Characteristics (cont.)
21AGBell – EECT 111
21
Work, Energy and Power
• Work: Expenditure of energy– Mechanical Work = Force × Distance– Electrical Work = Charge × Voltage
• Energy: Ability to do work– Energy = Power × Time
• Power: Rate of energy usage– Power = Energy/Time
22AGBell – EECT 111
22
Power and Ohm’s Law
Power = Current x Voltage
P (watts) = I (amps) x V (voltage)
If I = 25 mA and V = 10 V:
P = (25 mA) x (10 V) = 250 mW
23AGBell – EECT 111
23
Relationships
• Note that if the current in the last slide doubles, the power will also double.
P = IV = (50 mA)(10 V) = 500 mW
24AGBell – EECT 111
Power vs. Resistance
RIP
IRIP
IVP
2
)(
R
VP
VR
VP
IVP
2
25AGBell – EECT 111
25
Relationships Between Power and Resistance
• For a constant resistance:– Doubling the current or voltage will result in
a quadrupling of power.– Halving the current or voltage will result in
a reduction in power by one-fourth.
26AGBell – EECT 111
26
Equation Summary
27AGBell – EECT 111
27
Basic Circuit Examples
• R = 10 k, I = 10 mA, V = _____
V = 100 V
• V = 50 V, I = 2 mA, R = ______
R = 25 k
• V = 75 V, R = 10 k, I = ______
I = 7.5 mA
28AGBell – EECT 111
28
Power Calculations
• V = 100 V, I = 2 mA, R = _____, P = _____
R = 50 k, P = 200 mW
• V = 250 V, R = 25 k, I = _____, P = _____
I = 10 mA, P = 2.5 W
