ECE 102
Engineering Computation
Chapter 6
Circuit Elements
Dr. Herbert G. Mayer, PSU
Status 10/11/2015
For use at CCUT Fall 2015
Syllabus

Electric Charge and Field

Voltage

Current

Resistor

Inductor

Capacitor

Voltage and Current Sources

Voltage and Current Dividers

Examples
Electric Charge


Electric charge (q or Q) is an intrinsic property of
certain subatomic particles.
A particle’s charge affects its motion in the presence
of electric and magnetic fields.
The SI unit for electric charge is the coulomb (C).
1C=1A·s
 The amount of charge possessed by a single electron
is ≈ 1.602×10-19 C.


Charge can be positive or negative.
2
Electric Field

Lorentz Force Law → The force F acting on a charged
particle in the presence of an electric field and a
magnetic field is:
F  qE  qv  B

F, E, B, & v are vectors
q = electric charge of the particle
v = velocity of the particle
E = electric field, B = magnetic field
If the magnetic field is zero, then the electric field that
pervades the space around the charge is:
F
E= q

SI units for electric field: V/m (1 N/C = 1 V/m)
3
Voltage

If there is a difference in electric potential between
two spatial points, then a non-zero electric field will
exist between them.

An electric field causes charged particles within the
field to move.

The voltage V is the amount of work done in moving
a charge.

SI unit for voltage: volt (V) (1 V = 1 J/C)
4
Current

Current is the continuous movement of charge in a
particular direction.

Electric current I is the rate at which charge flows
through a cross-sectional area A.
A
I
Δq
Average: Iave =
Δt
dq
Instantaneous: I =
dt
5

Charge carriers may be:
 electrons (–)
 ions (+)
 holes (+)

Direct current (DC) → Charge carriers move in one
direction only (macroscopic behavior).

Alternating current (AC) → The charge carrier
direction varies periodically with time.

SI unit for current: ampere (A) (1 A = 1 C/s)
6
Conventional Current vs. Electron Current
Vhigh
Vlow
E
+
–
qVhigh
+
–
qVlow
Energy level representation

Conventional Current


Electron Current


Positive charge carriers flow from high to low potential
(same direction as the electric field).
Negative charge carriers flow from low to high potential
(opposite direction to the electric field).
In circuit analysis, conventional current is assumed, even
if electrons are the primary charge carriers.
7
+
–
+
–
+
–
From Vlow to Vhigh
→ Battery supplies energy
Vhigh
Battery
Circuit
Vlow
–
+
–
–
Negative charge moving:
From Vlow to Vhigh
→ Energy is dissipated
From Vhigh to Vlow
→ Battery supplies energy
+
+
Positive charge moving:
From Vhigh to Vlow
→ Energy is dissipated
8
Resistance

The resistance R is a measure of the opposition to
direct current through a material.

Interactions of charge carriers with the structure of
the material impedes the current.
Classes of materials:

 Conductor (low R : e.g., many metals)
 Insulator
(high R : e.g., ceramic)
 Semiconductor (intermediate R, e.g. doped Si)

SI unit for resistance: ohm (Ω)
9
Ohm’s Law

At a constant temperature, the current I through
certain materials is directly proportional to the
potential difference ΔV between its ends.
I  ΔV
V
R
I

The resistance R is defined as:

The general form of Ohm’s Law is:
V
V
I
R
V  IR
R
I
10
Example:
I
V1
R=2Ω
V2
V V1  V2
I

R
R
V1
V2
I
5V
0V
2.5 A
5V
2V
1.5 A
2V
5V
–1.5 A
1V
–3 V
2A
3V
3V
0A
If the potential difference
ΔV is zero, no current flows
through the resistor.
Note:
It is understood that Ohm’s Law refers to a potential
difference. The Δ is usually omitted.
V  IR
V
R
I
V
I
R
11
Application of Ohm’s Law

Given: Material of known resistance R
Voltage V is applied across the material
Result: Current I = V / R will flow through it.

Given: Material of known resistance R
Known current I flowing through it
Result: Voltage V = I · R exists across the
material (known as a “voltage drop”).

Given: Known voltage V across the material
Known current I through the material
Result: Resistance of the material is R = V / I.
12
Conductance

The conductance G is a measure of the ease with
which direct current flows through a material.
(reciprocal of resistance)
1
G
R

Ohm’s Law in terms of conductances:
I
I
 I  VG
G
 V
V
G

SI unit for conductance: siemens (S)
13
Power

Power is the rate at which energy is generated or
dissipated by an electrical element.
2
V
P  VI 
 I 2R
R
where
V = Voltage (V or J/C)
I = Current (A or C/s)
R = Resistance (Ω)

SI unit for power: watt (W) (1 W = 1 J/s)
14
Example:
I = 0.25 A
V1
R = 4.0 Ω
V2 = 6.0 V
What is the voltage drop across the resistor?
Vdrop  IR  0.25 A 4.0    1.0 V
What is the value V1 ?
V1  Vdrop  V2  1.0 V   6.0 V   7.0 V
Given current direction
implies that V1 > V2.
What is the power dissipated by the resistor?
P  I R  0.25 A  4.0    0.25 W
2
2
15
Resistor

A resistor is a passive electronic component that
obeys Ohm’s Law. It has resistance R

SI Unit: ohm ()

Symbol:

Impedance:

I-V relationship: v(t )  R i (t )
i(t)
v(t)
R
ZR
1
i (t )  v(t )
R
16
17
Resistors Connected in Series (end-to-end)

If N resistors are connected in series, with the
i-th resistor having a resistance Ri , then the
equivalent resistance Req is:
N
Req   Ri
i 1
R1
R2
R1
R2

R3

Req = R1 + R2
Req = R1 + R2 + R3
18
Properties of Series Resistances (DC)
Iin →
R1
R2
R3
→
→
→
I1
I2
I3
→ Iout

The amount of current Iin entering one end of a
series circuit is equal to the amount of current Iout
leaving the other end.

The current is the same through each resistor in the
series and is equal to Iin.
Iin = Iout = I1 = I2 = I3
19

The amount of voltage drop across each resistor in a
series circuit is given by Ohm’s Law.
I
→
R1
R2
–
+
V1
R3
–
+
V2
–
+
→I
V3
V1 = IR1 , V2 = IR2 , V3= IR3

By convention, the resistor terminal that the current
enters is labeled “+”, and the terminal the current
exits is labeled “–”
20
Example
I
A
2
6
3
a) Calculate the current I that
flows through the resistors.
b) Find the voltage drop across
the 6  resistor.
B
VAB = 3 V
Assume 3 significant figures.
Solution:
a) Approach – Use Ohm’s law: I 
VAB
Req
Calculate the equivalent resistance: Req  2  6  3   11 
3V
 0.273 A
Calculate the current: I 
11 
b) Use Ohm’s law
again:
V6  IR6  0.273 A 6    1.64 V
21
Resistors Connected in Parallel (side-by-side)
If N resistors
are in parallel, with the i-th resistor having a
resistance Ri , then the equivalent resistance is:

1
Req    
 i 1 Ri 
N
1
1
R1
R1
R2
R2
R3

1
1 
RR
Req      1 2
R1  R2
 R1 R2 

 1
1
1 

Req   
 
 R1 R2 R3 
1
22
Properties of Parallel Resistances (DC)
R1
+ V1 –
Iin →
R2
+ V2 –
→ Iout
R3
+ V3 –

The amount of current Iin entering one end of a
parallel circuit is equal to the amount of current Iout
leaving the other end

For parallel resistors, the voltage drop across each
resistor is the same
Iin = Iout and
V1 = V2 = V3
23

The amount of current through each resistor in a
parallel circuit is given by Ohm’s Law.
R1
I1
+ V –
R2
I→
I2
+ V –
R3
I3

→I
I1 = V / R1
I2 = V / R2
I3 = V / R3
+ V –
The sum of the currents through each resistor is
equal to the original amount of current entering or
leaving the parallel circuit
I1 + I 2 + I3 = I
24
Example
Ix
2
6
I =4A
A
B
3
VAB
a) Find the current Ix through
the 2  resistor.
b) What is the power dissipated
by the 3  resistor?
Assume 3 significant figures.
Solution:
a) Approach – Use Ohm’s law: VAB  IReq
1
1 1 1
Calculate the equivalent resistance: Req        1 
 2 6 3
Calculate the voltage drop: VAB  4 A1   4 V
Find the current: I x 
VAB
 2.00 A
2
2
VAB
 5.33 W
b) Use power equation: P 
3
25
Inductor & Capacitor

An inductor is a passive component that stores
energy within a magnetic field. It has inductance L
SI unit: henry (H)
Impedance: Z  jL
Symbol:

A capacitor is a passive component that stores
energy within an electric field. It has capacitance C
1
Impedance: Z   j C 
SI unit: farad (F)
Symbol:
non-polarized
+
polarized
26
DC Voltage & Current Sources


An ideal DC voltage source
outputs a constant voltage
regardless of the amount of
current through it
Vs +–
An ideal DC current source
outputs a constant current
regardless of the amount
of voltage across it
Is
I
I
+
Vs
–
V
27
Voltage & Current Dividers
I
+
V0
–
+
I
V1 –
R1
R2
+
V2
–
Voltage Divider
+
I1
I2
VS
G1
G2
–
Current Divider
R1
V1 
V0
R1  R2
G1
I1 
I
G1  G2
R2
 I1 
I
R1  R2
R2
V2 
V0
R1  R2
G2
I2 
I
G1  G2
R1
 I2 
I
R1  R2
28
Example Voltage Drop
Vs = 8 V
Ra = 1 Ω
Va
Rb = 3 Ω
Vb
+
–
What is the voltage drop across Ra?
Ra
Va 
VS
Ra  Rb
1
8 V   2 V
Va 
1 3 
What is the voltage drop across Rb?
Rb
Vb 
VS
Ra  Rb
3
8 V   6 V
Va 
1 3 
29
Example Resistor
The voltage Vg from a signal generator must be
reduced by a factor of three before entering an
audio amplifier. Determine the resistor values of a
voltage divider that performs this task
R1
Max input = 1 V
Peak output = 3 V
Signal
Generator
Vout
Vg
R2

Vg

R1  R2
3
Vg
R2
Vout
3R2  R1  R2
Audio
Amplifier
 R1  2R2
30
Example Resulting Resistance
R0
VS
+
–
R2
R1
R4
R3
R7
R5
R6
R0 = 5.0 
R1 = 2.0 
R2 = 8.0 
R3 = 6.0 
R4 = 1.0 
R5 = 3.0 
R6 = 6.0 
R7 = 8.0 
VS = 10.0 V
a)What is the equivalent resistance Req (in Ω) that is
“seen” by the voltage source?
b)What is the current (in A) in R6?
31