Ideal wires, Ideal device models, Ideal circuits. Today’s topics: Ideal models for circuit elements Wires Currents and Voltages Joints Resistors Voltage sources Current sources. Cast of Characters. Fundamental quantities Charge Current Voltage Power Fundamental concern
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Today’s topics:
EE 42 Lecture 2
EE 42 Lecture 2
1 proton = 1.6 x 1019 C
1 electron = 1.6 x 1019 C
1 C is a whole lot of protons!
6.25 x 1018 protons in 1 C.
EE 42 Lecture 2
Vector fields are like wind maps from your weather forecast.
+
Earth
EE 42 Lecture 2
+
b
a
EE 42 Lecture 2
saying that b is 5 V
higher than a.
saying that a is 5 V
higher than b.
+
b
a
 5 V +
+  5 V 
EE 42 Lecture 2
“Vab” means the potential at “a” minus the potential at “b” (that is, the potential drop from “a” to “b”).
a
convention to define a
voltage between two
labeled points:
names we wish, as long as weprovide
a reference frame (+ and ).
to right (or, the potential drop from right
to left, or the right potential minus the left).
b
VRoss

+
EE 42 Lecture 2
Examples
The flat end of the battery is at lower potential than the “bump” end.
B
C
A
D
9V
1.5V
1.5V
What is VAD ?
1.5 V + 1.5 V + 9 V = 6 V
Find V1 and Vx.
B
C
A
D
V1 = 1.5 V
VX = 6 V
1.5V
1.5V
9V

+
V1

VX
+
EE 42 Lecture 2
+
Vz

are often denoted using a
single subscript:
Also seen is
a
z
+
Va

EE 42 Lecture 2
where i is current in A, q is charge in C, and t is time in s
EE 42 Lecture 2
5 A
5 A
EE 42 Lecture 2
EE 42 Lecture 2
p = v i where p is power in W, v is voltage in V and i is current in A.
EE 42 Lecture 2
Power absorbed by device = (Vdevice) (i1)
Power generated by device = (Vdevice) (i2)
i1
i2
+ Vdevice 
EE 42 Lecture 2
Find the power absorbed by each element.
2

V
+
+
+
+
3 mA
2.5 mA
3
1 V
V
1 V
0.5 mA



Element :
(3 V)(3 mA) = 9 mW
Element :
(2 V)(3 mA) = 6 mW
(1 V)(0.5 mA) = 0.5 mW
Element :
(1 V)(2.5 mA) = 2.5 mW
Element :
EE 42 Lecture 2
EE 42 Lecture 2
EE 42 Lecture 2
EE 42 Lecture 2
i
voltage relationship called
Ohm’s law:
v = i R
where R is the resistance in Ω,
i is the current in A, and v is the
voltage in V, with reference
directions as pictured.
+
R
v

EE 42 Lecture 2
the voltage between its terminals.
Vs
EE 42 Lecture 2
value of the current running through it.
Is
EE 42 Lecture 2
EE 42 Lecture 2