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Engineering 17: Lecture 3
Dr. L. Lagerstrom, U.C. Davis
•
What about electrons?
•
Ch. 2 review (sample problem)
•
Resistors in series
•
Resistors in parallel
•
Applying seriesparallel simplification
•
Voltage division
•
Current division
What about electrons?
When we first learn about voltage and current in
physics, we often are concerned about which way the
actual charge carriers, i.e., electrons in most cases, are
flowing.
Consider a wire with some voltage across it (e.g., a wire
that connects the positive and negative terminals of a
battery). By
definition
, positive charge flows from the
positive terminal to the negative terminal.
From ht p:/ www.al aboutcircuits.com/vol_1/chpt_1/7.html.
What about electrons?
Since electrons, on the other hand, are negative charges,
they will flow from the negative to the positive terminal.
From ht p:/ www.al aboutcircuits.com/vol_1/chpt_1/7.html.
In general it doesn’t matter whether we use conventional flow
notation or electron flow notation. But most engineering
treatments of circuit analysis use conventional flow notation,
i.e., we assign a direction to positive current flow and don’t
worry about the actual direction the electrons are going.
Example 2.9
The voltage and current at the terminal of the device in (a)
below were measured, with results as shown in (b). Construct a
circuit model for the device and use it to predict the power the
device will deliver to a 10ohm resistor.
Example 2.9
The voltage and current at the terminal of the device in (a)
below were measured, with results as shown in (b). Construct a
circuit model for the device and use it to predict the power the
device will deliver to a 10ohm resistor.
Example 2.9, cont.
The voltage and current at the terminal of the device in (a)
below were measured, with results as shown in (b). Construct a
circuit model for the device and use it to predict the power the
device will deliver to a 10ohm resistor.
t
t
i
v
5
30

=
Plotting the data:
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Example 2.9, cont.
t
t
i
v
5
30

=
Remembering KVL, we see that the voltage equation
above is equivalent to a 30 V voltage source in series
with a 5 ohm resistor. So our circuit model is:
Example 2.9, cont.
To find the power this device will deliver to a 10ohm
resistor, we have:
KCL tells us that the current in the 5ohm resistor is
the same as the current in the 10ohm resistor. So
using KVL (with OL) around the circuit loop, we have:
i
i
i
i
10
5
30
or
,
0
10
5
30
+
=
=
+
+

So i = 2 A = the current through the 10ohm resistor.
And thus the power delivered to (or absorbed by) it is:
W
40
)
10
(
)
2
(
2
2
=
=
=
R
i
p
Resistors in Series
Consider the following “seriesconnected” resistors:
Applying KCL to each node in the circuit we find:
7
6
5
4
3
2
1
i
i
i
i
i
i
i
i
s
=

=

=
=
=

=
=
Thus we can redraw the circuit with just a single current:
Resistors in Series, cont.
To find the current i_s, apply KVL around the circuit
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This note was uploaded on 04/07/2008 for the course ENG 17 taught by Professor Lagerstrom during the Spring '08 term at UC Davis.
 Spring '08
 Lagerstrom
 Volt

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