Lecture 5: Chapter 2, Part 3
Matthew Shuman
1
Resistors in Series — Section 2.5 in Text
Figure 1: These resistors are in series
Figure 1 shows that resistors in series add their resistances together. The equivalent resistance
across these combined resistors would be 1kΩ + 3kΩ = 4kΩ. Remember that components are in
series if those two components are the only components connected to a single node.
REq = R1 + R2
2
(1)
Resistors in Parallel — Section 2.6 in Text
Figure 2: These resistors are in parallel
Figure 2 shows that two resistors in parallel add their conductances together.
1
1
1
1
1
=
+
=
+
= .0013333
REq
R1 R2
1000 3000
(2)
REq = 750Ω
(3)
1
The algebra required to compute inverses is clumsy and unecessary. Solving the resistor parallel
equation for REq yields the following equation.
REq =
R1 ∗ R2
R1 + R2
(4)
The equivalent resistance of two resistors in parallel can be found by taking the product of those
resistanced divided by the sum of the resistances. This simplification is only valid for two resistors
in parallel. If 3 1KΩ resistors were in parallel the product over sum shortcut could be used on
1000∗1000
two of them to get an equivalent resistance of 1000+1000
= 500 and then again with the remaining
1000∗500
resistor to get 1000+500 = 333.3Ω.
2.1
Example Problem
Find the equivalent resistance from A to B.
Figure 3: This is an example for problem for lecture
ENGR 201
3
3.1
The Voltage-Divider and Current-Divider Circuits — Section
2.5 & 2.6 in Text
The Voltage-Divider
A voltage source across two resistors in series is a common circuit in electronics, and is shown in
figure 4. A voltage divider is typically used to create a specified voltage between the two resistors.
Figure 4: This is a voltage-divider
Calculating the voltage between the resistors can be done by doing a KCL analysis on the node
between the two resistors as indicated in figure 5.
IR1 − IR2 = 0
(5)
Adding in the Ohm’s Law substitutions yields.
4 − VOut
VOut − 0
−
=0
1kΩ
3kΩ
(6)
Solving for VOut yields that VOut = 3V olts. This process of using KCL and Ohm’s Law substitutions
could be done for all voltage dividers, but there is a shortcut shown in equation 7. This enables a
quick circuit analysis to compute the output voltage of a voltage divider. Note that V1 is always the
voltage source across both resistors and R2 is always the resistor between VOut and ground.
Figure 5: This is a voltage-divider
Vout = VR2 = V 1 ∗
R2
R1 + R2
ENGR 201
(7)
3.2
The Current-Divider
Current dividers are not a common, but there is a shortcut equation for them as well. I1 is the
current split between two resistors in parallel and the resistor in the numerator of equation 8 is the
resistor of the branch you do not want to find the current.
Figure 6: This is a current-divider
IR2 = I1 ∗
4
4.1
R1
R1 + R2
(8)
Voltage Division and Current Division — Section 2.5 & 2.6 in
Text
Voltage Division
Voltage division can be used for dividers with several resistors.
Figure 7: This is voltage division
Vout = VR3 = V 1 ∗
R3
R1 + R2 + R3
ENGR 201
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4.2
Current Division
Current division can be done with more than 2 resistors in parallel as well.
Figure 8: This is current division
IR1 = I1 ∗
R1||R2||R3
R1
ENGR 201
(10)