Lecture #4
OUTLINE
• Resistors in series
– equivalent resistance
– voltage-divider circuit
– measuring current
• Resistors in parallel
– equivalent resistance
– current-divider circuit
– measuring voltage
• Examples
• Node Analysis
Reading
Chapter 2
EE 40 Fall 2004
Lecture 4, Slide 1
Prof. White
Resistors in Series
Consider a circuit with multiple resistors connected in series.
Find their “equivalent resistance”.
I
R1
R2
VSS
+

• KCL tells us that the same
current (I) flows through
every resistor
• KVL tells us
R3
R4
Equivalent resistance of resistors in series is the sum
EE 40 Fall 2004
Lecture 4, Slide 2
Prof. White
Voltage Divider
I
R1
+
– V1
I = VSS / (R1 + R2 + R3 + R4)
R2
VSS
+

R3
+
– V3
R4
EE 40 Fall 2004
Lecture 4, Slide 3
Prof. White
When can the Voltage Divider Formula be Used?
I
I
R1
R2
VSS
+

R1
+
– V2
R2
+

VSS
R3
R4
R
2
V =
V
2
SS
R +R +R +R
1 2 3 4
Correct, if nothing else
is connected to nodes
EE 40 Fall 2004
+
– V2
R3
R4
R5
R
2
V ≠
V
2
SS
R +R +R +R
1 2 3 4
because R5 removes condition
of resistors in series
Lecture 4, Slide 4
Prof. White
Measuring Voltage
To measure the voltage drop across an element in a
real circuit, insert a voltmeter (digital multimeter in
voltage mode) in parallel with the element.
Voltmeters are characterized by their “voltmeter input
resistance” (Rin). Ideally, this should be very high
(typical value 10 MW)
Ideal
Voltmeter
Rin
EE 40 Fall 2004
Lecture 4, Slide 5
Prof. White
Effect of Voltmeter
circuit with voltmeter inserted
undisturbed circuit
R1
VSS
+
_
R1
+
R2
–
 R2 
V2 = VSS 

R
+
R
 1
2
V2
VSS
+
_
+
R2
Rin
–
V2′
 R2 || Rin 

V2 = VSS 

R 2 || Rin + R1 
Example: VSS = 10V, R2 = 100K, R1 = 900K  V2 = 1V
If Rin = 10M, V2 = 0.991V,
EE 40 Fall 2004
Lecture 4, Slide 6
Prof. White
Resistors in Parallel
ISS
Consider a circuit with two resistors connected in parallel.
Find their “equivalent resistance”.
x
• KVL tells us that the
same voltage is dropped
I1
I2
across each resistor
Vx = I1 R1 = I2 R2
R1
R2
• KCL tells us
EE 40 Fall 2004
Lecture 4, Slide 7
Prof. White
General Formula for Parallel Resistors
What single resistance Req is equivalent to three resistors in parallel?
I
I
+
V
+
R1
R2
R3
eq


V
Req

Equivalent conductance of resistors in parallel is the sum
EE 40 Fall 2004
Lecture 4, Slide 8
Prof. White
Current Divider
x
ISS
EE 40 Fall 2004
I1
I2
R1
R2
Lecture 4, Slide 9
Vx = I1 R1 = ISS Req
Prof. White
Generalized Current Divider Formula
Consider a current divider circuit with >2 resistors in parallel:
+
I1
I
R1
V
I3
I2
R2
R3

V=
I
 1   1   1 

 +
 + 
 R1   R 2   R 3 


1/R 3
V
I3 =
= I

R3
1/R
+
1/R
+
1/R
 1
2
3
EE 40 Fall 2004
Lecture 4, Slide 10
Prof. White
Circuit w/ Dependent Source Example
Find i2, i1 and io
EE 40 Fall 2004
Lecture 4, Slide 11
Prof. White
Using Equivalent Resistances
Simplify a circuit before applying KCL and/or KVL:
Example: Find I
I
R1
7V
+

R2
R6
R4 = R5 = 5 kW
R6 = 10 kW
R4
R5
EE 40 Fall 2004
R3
R1 = R2 = 3 kW
R3 = 6 kW
Lecture 4, Slide 12
Prof. White
The Wheatstone Bridge
• Circuit used to precisely measure resistances in
the range from 1 W to 1 MW, with ±0.1% accuracy
 R1 and R2 are resistors with known values
 R3 is a variable resistor (typically 1 to 11,000W)
 Rx is the resistor whose value is to be measured
battery
R1
+
V
–
R2
current detector
R3
Rx
variable resistor
EE 40 Fall 2004
Lecture 4, Slide 13
Prof. White
Finding the value of Rx
• Adjust R3 until there is no current in the detector
Then, Rx =
R2
R1
R3
Derivation:
KCL => i1 = i3 and i2 = ix
R1
+
V
–
i1 i2
i3
R3
R2
KVL => i3R3 = ixRx and i1R1 = i2R2
i1R3 = i2Rx
ix
Rx
R3
Typically, R2 / R1 can be varied
from 0.001 to 1000 in decimal steps
EE 40 Fall 2004
Lecture 4, Slide 14
R1
=
Rx
R2
Prof. White
Identifying Series and Parallel Combinations
Some circuits must be analyzed (not amenable to simple inspection)
R1
R2
R1
V
+
I
R3

R4
R5
V
+
-
R2
R3
R4
Special cases:
R3 = 0 OR R3 = 
EE 40 Fall 2004
R5
Lecture 4, Slide 15
Prof. White
Resistive Circuits: Summary
• Equivalent resistance of k resistors in series:
k
Req = S Ri = R1 + R2 + ••• + Rk
i=1
• Equivalent resistance of k resistors in parallel:
k
1
1
1
1
1
Req
=S
i=1
Ri
=
R1
+
R2
+ ••• +
• Voltage divided between 2 series resistors:
R1
v1 =
vs
R1 + R2
• Current divided between 2 parallel resistors:
R2
i
i1 =
is
R1 + R2
S
EE 40 Fall 2004
Lecture 4, Slide 16
Rk
I
vs
+
v1
–
R1
+

R2
i1
i2
R1
R2
Prof. White
Node-Voltage Circuit Analysis Method
1. Choose a reference node (“ground”)
Look for the one with the most connections!
2. Define unknown node voltages
those which are not fixed by voltage sources
3. Write KCL at each unknown node, expressing
current in terms of the node voltages (using the
I-V relationships of branch elements)
Special cases: floating voltage sources
4. Solve the set of independent equations
N equations for N unknown node voltages
EE 40 Fall 2004
Lecture 4, Slide 17
Prof. White
Nodal Analysis: Example #1
R
3
R1
+
-
V1
R2
R4
IS
1. Choose a reference node.
2. Define the node voltages (except reference node and
the one set by the voltage source).
3. Apply KCL at the nodes with unknown voltage.
4. Solve for unknown node voltages.
EE 40 Fall 2004
Lecture 4, Slide 18
Prof. White
Nodal Analysis: Example #2
R1
Va
R3
V
1
EE 40 Fall 2004
R2
I1
R4
Lecture 4, Slide 19
R5
V2
Prof. White
Nodal Analysis w/ “Floating Voltage Source”
A “floating” voltage source is one for which neither side is
connected to the reference node, e.g. VLL in the circuit below:
VLL
Va
Vb
- +
I1
R2
R4
I2
Problem: We cannot write KCL at nodes a or b because
there is no way to express the current through the voltage
source in terms of Va-Vb.
Solution: Define a “supernode” – that chunk of the circuit
containing nodes a and b. Express KCL for this supernode.
Incorporate voltage source constraint into KCL equation.
EE 40 Fall 2004
Lecture 4, Slide 20
Prof. White
Nodal Analysis: Example #3
supernode
VLL
Va
-
I1
Vb
+
R2
R4
I2
Eq’n 1: KCL at supernode
Substitute property of voltage source:
EE 40 Fall 2004
Lecture 4, Slide 21
Prof. White