A circuit connects circuit elements together in a specific configuration
designed to transform the source signal (originating from a voltage or current source)
into another signal—the output—that corresponds to the current or
voltage defined for a particular circuit element. A simple resistive circuit is shown
in Figure 1. This circuit is the electrical
embodiment of a system having its input provided by a source system producing
v
in
t
v
in
t
.
To understand what this circuit accomplishes, we want to determine the voltage across
the resistor labeled by its value
R
2
R
2
. Recasting this problem mathematically, we need to solve some set of equations so that
we relate the output voltage
v
out
v
out
to the source voltage. It would be simple—a little too simple at this
point—if we could instantly write down the one equation that relates these two
voltages. Until we have more knowledge about how circuits work, we must write a set of
equations that allow us to find all the voltages and currents that
can be defined for every circuit element. Because we have a three-element circuit, we
have a total of six voltages and currents that must be either specified or determined.
You can define the directions for current flow and positive voltage drop any
way you like. When two people solve a circuit their own ways, the values of
their variables may not agree, but current and voltage values for each element will agree.
Do recall in defining your
voltage and current variables that the v-i relations for
the elements presume that positive current flow is in the same direction as positive
voltage drop. Once you define voltages and currents, we need six nonredundant equations
to solve for the six unknown voltages and currents. By specifying the source, we have
one; this amounts to providing the source's v-i relation. The
v-i relations for the resistors give us two more. We are only halfway
there.
What we need to solve every circuit problem are mathematical statements that express what
the interconnection of elements is. Said another way, we need the laws that govern the
electrical connection of circuit elements. First of all, the places where circuit elements
attach to each other are called nodes. Two nodes are explicitly indicated
in Figure 1; a third is at the bottom where the voltage
source and resistor
R
2
R
2
are connected. Electrical engineers tend to draw circuit diagrams—schematics—
in a rectilinear fashion. Thus the long line connecting the bottom of the voltage source
with the bottom of the resistor is intended to make the diagram look pretty. This line
simply means that the two elements are connected together. Kirchoff's
Laws, one for voltage
and one for current, determine
what a connection among circuit elements means. These laws can help us analyze this circuit.
At every node, the sum of all currents entering a node must equal zero. What
this law means physically is that charge cannot accumulate in a node; what
goes in must come out. In the example, Figure 1,
below we have a three-node circuit and thus have three KCL equations.
i
1
−
i
2
=0
i
1
i
2
0
(2)
Note that the current entering a node is the negative of the current leaving
the node.
Given any two of these KCL equations, we can find the other by adding or subtracting
them. Thus, one of them is redundant and, in mathematical terms, we can discard any one
of them. The convention is to discard the equation for the (unlabeled) node at the bottom of the
circuit.
In writing KCL equations, you will find that in an
nn-node circuit, exactly one of them is always
redundant. Can you sketch a proof of why this might be true? Hint: It has to
do with the fact that charge won't accumulate in one place on its own.
KCL says that the sum of currents entering or leaving a node must be zero. If
we consider two nodes together as a "supernode", KCL applies as well to currents
entering the combination. Since no currents enter an entire circuit, the sum of
currents must be zero. If we had a two-node circuit, the KCL equation of one
must be the negative of the other, We can combine all but one
node in a circuit into a supernode; KCL for the supernode must be the negative of
the remaining node's KCL equation. Consequently, specifying
n−1
n
1
KCL equations always specifies the remaining one.
The voltage law says that the sum of voltages around every closed loop in the circuit
must equal zero. A closed loop has the obvious definition: Starting at a node, trace
a path through the circuit that returns you to the origin node. KVL expresses the fact
that electric fields are conservative: The total work performed in moving a test charge
around a closed path is zero. The KVL equation for our circuit is
v
1
+
v
2
−v=0
v
1
v
2
v
0
(4)
In writing KVL equations, we follow the convention that an element's voltage enters
with a plus sign if traversing the closed path, we go from the positive to the
negative of the voltage's definition.
For the example circuit above, we have three v-i relations,
two KCL equations, and one KVL equation for solving for the circuit's six voltages
and currents.
v-i:
v
1
=
R
1
i
1
v
1
R
1
i
1
(6)
v
out
=
R
2
i
out
v
out
R
2
i
out
(7)
KCL:
i
1
−
i
out
=0
i
1
i
out
0
(9)
KVL:
−v+
v
1
+
v
out
=0
v
v
1
v
out
0
(10)
We have exactly the right number of equations! Eventually, we will discover shortcuts
for solving circuit problems; for now, we want to eliminate all the variables but
v
out
v
out
. The KVL equation can be rewritten as
v
in
=
v
1
+
v
out
v
in
v
1
v
out
.
Substituting into it the resistor's v-i relations, we have
v
in
=
R
1
i
1
+
R
2
i
out
v
in
R
1
i
1
R
2
i
out
.
Yes, we temporarily eliminate the quantity we seek. Though not obvious, it is the
simplest way to solve the equations. One of the KCL equations says
i
1
=
i
out
i
1
i
out
,
which means that
v
in
=
R
1
i
out
+
R
2
i
out
=(
R
1
+
R
2
)
i
out
v
in
R
1
i
out
R
2
i
out
R
1
R
2
i
out
.
Solving for the current in the output resistor, we have
i
out
=
v
in
R
1
+
R
2
i
out
v
in
R
1
R
2
. We have now solved the circuit: We have expressed one voltage or
current in terms of sources and circuit-element values. To find any other circuit
quantities, we can back substitute this answer into our original equations or ones we
developed along the way. Using the v-i relation for the
output resistor, we obtain the quantity we seek.
V
out
=
R
2
R
1
+
R
2
v
in
V
out
R
2
R
1
R
2
v
in
(11)
Referring back to Figure 1, a circuit should
serve some useful purpose. What kind of system does our circuit realize and, in
terms of element values, what are the system's parameter(s)?
The circuit serves as an amplifier having a gain of
R
2
R
1
+
R
2
R
2
R
1
R
2
.
The results shown in other modules (circuit
elements,KVL and KCL,
interconnection laws) with regard to the
circuit above (Figure 1), and the values of other
currents and voltages in this circuit as well, have profound implications.
Resistors connected in such a way that current from one must flow only
into another—currents in all resistors connected this way have the same
magnitude—are said to be connected in series. For the two
series-connected resistors in the example, the voltage across one resistor
equals the ratio of that resistor's value and the sum of resistances times the
voltage across the series combination. This concept is so pervasive it has
a name: voltage divider.
The input-output relationship for this system, found in this particular
case by voltage divider, takes the form of a ratio of the output voltage to the input
voltage.
v
out
v
in
=
R
2
R
1
+
R
2
v
out
v
in
R
2
R
1
R
2
(12)
In this way, we express how the components used to build the system affect the
input-output relationship. Because this analysis was made with ideal circuit
elements, we might expect this relation to break down if the input amplitude is
too high (Will the circuit survive if the input changes from 1 volt to one million
volts?) or if the source's frequency becomes too high. In any case, this important
way of expressing input-output relationships—as a ratio of output to
input—pervades circuit and system theory.
The current
i
1
i
1
is the current flowing out of the voltage source. Because it equals
i
2
i
2
, we have that
v
in
i
1
=
R
1
+
R
2
v
in
i
1
R
1
R
2
:
The series combination of two resistors acts, as far as the voltage
source is concerned, as a single resistor having a value equal to the sum of
the two resistances.
This result is the first of several
equivalent circuit ideas: In many cases, a complicated
circuit when viewed from its terminals (the two places to which you might attach
a source) appears to be a single circuit element (at best) or a simple combination
of elements at worst. Thus, the equivalent circuit for a series combination of
resistors is a single resistor having a resistance equal to the sum of its
component resistances.
Thus, the circuit the voltage source "feels" (through the current drawn from it)
is a single resistor having resistance
R
1
+
R
2
R
1
R
2
. Note that in making this equivalent circuit, the output voltage can no longer be
defined: The output resistor labeled
R
2
R
2
no longer appears. Thus, this equivalence is made strictly from the voltage source's
viewpoint.
One interesting simple circuit (Figure 6)
has two resistors connected side-by-side, what we will term a parallel
connection, rather than in series.
Here, applying KVL reveals that all the voltages are identical:
v
1
=v
v
1
v
and
v
2
=v
v
2
v
. This result typifies parallel connections. To write the KCL equation, note that
the top node consists of the entire upper interconnection section. The KCL equation is
i
in
−
i
1
−
i
2
=0
i
in
i
1
i
2
0
. Using the v-i relations, we find that
i
out
=
R
1
R
1
+
R
2
i
in
i
out
R
1
R
1
R
2
i
in
(13)
Suppose that you replaced the current source in Figure 6 by a voltage source. How would
i
out
i
out
be related to the source voltage? Based on this result, what purpose does
this revised circuit have?
Replacing the current source by a voltage source does not change the fact that
the voltages are identical. Consequently,
v
in
=
R
2
i
out
v
in
R
2
i
out
or
i
out
=
v
in
R
2
i
out
v
in
R
2
. This result does not depend on the resistor
R
1
R
1
, which means that we simply have a resistor (
R
2
R
2
)across a voltage source. The two-resistor circuit has no apparent use.
This circuit highlights some important properties of parallel circuits. You can
easily show that the parallel combination of
R
1
R
1
and
R
2
R
2
has the v-i relation of a resistor having resistance
1
R
1
+1
R
2
−1=
R
1
R
2
R
1
+
R
2
1
R
1
1
R
2
1
R
1
R
2
R
1
R
2
. A shorthand notation for this quantity is
R
1
∥
R
2
R
1
∥
R
2
. As the reciprocal of resistance is
conductance, we can say that for a parallel combination of resistors,
the equivalent conductance is the sum of the conductances.
Similar to voltage divider
for series resistances, we have current divider for parallel resistances.
The current through a resistor in parallel with another is the ratio of the conductance
of the first to the sum of the conductances. Thus, for the depicted circuit,
i
2
=
G
2
G
1
+
G
2
i
i
2
G
2
G
1
G
2
i
. Expressed in terms of resistances, current divider takes the form of the resistance
of the other resistor divided by the sum of resistances:
i
2
=
R
1
R
1
+
R
2
i
i
2
R
1
R
1
R
2
i
.
Suppose we want to pass the output signal into a voltage measurement device,
such as an oscilloscope or a voltmeter. In system-theory terms, we want to
pass our circuit's output to a sink. For most applications, we can represent
these measurement devices as a resistor, with the current passing through it
driving the measurement device through some type of display.
In circuits, a sink is called a load; thus, we describe a system-theoretic
sink as a load resistance
R
L
R
L
. Thus, we have a complete system built from a cascade of three systems: a
source, a signal processing system (simple as it is), and a sink.
We must analyze afresh how this revised circuit, shown in Figure 9, works. Rather than defining eight variables and solving for the current in
the load resistor, let's take a hint from other
analysis (series rules,
parallel rules). Resistors
R
2
R
2
and
R
L
R
L
are in a parallel configuration: The voltages across each resistor
are the same while the currents are not. Because the voltages are the same, we can find
the current through each from their v-i relations:
i
2
=
v
out
R
2
i
2
v
out
R
2
and
i
L
=
v
out
R
L
i
L
v
out
R
L
.
Considering the node where all three resistors join, KCL says that the sum of the
three currents must equal zero. Said another way, the current entering the node through
R
1
R
1
must equal the sum of the other two currents leaving the node. Therefore,
i
1
=
i
2
+
i
L
i
1
i
2
i
L
, which means that
i
1
=
v
out
(1
R
2
+1
R
1
)
i
1
v
out
1
R
2
1
R
1
.
Let
R
eq
R
eq
denote the equivalent resistance of the parallel combination of
R
2
R
2
and
R
L
R
L
. Using
R
1
R
1
's v-i relation, the voltage across it is
v
1
=
R
1
v
out
R
eq
v
1
R
1
v
out
R
eq
.
The KVL equation written around the leftmost loop has
v
in
=
v
1
+
v
out
v
in
v
1
v
out
; substituting for
v
1
v
1
,
we find
v
in
=
v
out
(
R
1
R
eq
+1)
v
in
v
out
R
1
R
eq
1
(14)
or
v
out
v
in
=
R
eq
R
1
+
R
eq
v
out
v
in
R
eq
R
1
R
eq
(15)
Thus, we have the input-output relationship for our entire system having the form
of voltage divider, but it does not equal the input-output
relation of the circuit without the voltage measurement device. We can not measure
voltages reliably unless the measurement device has little effect on what we are
trying to measure. We should look more carefully to determine if any values for the
load resistance would lessen its impact on the circuit. Comparing the input-output
relations before and after, what we need is
R
eq
≃
R
2
R
eq
R
2
. As
R
eq
=1
R
2
+1
R
L
−1
R
eq
1
R
2
1
R
L
1
, the approximation would apply if
1
R
2
≫1
R
L
1
R
2
≫
1
R
L
or
R
2
≪
R
L
R
2
≪
R
L
.
This is the condition we seek:
Voltage measurement devices must have large
resistances compared with that of the resistor across which the voltage is to be
measured.
Let's be more precise: How much larger would a load resistance need to be to
affect the input-output relation by less than 10%? by less than 1%?
R
eq
=
R
2
1+
R
2
R
L
R
eq
R
2
1
R
2
R
L
. Thus, a 10% change means that the ratio
R
2
R
L
R
2
R
L
must be less than 0.1. A 1% change means that
R
2
R
L
<0.01
R
2
R
L
0.01
.
We want to find the total resistance of the example circuit. To apply the series
and parallel combination rules, it is best to first determine the circuit's structure:
What is in series with what and what is in parallel with what at both small-
and large-scale views. We have
R
2
R
2
in parallel with
R
3
R
3
; this combination is in series with
R
4
R
4
. This series combination is in parallel with
R
1
R
1
. Note that in determining this structure, we started away from
the terminals, and worked toward them. In most cases, this approach works well; try it
first. The total resistance expression mimics the structure:
R
T
=
R
1
∥(
R
2
∥
R
3
+
R
4
)
R
T
R
1
∥
R
2
∥
R
3
R
4
(16)
R
T
=
R
1
R
2
R
3
+
R
1
R
2
R
4
+
R
1
R
3
R
4
R
1
R
2
+
R
1
R
3
+
R
2
R
3
+
R
2
R
4
+
R
3
R
4
R
T
R
1
R
2
R
3
R
1
R
2
R
4
R
1
R
3
R
4
R
1
R
2
R
1
R
3
R
2
R
3
R
2
R
4
R
3
R
4
(17)
Such complicated expressions typify circuit "simplifications". A simple check for
accuracy is the units: Each component of the numerator should have the same units
(here
Ω3
Ω
3
) as well as in the denominator (
Ω2
Ω
2
). The entire expression is to have units of resistance; thus, the ratio of the
numerator's and denominator's units should be ohms. Checking units does not guarantee
accuracy, but can catch many errors.
Another valuable lesson emerges from this example concerning the difference between
cascading systems and cascading circuits. In system theory, systems can be cascaded
without changing the input-output relation of intermediate systems. In cascading circuits,
this ideal is rarely true unless the circuits are so
designed. Design is in the hands of the engineer; he or she must
recognize what have come to be known as loading effects. In our simple circuit, you
might think that making the resistance
R
L
R
L
large enough would do the trick. Because the resistors
R
1
R
1
and
R
2
R
2
can have virtually any value, you can never make the resistance of your voltage
measurement device big enough. Said another way, a circuit cannot be designed
in isolation that will work in cascade with all other circuits. Electrical
engineers deal with this situation through the notion of specifications
: Under what conditions will the circuit perform as designed? Thus, you
will find that oscilloscopes and voltmeters have their internal resistances clearly
stated, enabling you to determine whether the voltage you measure closely equals what
was present before they were attached to your circuit. Furthermore, since our resistor
circuit functions as an attenuator, with the attenuation (a fancy word for gains less
than one) depending only on the ratio of the two resistor values
R
2
R
1
+
R
2
=1+
R
1
R
2
−1
R
2
R
1
R
2
1
R
1
R
2
1
, we can select any values for the two resistances we want to
achieve the desired attenuation. The designer of this circuit must thus specify not
only what the attenuation is, but also the resistance values employed so that
integrators—people who put systems together from component systems—can
combine systems together and have a chance of the combination working.
Figure 11 summarizes the series and parallel combination results. These results are easy to remember and very useful. Keep in mind that for series combinations, voltage and resistance are the key quantities, while
for parallel combinations current and conductance are more important. In series combinations, the
currents through each element are the same; in parallel ones, the voltages are the same.
Contrast a series combination of resistors with a parallel one. Which variable (voltage or current) is the same for each and which differs? What are the equivalent resistances? When resistors are placed in series, is the equivalent resistance bigger, in between, or smaller than the component resistances? What is this relationship for a parallel combination?
In a series combination of resistors, the current is the same in each; in a parallel
combination, the voltage is the same. For a series combination, the equivalent resistance
is the sum of the resistances, which will be larger than any component resistor's
value; for a parallel combination, the equivalent conductance is the sum of the
component conductances, which is larger than any component conductance.
The equivalent resistance is therefore smaller than any component resistance.
