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 positive current flow
and positive voltage drop any way you
like.
Once the values for the voltages and currents are calculated, they may be positive or negative according to your definition.
When two people define variables according to their individual preferences,
the signs of their variables may not agree, but current flow and
voltage drop 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;
where do we get the other three equations we need?
What we need to solve every circuit problem are
mathematical statements that express how the circuit elements are interconnected. 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. Kirchhoff's Laws,
one for voltage and
one for current,
determine what a connection among circuit elements means. These
laws are essential to analyzing this and any circuit. They are named for Gustav Kirchhoff, a nineteenth century German physicist.
At every node, the sum of all currents entering or leaving 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)−i1=0
i1−i2=0
i+i2=0
i
i1
0
i1
i2
0
i
i2
0
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
In writing KVL equations, we follow the convention that an
element's voltage enters with a plus sign when traversing the
closed path, we go from the positive to the negative of the
voltage's definition.
For the example
circuit, 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=vin
v1=R1i1
vout=R2iout
KCL:
(−i)−i1=0
i1−iout=0
KVL:
−v+v1+vout=0
v-i:
v
vin
v1
R1
i1
vout
R2
iout
KCL:
i
i1
0
i1
iout
0
KVL:
v
v1
vout
0
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
and determine how it depends on
v
in
v
in
and on resistor values.
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
relation, 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
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
.
"Electrical Engineering Digital Processing Systems in Braille."