OpenStax-CNX module: m10674
1
Series and Parallel Circuits
∗
Don Johnson
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0†
Abstract
A brief look at series and parallel circuits. Also denes voltage and current divider.
i1 + v1 –
vin
+
–
R1
i
+
R2
vout
vin
+
–
–
(a)
+
v
–
iout
R1
+
R2
vout
–
(b)
Figure 1: The circuit shown is perhaps the simplest circuit that performs a signal processing function.
The input is provided by the voltage source vin and the output is the voltage vout across the resistor
labelled R2 .
The results shown in other modules (circuit elements , KVL and KCL , interconnection laws ) with
regard to this circuit (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 ow only into anothercurrents in all
resistors connected this way have the same magnitudeare 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.
1
2
R2
vout
=
vin
R1 + R2
∗ Version
2.10: May 15, 2013 11:06 pm -0500
† http://creativecommons.org/licenses/by/3.0/
1 "Electric Circuits and Interconnection Laws" <http://cnx.org/content/m0014/latest/>
2 "Electric Circuits and Interconnection Laws" <http://cnx.org/content/m0014/latest/>
3 "Electric Circuits and Interconnection Laws" <http://cnx.org/content/m0014/latest/>
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3
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In this way, we express how the components used to build the system aect 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
relationshipsas a ratio of output to inputpervades circuit and system theory.
The current i is the current owing out of the voltage source. Because it equals i , we have that
=R +R :
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 rst 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.
1
vin
i1
1
2
2
Resistors in series:
vin
R1
+
–
R2
vin
+
–
R1+R2
Figure 2: The resistor (on the right) is equivalent to the two resistors (on the left) and has a resistance
equal to the sum of the resistances of the other two resistors.
Thus, the circuit the voltage source "feels" (through the current drawn from it) is a single resistor having
resistance R + R . Note that in making this equivalent circuit, the output voltage can no longer be dened:
The output resistor labeled R no longer appears. Thus, this equivalence is made strictly from the voltage
source's viewpoint.
1
2
2
iout
iin
R1
Figure 3: A simple parallel circuit.
http://cnx.org/content/m10674/2.10/
R2
iin
i1
+
+
v R1 v1 R2
iout
+
v2
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One interesting simple circuit (Figure 3) has two resistors connected side-by-side, what we will term a
rather than in series. Here, applying KVL reveals that all the voltages are identical:
v = v and v = v . This result typies parallel connections. To write the KCL equation, note that the top
node consists of the entire upper interconnection section. The KCL equation is i − i − i = 0. Using the
v-i relations, we nd that
R
parallel connection,
1
2
in
iout =
1
R1 + R2
1
2
iin
Exercise 1
(Solution on p. 7.)
Suppose that you replaced the current source in Figure 3 by a voltage source. How would i be
related to the source voltage? Based on this result, what purpose does this revised circuit have?
This circuit highlights some important properties of parallel circuits. You can easily
showthat the parallel
.A
=
combination of R and R has the v-i relation of a resistor having resistance +
shorthand notation for this quantity is R k R . 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.
out
1
1
R1
2
1
R1
1
R2
−1
R1 R2
R1 +R2
4
2
R 1R 2
R1+R2
R2
Figure 4
Similar to voltage divider (p. 1) 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 rst to the sum
of the conductances. Thus, for the depicted circuit, i =
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 =
i.
2
G2
G1 +G2
2
4 "Ideal Circuit Elements": Section Resistor <http://cnx.org/content/m0012/latest/#res>
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R1
R1 +R2
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i
i2
R1
R2
Figure 5
R1
+
vin
+
R2
–
vout
RL
–
source
system
sink
Figure 6: The simple attenuator circuit (Figure 1) is attached to an oscilloscope's input. The input2
output relation for the above circuit without a load is: vout = R1R+R
vin .
2
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 . 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 6, works. Rather than dening eight
variables and solving for the current in the load resistor, let's take a hint from other analysis (series rules
(p. 1), parallel rules (p. 3)). Resistors R and R are in a parallel conguration: The voltages across each
resistor are the same while the currents are not. Because the voltages are the same, we can nd the current
through each from their v-i relations: i = and i = . 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 must equal
the sum
of the other two currents leaving the node. Therefore, i = i + i ,
which means that i = v
+
.
Let R denote the equivalent resistance of the parallel combination of R and R . Using R 's v-i
relation, the voltage across it is v =
. The KVL equation written around the leftmost loop has
v = v + v ; substituting for v , we nd
L
2
2
L
vout
R2
L
vout
RL
1
1
1
out
1
R2
eq
2
1
in
1
out
R1 vout
Req
1
vin = vout
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2
1
RL
R1
+1
Req
L
1
L
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or
Req
vout
=
vin
R1 + Req
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 eect 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 onthe circuit. Comparing the input-output relations before and after, what we need is R ' R .
, the approximation would apply if or R R . This is the condition we
As R = +
seek:
Voltage measurement devices must have large resistances compared
with that of the resistor across which the voltage is to be measured.
eq
eq
1
R2
1
RL
−1
1
R2
1
RL
2
L
Voltage measurement:
Exercise 2
(Solution on p. 7.)
Let's be more precise: How much larger would a load resistance need to be to aect the inputoutput relation by less than 10%? by less than 1%?
Example 1
R2
R3
R1
R4
Figure 7
We want to nd the total resistance of the example circuit. To apply the series and parallel
combination rules, it is best to rst 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 in parallel
with R ; this combination is in series with R . This series combination is in parallel with R . 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 rst. The total resistance expression mimics the
structure:
2
3
4
1
RT = R1 k (R2 k R3 + R4 )
RT =
R1 R2 R3 + R1 R2 R4 + R1 R3 R4
R1 R2 + R1 R3 + R2 R3 + R2 R4 + R3 R4
Such complicated expressions typify circuit "simplications." A simple check for accuracy is the
units: Each component of the numerator should have the same units (here Ω ) as well as in the
denominator (Ω ). 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.
3
2
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Another valuable lesson emerges from this example concerning the dierence 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
eects. In our simple circuit, you might think that making the resistance R large enough would do the trick.
Because the resistors R and R 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 specications: Under what conditions will the circuit perform as designed? Thus, you will
nd 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
= 1+
, we can
gains less than one) depending only on the ratio of the two resistor values
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
integratorspeople who put systems together from component systemscan combine systems together and
have a chance of the combination working.
Figure 8 (series and parallel combination rules) 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.
L
1
2
R2
R1 +R2
R1
R2
−1
series and parallel combination rules
+
R1
+
RT
R2
v2
v
i
–
…
RN
GT
G1
G2
i2
… GN
–
(a) series combination rule
(b) parallel combination rule
Figure 8: Series and parallel combination rules. (a) RT =
in =
Gn
GT
PN
n=1
Rn vn =
Rn
RT
v (b) GT =
PN
n=1
Gn
i
Exercise 3
(Solution on p. 7.)
Contrast a series combination of resistors with a parallel one. Which variable (voltage or current)
is the same for each and which diers? 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?
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OpenStax-CNX module: m10674
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Solutions to Exercises in this Module
Solution to Exercise (p. 3)
Replacing the current source by a voltage source does not change the fact that the voltages are identical.
Consequently, v = R i or i = . This result does not depend on the resistor R , which means that
we simply have a resistor (R ) across a voltage source. The two-resistor circuit has no apparent use.
Solution to Exercise (p. 5)
. Thus, a 10% change means that the ratio must be less than 0.1. A 1% change means that
R =
< 0.01.
Solution to Exercise (p. 6)
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.
in
2 out
out
vin
R2
1
2
eq
R2
RL
R2
R
1+ R 2
L
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R2
RL