Lesson 4: Series Circuits
Learning Objectives





Identify elements that are connected in series.
State and apply KVL in analysis of a series circuit.
Determine the net effect of series-aiding and seriesopposing voltage sources.
Compute the power dissipated by each element and
the total power in a series circuit.
Compute voltage drops across resistors using the
voltage divider formula.
Learning Objectives

Apply concept of voltage potential between two
points to the use of subscripts and the location of
the reference voltage.

Analyze a series resistive circuit with the ground
placed at various points
SERIES RESISTORS

Before the series
connection is described,
first recognize that every
fixed resistor has only two
terminals to connect in a
configuration—it is
therefore referred to as a
two-terminal device.
FIG. 5.4 Series connection
of resistors.
SERIES RESISTORS
FIG. 5.6 Series connection of
resistors.
FIG. 5.7 Series connection of
four resistors of the same
value
Series Circuits

Two elements in a series
 Connected
at a single point
(node)
 No
other current-carrying
connections at this node
 The
same current flows
through series connected
circuits.
Series Circuits

Normally
 Current
will leave the positive terminal of a
voltage source and move through the resistors
 Return to negative terminal of the source
 Current is the same everywhere in a series
circuit
Series Circuits

Current is similar to water flowing through a
pipe
 Current
leaving the element must be the same as
the current entering the element
 Current = water flow rate
 Pressure = potential difference = voltage

Same current passes through every element of
a series circuit
Kirchhoff’s voltage law (1)



Kirchhoff’s voltage law (KVL) states that the
algebraic sum of all voltages around a closed
path (or loop) is zero.
M
vm  0

Mathematically, KVL implies
m 1
ET - V1 - V2 - V3 - ∙∙∙ - Vn = 0
Kirchhoff’s voltage law (2)
Another way of stating KVL is:
Summation of voltage rises is equal
to the summation of voltage drops
around a closed loop
V1 + V2 + V3 + ∙∙∙ + Vn = ET
being
ET= E1+E2+E3+…+En
Kirchhoff’s Voltage Law (KVL) (3)

A closed loop is any
path that:
 Originates at a point
 Travels around a
circuit
 Returns to the original
point without retracing
any segments

V  0 ( c lo s e d lo o p )
Kirchhoff’s Voltage Law (4)

Summation of voltage rises is equal to the
summation of voltage drops around a closed
loop

E r is e s 

V drops
E1 + E2 = v1 + v2 + v3
Example Problem 1
Determine the unknown voltages in the network
below:
Example Problem 2
Use Kirchhoff’s Voltage Law to determine the
magnitude and polarity of the unknown voltage ES
in the circuit below:
Resistors in Series
Most complicated circuits can be simplified
 For a series circuit

 V1
+ V2 + V3 = E
 IR1 + IR2 + IR3 = E
 I(R1 + R2 + R3 )= E
 I(R1 + R2 + R3 )= IRtotal (Note: I’s cancel)
Two resistors in series

Two resistors in series can be replaced by an
equivalent resistance Req.
Req  R1  R2
N resistors in series

The equivalent resistance Req of any number of
resistors in series is the sum of the individual
resistances.
Req  R1  R2 
N
 RN   Rn
n 1
The Voltage Divider Rule

Voltage applied to a series circuit:
 Voltage
drop across any series resistor is
proportional to the magnitude of the resistor.
 The voltage divider rule allows us to calculate the
voltage across any series resistance in a single step,
without first calculating the current.
VOLTAGE DIVISION IN A SERIES CIRCUIT
Voltage Divider Rule (VDR)

The voltage divider rule states that the
voltage across a resistor in a series
circuit is equal to the value of that
resistor times the total applied voltage
divided by the total resistance of the
series configuration.
The Voltage Divider Rule

Voltage applied to a series circuit:
 Voltage
drop across each resistor may be
determined by the proportion of its resistance to the
total resistance:
I Total 
E
RTotal
 Total Current

 in the Circuit
Rx
Vx  I Total Rx  Vx 
E
RTotal
Voltage Divider Rule Application


If a single resistor is very large (~100x) compared to the
other series resistors, the voltage across that resistor will be
the source voltage
Voltage across the small resistors will be essentially zero


R2
V2  
E
 R1  R2  R3  R4 
600



 120V 
 2  600  3  1 
 119V
Voltage Divider Application:
Simple Power Supply
= 75 Ω
= 25 Ω
 R2 
Vab  E 

 R1  R2 
 25 
 12 
  3V
 100 
Resistors in Series
Example Problem 3
For the circuit below (with Rtot =800Ω), determine:



Direction and magnitude of current
Voltage drop across each resistor
Value of the unknown resistance
Power in a Series Circuit

Power dissipated by each resistor is
determined by the power formulas:
P = VI = V2/R = I2R
Power in a Series Circuit

Since energy must be conserved, power
delivered by voltage source is equal to
total power dissipated by resistors
PT = P1 + P2 + P3 + ∙∙∙ + Pn
Example Problem 4
For the circuit below, determine:


Power dissipated by each resistor and total power
dissipated by the circuit.
Verify that the summation of the power dissipated by
each resistor equals the total power delivered by the
voltage source.
Voltage Sources in Series


In a circuit with more than one source in series
 Sources can be replaced by a single source having
a value that is the sum or difference of the
individual sources
Polarities must be taken into account
Voltage Sources in Series

Resultant source
 Sum
of the rises in one direction minus the
sum of the voltages in the opposite direction
Simplifying Sources
Interchanging Series
Components

Order of series components
 May
be changed without affecting operation of
circuit
Sources may be interchanged, but their
polarities can not be reversed
 After circuits have been redrawn, it may
become easier to visualize circuit operation

Interchanging Series
Components
Example Problem 5
Redraw the circuit below, showing a single voltage
source and single resistor. Solve for the current in
the circuit.
Example Problem 6
Find the output voltage Vab across the R2 resistor.
= 50 Ω
= 30 Ω
= 20 Ω
Circuit Ground

Ground
 Point
of reference or a
common point in a circuit
for making measurements

Chassis ground:
 Common
point of circuit is
often the metal chassis of
the piece of equipment
Circuit Ground

Earth ground
 Physically
connected to the
earth by a metal pipe or rod

Chassis ground
 Often
connected to Earth
Ground
 If a fault occurs within a
circuit, the current is
redirected to the earth
NOTATION
Voltage Sources and Ground
FIG. 5.46 Three ways to sketch the same
series dc circuit.
NOTATION
Voltage Sources and Ground
FIG. 5.47 Replacing the
special notation for a dc
voltage source with the
standard symbol.
FIG. 5.48 Replacing the
notation for a negative dc
supply with the standard
notation.
NOTATION
Double-Subscript Notation

The fact that voltage is an across variable and
exists between two points has resulted in a
double-subscript notation that defines the first
subscript as the higher potential.
FIG. 5.50 Defining the sign for doublesubscript notation.
NOTATION
Double-Subscript Notation
The double-subscript notation Vab
specifies point a as the higher
potential.
 If this is not the case, a negative sign
must be associated with the magnitude
of Vab.
 In other words, the voltage Vab is the
voltage at point a with respect to (w.r.t.)
point b.
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NOTATION
General Comments
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A particularly useful relationship can now be established
that has extensive applications in the analysis of
electronic circuits.
For the above notational standards, the following
relationship exists:
NOTATION
General Comments
FIG. 5.52
Example 5.21.
FIG. 5.53
Example 5.22.
NOTATION
General Comments
FIG. 5.54
Example 5.23.
FIG. 5.55 The impact of
positive and negative voltages
on the total voltage drop.
NOTATION
Single-Subscript Notation

If point b of the notation Vab is specified as
ground potential (zero volts), then a singlesubscript notation can be used that provides the
voltage at a point with respect to ground.
FIG. 5.51 Defining
the use of singlesubscript notation
for voltage levels.
Single Subscripts
If voltages at
various points in a
circuit are known
with respect to
ground, then the
voltage between
points is easily
determined
 Vab = Va – Vb

Example Problem 4
Find Va, Vb, Vc, Vd