Download Series Circuits

Lesson 4: Series DC Circuits
and Kirchhoff’s Voltage Law
(KVL)
1
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 series-opposing 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.
• Describe the basic function of a fuse and a switch.
• Draw a schematic of a typical electrical circuit, and explain the purpose of
each component and indicate the polarity and current direction.
• Explain and compute how voltage divides between elements in a series
circuit.
• Compute voltage drops across resistors using the voltage divider formula.
• 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.
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Series Circuits
• A series circuit is defined by 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.
3
Series Circuits
• Normally;
− Current will leave the positive terminal of a voltage source
and move through the resistors.
− Current will return to the negative terminal of the voltage
source.
− Current is the same everywhere in a series circuit.
Is
Is
Is
Load
4
Is
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.
5
Series Resistors
• Now that the series connection has been described, yo
can now recognize that every fixed resistor has only
two terminals to connect in a configuration
− referred to as a two-terminal device.
Series connection of resistors.
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Series Resistors
Series connection of resistors.
Series connection of four
resistors of the same value
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Series Resistors Example
• In each circuits below, list the resistors in series
with R2.
R1 and R3
None
Here is a look at the current
R
flow,
1 which we will talk about
later
8
Two Resistors in Series
• Two resistors in series can be replaced by an
equivalent resistance Req.
R1  R2  Req
9
N Resistors in Series
• The equivalent resistance Req of any number of
resistors in series is the sum of the individual
resistances.
N
 RN   Rn
Req  R1  R2 
n 1
10
Resistors in Series
• The equivalent resistance Req of any number of
resistors in series is the sum of the individual
resistances.
R R R R R R R
eq
1
2
3
4
5
6
Req  2  4  6  3  1  2  18
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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.
• Mathematically, KVL implies:
ET - V1 - V2 - V3 - ... - Vn  0
Where ET => ET= E1+E2+E3+…+En
ET  V1  V2  V3  ...  Vn
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Kirchhoff’s Voltage Law (2)
• Another way of stating KVL is: KVL is the
summation of voltage rises is equal to the
summation of voltage drops around a closed
E1  E 2  V1  V2  V3
loop.
E1  V1  V2  E 2  V3  0
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Kirchhoff’s Voltage Law (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.
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Kirchhoff’s Voltage Law (4)
• Summation of voltage rises is equal to the summation
of voltage drops around a closed loop.
E1  E 2  V1  V2  V3
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Example Problem 1
Determine the unknown voltages in the network below:
First, determine the value of V3:
V3  I * R3
V3  40mA *1.5k   60V
Next, determine the value of V1:
E  V1  V2  V3
100V  V1  60V  30V
V1  100V  60V  30V  10V
Verify:
E  I * RT
E  40mA *(250  750  1.5k )  100V
Because you know I, V1 and V2 we can also
find R1 and R1:
V
V
R2  2
R1  1
I
I
30V
10V
R2 
 750
R1 
 250
40mA
40mA
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Example Problem 2
Use Kirchhoff’s Voltage Law to determine the
magnitude and polarity of the unknown voltage ES in
the circuit below:
KVL States : E s  E 2  V1  V2  V3
Knowns: R1 = 12kΩ, Is = 2 mA, E2 = 62 V,
R2 = 6k Ω, and R3 = 3 k Ω
First, determine the value of v1, v2 and v3:
V1  I * R1  2mA *12k   24V
V2  2mA * 6k   12V
V3  2mA * 3k   6V
Using KVL, find Es:
E s  E 2  V1  V2  V3
E s  62V  24V  12V  6V
E s  24V  12V  6V  62V  20V
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VOLTAGE DIVISION IN A SERIES CIRCUIT
Voltage Divider Rule (VDR)
• The voltage divider rule states that:
− The voltage across a resistor (Vx) in a series circuit is
equal to the value of that resistor (Rx) times the total
applied voltage divided (E) by the total resistance (RT) of
the series configuration.
E
Vx  Rx *
RT
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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.
Vx  Rx *
VR1  R1 *
E
RT
E
10V
 2k  *
 2V
RT
10k 
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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:
ITotal 
E
RTotal
Total Current

 in the Circuit
Vx  ITotal Rx  Vx  Rx *
E
RTotal
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Voltage Divider Rule
• 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 
V2  119V
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Example Problem 3
For the circuit below (with Rtot =800Ω), determine:
E Total Current
I


a) Direction and magnitude of current.
Total
RTotal  in the Circuit
b) Voltage drop across each resistor.
E
Vx  ITotal Rx  Vx  Rx *
c) Value of the unknown resistance.
RTotal
a ) ITotal 
E
RTotal

25V
 31.25mA
800
25V
 3.125V
800
25V
V2  150 *
 4.688V
800
25V
V3  250 *
 7.813V
800
25V
V4  300 *
 9.375V
800
V1  100 *
b)Vx  Rx *
E
RTotal
Verify:
E  V1  V2  V3  V4
E  3.125V  4.687V  7.813V  9.375V  25V
c) Req  R1  R2  R3  R4
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R3  800  100  150  300  250
Power in a Series Circuit
• Power dissipated by each resistor is determined by
the power formulas:
P = VI = V2/R = I2R
• Since energy must be conserved, power delivered by
voltage source is equal to total power dissipated by
resistors.
PT = P1 + P2 + P3 + ∙∙∙ + Pn
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Example Problem
Using VDR to Solve For Power
For the circuit below, determine:
a)
b)
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.
a ) ITotal 
E
RTotal

63V
 3mA
21k 
P1  (3mA) 2 * 12k   108mW
Px  I 2 R x
P2  (3mA) 2 * 6k   54mW
PT  P1 +P2 +P3 =189mW
P3  (3mA) 2 *3k   27 mW
b) PT  V*I = 63V*3mA=189mW
Another way to solve would have been
to use the voltage divider rule if you
wanted…
Vx  Rx *
E
RTotal
V1  12k  *
63V
 36V
21k 
V2  6k  *
63V
 18V
21k 
V3  3k  *
24
25V
 9V
21k 
Px 
Vx 2
Rx
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.
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Interchanging Series Components
Be cautious that you recognize the
polarity of the source calculated:
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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.
ET= E1+E2+E3+…+En
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Simplifying Voltage Sources
Polarities must be taken into account.
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Voltage Sources in Series
• Resultant source
− Sum of the rises in one direction minus the sum of the
voltages in the opposite direction.
ET = 5V
ET= E1+E2+E3+…+En
ET = 2V - 6V - 1V = 5V
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16Ω
Example Problem 4
Redraw the circuit below, showing a single voltage
source and single resistor. Solve for the current in the
circuit.
ET= E1+E2+E3+…+En
ET = 12V + 3V - 6V = 9V
RT= R1+R2+R3+…+Rn
RT = 27kΩ + 33kΩ + 18kΩ = 78kΩ
= 9V
IT 
ET
9V

 115.4  A
RT
78k 
30
= 78kΩ
Notation
• How circuits are drawn and how they are referenced
is an important part of being able to analyze a circuit.
• Some standard industry notation follows:
31
Grounds
Ground: Is a point of reference or a
common point in a circuit for making
measurements.
COMMON TYPES:
• Chassis ground
− Common point of circuit is often the metal
chassis of the piece of equipment.
− Often connected to Earth Ground.
− If a fault occurs within a circuit, the current
is redirected to the earth.
• Earth ground
− Physically connected to the earth by a metal
pipe or rod.
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NOTATION
Voltage Sources and Grounds
Three ways to sketch the same series dc circuit.
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NOTATION
Voltage Sources
Replacing the notation for a
negative dc supply with the
standard notation.
Replacing the special notation
for a dc voltage source with
the standard symbol.
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NOTATION
Single-Subscript Notation
• If point b of the notation Vab is specified as ground
with potential (zero volts), then a single-subscript
notation can be used that provides the voltage at a
point with respect to ground.
Defining the use of
single-subscript notation
for voltage levels.
35
NOTATION
Double-Subscript Notation
• The fact that voltage is across a variable and exists
between two points has resulted in a doublesubscript notation that defines the first subscript as
the higher potential.
Defining the sign for double-subscript notation.
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NOTATION
Double-Subscript Notation
• The double-subscript notation Vab specifies point a as the
higher potential than point b.
• 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 point b.
• 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:
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NOTATION
General Comments
Another way to view the voltage
The impact of positive and
negative voltages on the total
voltage drop.
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NOTATION
Examples
Vab  Va  Vb
Vab  Va  Vb  16V  20V  4V
Va  Vab  Vb  5V  4V  9V
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Voltage Divider Rule Application
Using Notation
Given the circuit below, find the output voltage Vab
across the R2 resistor.
= 75 Ω
 R2 
Vab  E 

R

R
 1
2 
25


Vab  12V 
  3V
 75  25 
= 25 Ω
Incidentally; how would you find the current (I)
for this circuit using the voltage divider rule?
I  E /  RT 
I  12V /  75  25   120mA
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Example Problem 5
Find the output voltage Vab across the R2 resistor.
= 50 Ω
= 30 Ω
= 20 Ω
Using the voltage divider rule:
E
Vx  ITotal Rx  Vx  Rx *
RTotal
Vab  R2
Vab
*
E
RT
110V
 30 *
 33V
(50  30  20)
41
Example Problem 6
Use the Voltage Divider rule to find ES.
Recall the voltage divider rule:
E
Vx  ITotal Rx  Vx  Rx *
RTotal
However, first you need to find voltage across one of the
resistors and because power was givn for R 2 (P2 =500mW) start there:
V 2R2
P2 
 VR 2 
R2
P2 * R2 
500mW * 200  10V
Now, solve for E:
VR 2  R 2 *
V *RT
E
10V *1.6k 
 E  R 2

 80V
RT
R2
200
42
Example Problem 7
Find Va, Vb, Vc, Vd .
Va  0V because it is a source referenced to ground
Vb  100V because it is a source referenced to E
VR1  Vb  Vc
Vc  Vb  VR1  100  10  90V
VR 3  Vd  Va
Vd  VR 3  Va  30  0  30V
Incidentally, we could also find v2:
ET= V1+V2+V3
V2 = ET - (V1+V3) = 100V-10V-30V = 60V
43
Other Basic Circuit Components
• Beyond the resistor we have discussed extensively,
there are some other circuit components that you may
encounter:
− Switch
− Fuse
− Circuit Breaker
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Switches
• The most basic circuit components is a switch.
• The switch below is known as a single-pole, singlethrow (SPST) switch.
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Fuses
• A fuse is a device that prevents excessive current to
protect against overloads or possible fires.
• A fuse literally “blown” and can not be reset and
must be replaced.
46
Circuit breakers
• A circuit breaker (CB), like a fuse, will prevent
excessive current from damaging circuits.
• However, a CB uses an electro-mechanical
mechanism that opens a switch.
• A big difference between a fuse and CB is that a
“popped” CB can be reset.
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Consolidated Schematic
Circuit Breaker
Ammeter
Lamp
Battery
Fuse
Voltmeter
48
QUESTIONS?
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