EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Learning Objectives
a. Identify elements that are connected in series
b. State and apply KVL in analysis of a series circuit
c. Determine the net effect of series-aiding and series-opposing voltage sources
d. Compute the power dissipated by each element and the total power in a series circuit
e. Compute voltage drops across resistors using the voltage divider formula
f. Apply concept of voltage potential between two points to the use of subscripts and the location of the
reference voltage
g. Analyze a series resistive circuit with the ground placed at various points
h. Explain and compute how voltage divides between elements in a series circuit
Series Circuits
Two elements are in series if:
 They are connected at a single point (termed a node)
 No other current-carrying connections exist at this node
So, in the picture on the right, R1 and R2 are in series.
Current is similar to water flowing through a pipe. Current leaving the element
(e.g., a resistor) must be the same as the current entering the element.
If two elements are in series, the same current passes through them. A series circuit is constructed by
connecting elements in series. The same current passes through every element of a series circuit.
Example: In each circuit below, list the resistors in series with resistor R2 .
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Resistors in Series Most complicated circuits can be simplified. For example, two resistors in series
can be replaced by an equivalent resistance RT.
Req  R1  R2
The equivalent resistance Req of any number of resistors in series is the sum of the individual
resistances.
It is worth pausing to remind you that THE POLARITY OF THE VOLTAGE ACROSS A RESISTOR
IS DETERMINED BY THE DIRECTION OF CURRENT FLOW!!!
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Determine RTOT in the circuit below.
Solution:
Example: Determine the ohmmeter reading in the circuit below. Determine the ohmmeter reading in
the circuit below if the leads were reversed.
Solution:
Power in a Series Circuit The 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
the resistors:
PT = P1 + P2 + P3 + ∙∙∙ + Pn
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Multiple Voltage Sources in Series Sources can be replaced by a single source having a value that is
the sum or difference of the individual sources. The source polarities must be taken into account. The
resultant source will be the sum of the rises in one direction minus the sum of the voltages in the
opposite direction. For example, the collection of four sources shown below on the left can be
combined into the single 3V source shown on the right. Again, note the polarity.
Interchanging Series Components
The order of series components may be changed without affecting the operation of circuit. Sources may
be interchanged, but their polarities cannot be reversed. After circuits have been redrawn, it may
become easier to visualize circuit operation. The circuit shown below on the left can be simplified,
resulting in the circuit shown on the right (where you should convince yourself that the total resistance
RTOT is 10  and the total voltage is 5 V with the polarity shown.
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Switches A basic circuit component you will see as the course progresses is a switch. The switch
shown below is known as a single-pole, single-throw (SPST) switch.
Fuses A fuse is a device that prevents excessive current to protect against overloads or possible fires.
A fuse literally “blows” and cannot be reset
Circuit Breakers A circuit breaker also prevents excessive current in circuits; however it uses an
electro-mechanical mechanism that opens a switch. A “popped” circuit break can be reset.
Symbol for a circuit breaker
Symbol for an ammeter
Symbol for a lamp
Symbol for a battery
Symbol for a fuse
Symbol for a voltmeter
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Kirchhoff’s Voltage Law (KVL) Kirchhoff’s voltage law (KVL) states that the algebraic sum of all
voltages around a closed loop is zero.
M
Mathematically, KVL implies
vm  0

A closed loop is any path that:
m 1
 Originates at a point
 Travels around a circuit
 Returns to the original point without retracing any segments
The algebraic sum of the voltage that rises and drops around a closed loop is equal to zero.
E1  v1  v2  E2  v3  0
Another way of expressing KVL: Summation of voltage rises is
equal to the summation of voltage drops around a closed loop.
E1  E2  v1  v2  v3
Example: Determine the voltage V1 in the circuit below using KVL.
Solution:
Example: Determine the voltage Vx in the circuit below using
KVL.
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Determine the unknown voltages V1 and V3 in the circuit below.
Solution:
Example: Determine the unknown voltage ES shown in the circuit below.
Solution:
Example: Given the circuit:
Determine:
a. The direction and magnitude of current
b. The voltage drop across each resistor
c. The value of the unknown resistance
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Given the circuit:
Determine:
a. The power dissipated by each resistor and total power dissipated
by the circuit.
b. Verify that the summation of the powers dissipated by the
resistors equals the total power delivered by the voltage source.
Solution:
Example: Given the circuit:
a. Redraw the circuit with a single voltage source and single resistor.
b. Determine the current.
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Your roommate has hooked up the circuit shown below.
a. Is the voltage reading across R2 correct?
b. Is the voltage reading across R3 correct?
Solution:
The Voltage Divider Rule For the voltage applied to a series circuit:
The voltage drop across each resistor may be determined by the proportion
of its resistance to the total resistance:
Vx 
Rx
E
RTotal
If a single resistor is very large compared to the other series resistors (say,
100 times larger), the voltage across that resistor will be the source voltage.
Put another way, the voltage across the small resistors will be essentially
zero.


R2
V2  
E
 R1  R2  R3  R4 
600



 120V 
 2  600  3  1 
 119V
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Determine the voltage from a to b in the circuit
shown below.
= 75 Ω
Solution:
= 25 Ω
Circuit Ground A circuit ground (shown to the right) is a point of reference, or a
common point in a circuit for making measurements. A circuit ground is usually
physically connected to the earth by a metal pipe or rod. The idea is that if a fault
occurs within a circuit, the current is redirected to the earth.
So, ground represents a point of zero reference potential. Ground is 0 volts.
Single subscript notation for voltage: In a circuit with a ground reference point
voltages may be expressed with respect to that reference point (e.g. Vc is the
voltage at node c with respect to the 0 volts found at the ground).
Your friend says that no current flows in the circuit shown on the right, since there
is no loop for current to flow. Is your friend correct?
Double subscript notation for voltage: Voltage between any two node points (a and b) can be written as
Vab.
Vab  Va  Vb 
If b is at a higher potential than c, then Vbc is positive. If a is at a lower potential than b, then Vab is
negative.
Example: Determine Vab and Vbc in the circuit shown.
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Voltages between two points can be determined two ways:
Using known node voltages
Vbd = Vb – Vd = 100 – 30 = 70 V
Summing voltage drops or gains across components
Vbd = v2+v1 = 60 + 10 = 70 V
v2= 60𝑉
Point Sources If the voltage at two nodes is known, then you can simplify a circuit for easier analysis
by creating a circuit with a point source as shown in the figure below:
v2= 60𝑉
Example: Determine the voltage Vab in the circuit below.
= 50 Ω
Solution:
= 30 Ω
= 20 Ω
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Determine the voltage labeled V  shown in the
circuit below.
Solution:
Example: Determine the value of V3 in the circuit shown below. Note that the voltmeter reads +5.6 V.
Solution:
Example: Determine the voltage Vab in the circuit below.
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Determine the voltage Va in the circuit below.
Solution:
Example: Use the Voltage Divider Rule to find ES in the circuit below.
Solution:
Example: Find Va, Vb, Vc, Vd in the circuit below.
Solution:
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EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW
Example: Find Vbc, Vbe, Vda, Vab, R1 in the circuit below.
Solution:
Example: In the circuit below, determine Va, I, and voltage drop across each resistor.
Solution:
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