EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW Notes/EE301... · EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW 1 Learning Objectives a. Identify elements that are connected

EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW

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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, 1R and 2R 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 2R .

Solution:


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Resistors in Series Most complicated circuits can be simplified. For example, two resistors in series

can be replaced by an equivalent resistance RT.

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!!!

1 2eqR R R


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Example: Determine TOTR 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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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

TOTR is 10 and the total voltage is 5 V with the polarity shown.


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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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Kirchhoff’s Voltage Law (KVL) Kirchhoff’s voltage law (KVL) states that the algebraic sum of all

voltages around a closed loop is zero.

Mathematically, KVL implies

A closed loop is any path that:

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.

Another way of expressing KVL: Summation of voltage rises is

equal to the summation of voltage drops around a closed loop.

Example: Determine the voltage 1V in the circuit below using KVL.

Solution:

Example: Determine the voltage xV in the circuit below using

KVL.

Solution:

1

0M

m

m

v

1 1 2 2 3 0E v v E v

1 2 1 2 3E E v v v


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Example: Determine the unknown voltages 1V and 3V 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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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:


a. Redraw the circuit with a single voltage source and single resistor.

b. Determine the current.

Solution:


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Example: Your roommate has hooked up the circuit shown below.

a. Is the voltage reading across 2R correct?

b. Is the voltage reading across 3R 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:

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.

x

x

Total

RV E

R

22

1 2 3 4

600 120

2 600 3 1

119

RV E

R R R R

V

V


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Example: Determine the voltage from a to b in the circuit

shown below.

Solution:

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.

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 abV and bcV in the circuit shown.

Solution:

= 75 Ω

= 25 Ω

ab a bV V V


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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

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:

Example: Determine the voltage Vab in the circuit below.

Solution:

= 30 Ω

= 50 Ω

= 20 Ω

v2= 60𝑉

v2= 60𝑉


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Example: Determine the voltage labeled V shown in the

circuit below.

Solution:

Example: Determine the value of 3V in the circuit shown below. Note that the voltmeter reads +5.6 V.

Solution:

Example: Determine the voltage abV in the circuit below.

Solution:


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Example: Determine the voltage aV 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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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:

EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW Notes/EE301... · EE301 - SERIES CIRCUITS, KIRCHHOFF’S VOLTAGE LAW 1 Learning Objectives a. Identify elements that are connected

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