Electronics Series Resistive Circuits 1 Copyright © Texas Education Agency, 2014. All rights reserved

Electronics

Series Resistive

Circuits

1

Copyright © Texas Education Agency, 2014. All rights reserved.

What is a Series Circuit?

A series circuit is one of the simplest electrical circuits.

Because of this, it is the type of circuit used for an introduction to problem solving and circuit analysis.

Problem solving is a technique developed through practice.

It uses math to calculate electrical values. This is called applied math (based on Ohm’s

Law). 2


Pre-Requisites

You have to know some basics: What an electrical circuit is Common electrical components and their schematic

symbols Resistors, batteries, ground Current, voltage, resistance Switches, fuses, wires

You should gain this knowledge by completing the prior electronics lessons in this sequence

3


An Example Circuit

4


An Example Circuit

5

Wires


An Example Circuit

6

Battery(DC voltagesource)


An Example Circuit

7

Fuse


An Example Circuit

8

Switch


An Example Circuit

9

Load(uses a resistorsymbol)


Typical Circuit Labels

10

VS

F1 S1

R1


Open switch, no current Resistance is infinite Voltage is dropped across the switch

Circuit Operation


Circuit Operation

Closed switch, current flows Current flows from negative to positive Amount of current determined by Ohm’s Law


Back to Circuit Analysis

Circuit analysis requires use of some fundamental electrical laws

Ohm’s Law You should know this law by now There are three forms of Ohm’s Law

Kirchhoff’s Law There are two parts to Kirchhoff’s Law You need to know both

13


Kirchhoff’s Laws

Voltage law: the sum of all voltages in a closed loop is equal to zero The sum of the voltage “drops” equals the sum of

the voltage “sources” All of the voltage is always used in a loop

Current law: the sum of the currents “into” a node is equal to the sum of the currents “leaving” the node The current into a conductor is the same as the

current out of the conductorCopyright © Texas Education Agency, 2014. All rights reserved.

Where do the laws apply?

These rules always apply to every DC circuit with a resistive load

For AC circuits and active loads, the rules generally apply, but not always

AC circuits have changing voltage and current Active loads do not have constant resistance An active load is like a variable resistor A load is any device that the circuit is designed

to deliver power to


16

The Simplest Circuit

This is a trivial example of a series circuit

VS R1


A Series Circuit

A series circuit has only one path for current flow There are no branches going to another circuit

For every point in the circuit, the current “in” equals the current “out” Kirchhoff’s current law

This means, in a series circuit, the current has the same value everywhere Current is constant everywhere


Using Kirchhoff’s Voltage Law

To do Kirchhoff’s Law right, you must have correct polarities

Make current loops going from the negative side of the battery to the positive side

Current goes from negative to positive everywhere outside the battery

Use arrows to indicate the direction of current flow

The arrows point from negative to positive


19

Using Arrows

Use arrows to indicate the direction of current flow and the polarity of voltage

The red line represents a closed loop and shows the path for current flow

+VS R1


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Use the polarity that the arrow points to The top arrow points to the positive side of

the battery

The bottom arrow points to the negative side of the resistor

+VS R1


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Kirchhoff’s voltage law states that the sum of the voltages in a closed loop equals zero

The voltage used equals the supply voltage This is actually a lot easier to use than it looks


- = 0or

=


Calculate Current

The only voltage in this circuit is the supply voltage, and it is dropped across R1

R1 is the only resistor so its value is the value of total resistance

Use Ohm’s Law to calculate current:

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= = = VS

R1


Here is a slightly more complicated circuit:

Apply Kirchhoff’s Law by drawing a current loop with arrows

Adding Circuit Components

23

VS

R3

R2

R1


Draw the Current Loop

Place arrows according to the direction of current flow

Current flows from the negative terminal of the battery to the positive terminal

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VS

R3

R2

R1


Place Polarity on Components

Current flows from negative to positive outside the battery

25

VS

R3

R2

R1+

+


Start with the arrow above the battery

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Write Kirchhoff’s Voltage Law

VS

R3

R2

R1+

+

+ VS – VR3 – VR2 – VR1 = 0


Rearrange the Formula

Each resistor drops some of the source voltage; the three resistors together drop all of the source voltage.

27

VS

R3

R2

R1+

+

VS = VR1 + VR2 + VR3


Partial Summary

A partial summary of what we have learned so far: Current is the same everywhere in a series

circuit

Voltage drops add to equal the source voltage

We have formulas for voltage and current— now we need a formula for resistance

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= I1 = I2 = I3V = V1 + V2 + V3


Solve for Resistance

From Ohm’s Law, V = I R Substitute into the series voltage formula:

Current is the same everywhere so it divides out:

Resistance adds in a series circuit

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IRT = I1R1 + I2 R2+ I3R3RT = R1 + R2+ R3


Series Circuit Tool Kit

Here are the three equations for a series circuit:

These three equations, plus Ohm’s Law, form a “tool kit” to analyze series circuits

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= I1 = I2 = I3V = V1 + V2 + V3RT = R1 + R2+ R3


Example Problem One

Solve for each of the voltage drops in this circuit

In other words, solve for V1, V2, and V3

31

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Problem One Solution

First, write the equation(s) that solve the problem:

V1 = I1R1, V2 = I2R2, and V3 = I3R3

32



First, write the equation(s) that solve the problem:

V1 = I1R1, V2 = I2R2, and V3 = I3R3

Second, look for what is needed to solve those equations Sometimes the information needed is given Sometimes, like in this case, it is not

To solve for voltage, we need to know current

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Follow the Logic

Now we are looking to solve this equation:

Write down what we know in this formula We know VT = 12 V

Pay attention to the subscript, to use VT we need RT

Solving this will give us IT (which is also I1, etc.)

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Example Problem One

RT = R1 + R2 + R3

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VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Example Problem One

RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω

36

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Example Problem One

RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω

RT = 600 Ω

37

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Example Problem One

RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω

RT = 600 Ω

38

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Example Problem One

RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω

RT = 600 Ω

= 39

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω


Example Problem One

RT = R1 + R2 + R3 = 200 Ω + 200 Ω + 200 Ω

RT = 600 Ω

= = 0.02 A = 20 mA40

VS =12 V

R3= 200 Ω

R2= 200 Ω

R1 = 200 Ω



41

= I1 = I2 = I3 = 20 mAPlug these into the first set of equations:V1 = I1R1 = 20 mA • 200 Ω = 4 VV2 = I2R2 = 20 mA • 200 Ω = 4 VV3 = I3R3 = 20 mA • 200 Ω = 4 VIn a series circuit, equal resistances drop

equal amounts of voltage


How to Develop a Skill

This may seem hard, but the point is to have a systematic, step-by-step method

Problem solving is not magic; it is a skill developed by following a procedure A valuable skill

Problem solving and troubleshooting requires a logical, systematic, and consistent step-by- step process

Any skill also requires practice42


Circuit Example Two

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VS =12 V

R1 = 200 Ω

R2= 600 Ω

Solve for the voltage drops across R1 and R2


1. Write the equations for VR1 and VR2VR1 = I1 • R1 VR2 = I2 • R2

2. To use these equations to solve for voltage we need current. Write the equation for currentI = 3. Looking for known values in this equation, we have VT, we need RT RT = R1 + R2 = 200 Ω + 600 Ω = 800 Ω4. Substitute this into the formula from step two

44

IT = = 0.015 A = 15 mACopyright © Texas Education Agency, 2014. All rights reserved.

Problem Two Solution

IT = I1 = I2 = 15 mA5. Now solve for voltage drops from step oneVR1 = I1 • R1 = 15 mA • 200 Ω = 3 V VR2 = I2 • R2 = 15 mA • 600 Ω = 9 VNote that R2 has three times the resistance of R1 and that VR2 has three times the voltage of VR1. There is a rule for that; it’s called the voltage divider rule.

45


The Voltage Divider Rule

The ratio of the voltages equals the ratio of the resistances in a series circuit

This rule is true because the current is the same everywhere in a series circuit

This rule is typically expressed as a formula:

This formula applies to any ratio of voltage and resistance in a series circuit as long as the equivalent values are used properly

46

=


Example Three

Let’s work an example from the very beginning

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VS

R1

R2

R3

R4


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Draw the current loop

Place arrows negative to positive

VS

R1

R2

R3

R4

Example Three


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VS

R1

R2

R3

R4

+

++

+

+ VS – V4 – V3 – V2 – V1 = 0

VS = V1 + V2 + V3 + V4

Example Three


Example Three

VS = 9 V, R1 = 180 Ω, R2 = 330 Ω, R3 = 470 Ω, R4 = 150 Ω

Solve for each voltage drop

Go through the step-by-step process

50

VS

R1

R2

R3

R4


1. Write the equations we need:VR1 = I1 • R1 ; VR2 = I2 • R2 VR3 = I3 • R3 ; VR4 = I4 • R4

2. Write the equation for current:I = 3. We have VT, we need RT RT = R1 + R2 + R3 + R4 = 180 Ω + 330 Ω + 470 Ω + 150 Ω = 1130 Ω4. Substitute this into the formula from step two

51

IT = = 0.008 A = 8 mACopyright © Texas Education Agency, 2014. All rights reserved.

IT = I1 = I2 = I3 = I4 = 8 mA5. Now solve for voltage drops from step oneVR1 = I1 • R1 = 8 mA • 180 Ω = 1.4 V VR2 = I2 • R2 = 8 mA • 330 Ω = 2.64 VVR3 = I3 • R3 = 8 mA • 470 Ω = 3.76 V VR4 = I4 • R4 = 8 mA • 150 Ω = 1.2 V Note how the voltage drops are proportional to the resistance values. 52

Problem Three Solution


Example Four

Let’s try something a little different

Solve for VS

53

VS

R3= 2.5 kΩ

R2= 3.5 kΩI2 = 2.5 mA

R1 = 2 kΩ


1. Write the equation we need:VS = VT = IT • RT

2. Look for what is needed to solve the equation: We need IT

We need RT

3. Solve for IT

IT = I1 = I2 = I3 = 2.5 mA4. Solve for RTRT = R1 + R2 + R3= 2 kΩ + 3.5 kΩ + 2.5 kΩ = 8 kΩ

54


Problem Four Solution

5. Plug these into the equation from step one:

VT = IT • RT = 2.5 mA • 8 kΩ = 20 V Note: 1 mA = 0.001 A, 1 kΩ = 1000 Ω

55


Example Five

Now let’s try something a little harder

Solve for VS

56

VS

R2= 1.5 kΩI2 = 3 mA

R1 = 1.2 kΩ


V3 = 6.9 V

1. Write the equation we need:VS = VT = IT • RT

2. Look for what is needed to solve the equation: We need IT

We need RT

3. Solve for IT

IT = I1 = I2 = I3 = 3 mA4. Solve for RTRT = R1 + R2 + R3

We don’t know R3 so we need to solve for it57


Problem Five Solution

5. Write the formula for R3:R3 = = = 2.3 kΩ

6. Plug this into the equation from step four:RT = R1 + R2 + R3 = 1.2 kΩ + 1.5 kΩ + 2.3 kΩ = 5 kΩ

7. Plug this into the equation from step one:VT = IT • RT = 3 mA • 5 kΩ = 15 V 58


Summary

There is a logical, step-by-step process to solve a problem

Know the equations that form the tool kit for series circuit analysis

59

I =


What’s Next?

Practice Practice Practice

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Electronics Series Resistive Circuits 1 Copyright © Texas Education Agency, 2014. All rights reserved