## Introduction: Electronics for Absolute Beginners, Chapter 3

Welcome to Chapter 3!

In this chapter we will go through:

1. Series Circuits

2. Voltage Divider Rule

3. Applied Voltage

4. Kirchoff's Voltage Law

5. Power in Series Circuit

6. Effects of Open in Series Circuit

7. Effects of Close in Series Circuit

8. Designing a Series Circuit

In this chapter we will go through:

1. Series Circuits

2. Voltage Divider Rule

3. Applied Voltage

4. Kirchoff's Voltage Law

5. Power in Series Circuit

6. Effects of Open in Series Circuit

7. Effects of Close in Series Circuit

8. Designing a Series Circuit

## Step 1: Series Circuit

A series circuit is a a circuit with just one path for current to flow.

In a series circuit:

1. Resistance is additive

2. Current is constant

3. Voltage drops are additive

Click on the images attached to this lecture to see what a series circuit looks like.

In a series circuit:

1. Resistance is additive

2. Current is constant

3. Voltage drops are additive

Click on the images attached to this lecture to see what a series circuit looks like.

## Step 2: Series Circuit 2

So if we have a circuit with 3 resistors, we want to add them to get the total resistance (RT).

Exercise 1:

R1 = 10 Ω

R2 = 50 Ω

R3 = 60 Ω

Answer:

RT = R1+R2+R3

RT = 10+50+60

RT = 120 Ω

Exercise 1:

R1 = 10 Ω

R2 = 50 Ω

R3 = 60 Ω

Answer:

RT = R1+R2+R3

RT = 10+50+60

RT = 120 Ω

## Step 3: Series Circuit 3

Another way of finding total resistance RT, is by using Ohm's Law. That is by dividing total voltage by total current.

RT = VT / IT

Exercise 2: A circuit has a total voltage VT of 5v and a total current IT of 2A. Find the total resistance RT.

Answer:

RT = VT / IT

RT = 5 / 2

RT = 2.5 Ω

Because current is constant or the same throughout the circuit, all we have to do is find one value of current and we can find the current for every part of the circuit.

RT = VT / IT

Exercise 2: A circuit has a total voltage VT of 5v and a total current IT of 2A. Find the total resistance RT.

Answer:

RT = VT / IT

RT = 5 / 2

RT = 2.5 Ω

Because current is constant or the same throughout the circuit, all we have to do is find one value of current and we can find the current for every part of the circuit.

## Step 4: Series Circuit 4

Exercise 3: A circuit has 5V, three resistors connected in series with a value of 10 Ω, 5 Ω, and 1 Ω respectively. Find the current passing through resistor one (IR1), the current passing through resistor two (IR2), the current passing through resistor three (IR3), and the total current.

Answer:

IT = VT / RT

IT = 5 / 16

IT = .3125A or 312.5mA

IR1 = 312.5mA

IR2 = 312.5mA

IR3 = 312.5mA

Answer:

IT = VT / RT

IT = 5 / 16

IT = .3125A or 312.5mA

IR1 = 312.5mA

IR2 = 312.5mA

IR3 = 312.5mA

## Step 5: Series Circuit 5

We mentioned that in a series circuit, voltage drops are additive. A voltage drop is a voltage that is dropped across a resistor. When the resistor has a higher resistance, the voltage drop across that resistor is also higher. When the resistor has a smaller resistance, the voltage drop across that resistor is also smaller.

The voltage drop of across a resistor (VR) can be calculated by multiplying the resistor by the total voltage (VT) and dividing the result by the total resistance (RT).

VR = (R x VT) / RT

Exercise 4: A circuit with a total voltage of 12V and three resistors 5k Ω, 10k Ω, 1k Ω respectively. Find the voltage drop across the first resistor (VR1), the voltage drop across the second resistor (VR2), and the voltage drop across the third resistor (VR3).

RT = R1+R2+R3

RT = 5000 + 10000 + 1000

RT = 16000

RT = 16k Ω

VR1 = (R1 x VT) / RT

VR1 = (5000 x 12) / 16000

VR1 = 60000 / 16000

VR1 = 3.75V

VR2 = (R2 x VT) / RT

VR2 = (10000 x 12) / 16000

VR2 = 120000 / 16000

VR2 = 7.5V

VR3 = (R3 x VT) / RT

VR3 = (1000 x 12) / 16000

VR3 = 12000 / 16000

VR3 = .75 or 750mV

Most importantly, when we add all of the voltage drops together, we can find the total voltage VT.

VT = VR1+VR2+VR3

VT = 3.75 + 7.5 + .75

VT = 12V

The voltage drop of across a resistor (VR) can be calculated by multiplying the resistor by the total voltage (VT) and dividing the result by the total resistance (RT).

VR = (R x VT) / RT

Exercise 4: A circuit with a total voltage of 12V and three resistors 5k Ω, 10k Ω, 1k Ω respectively. Find the voltage drop across the first resistor (VR1), the voltage drop across the second resistor (VR2), and the voltage drop across the third resistor (VR3).

RT = R1+R2+R3

RT = 5000 + 10000 + 1000

RT = 16000

RT = 16k Ω

VR1 = (R1 x VT) / RT

VR1 = (5000 x 12) / 16000

VR1 = 60000 / 16000

VR1 = 3.75V

VR2 = (R2 x VT) / RT

VR2 = (10000 x 12) / 16000

VR2 = 120000 / 16000

VR2 = 7.5V

VR3 = (R3 x VT) / RT

VR3 = (1000 x 12) / 16000

VR3 = 12000 / 16000

VR3 = .75 or 750mV

Most importantly, when we add all of the voltage drops together, we can find the total voltage VT.

VT = VR1+VR2+VR3

VT = 3.75 + 7.5 + .75

VT = 12V

## Step 6: Voltage Divider Rule

Voltage drop further explained.

Look at voltage drop as a reduction, and the voltage is reduced/dropped by the resistance of the resistor. In the same analogy as the amount of water allowed to go through is reduced depending on how much resistance is present.

So by considering how much total voltage is going through a circuit and by how much is it reduced, we can find out the voltage at a certain point of a circuit.

To find the voltage in certain point of circuit consisting of resistors, then we have to take into account how much is the voltage dropped by the particular resistor. For a point in a circuit after the particular resistor we must subtract the voltage drop of that resistor to the voltage total of the circuit;

V = VT - VR

Look at voltage drop as a reduction, and the voltage is reduced/dropped by the resistance of the resistor. In the same analogy as the amount of water allowed to go through is reduced depending on how much resistance is present.

So by considering how much total voltage is going through a circuit and by how much is it reduced, we can find out the voltage at a certain point of a circuit.

To find the voltage in certain point of circuit consisting of resistors, then we have to take into account how much is the voltage dropped by the particular resistor. For a point in a circuit after the particular resistor we must subtract the voltage drop of that resistor to the voltage total of the circuit;

V = VT - VR

## Step 7: Voltage Divider Rule 2

Exercise 1: A circuit has a voltage of 10V and two resistors, 20 Ω and 10 Ω respectively. Find the voltage on the point after the first resistor V1 and the voltage on the point after the second resistor V2.

V1 = (R2 / (R1 + R2)) x VT

V1 = (10 / (20 + 10)) x 10

V1 = (10 / 30) X 10

V1 = .33 x 10

V1 = 3.3V

V2 = (R1 / (R1 + R2)) x VT

V2 = (20 / (20 + 10)) x 10

V2 = (20 / 30) x 10

V2 = .67 x 10

V2 = 6.7V

This means that the point after R1, there is 3.3V available, and after the point in R2 6.7V is available. Also when we combine V1 and V2, we should get total voltage VT.

VT = V1 + V2

VT = 3.3 + 6.7

VT = 10V

V1 = (R2 / (R1 + R2)) x VT

V1 = (10 / (20 + 10)) x 10

V1 = (10 / 30) X 10

V1 = .33 x 10

V1 = 3.3V

V2 = (R1 / (R1 + R2)) x VT

V2 = (20 / (20 + 10)) x 10

V2 = (20 / 30) x 10

V2 = .67 x 10

V2 = 6.7V

This means that the point after R1, there is 3.3V available, and after the point in R2 6.7V is available. Also when we combine V1 and V2, we should get total voltage VT.

VT = V1 + V2

VT = 3.3 + 6.7

VT = 10V

## Step 8: Kirchoff's Voltage Law

Kirchoff''s voltage law states

1. that the sum of all voltages around the circuit is equal to total voltage applied,

VT = V1 + V2 + .......

2. and the algebraic sum of all voltage in the circuit and the applied total voltage is zero.

Exercise 1: The voltage at point one V1 is 10V, the voltage at point two V2 is 5V, and the voltage at point three V3 is 20V. Find the total applied voltage VA. During Chapter 3, Voltage Total and Voltage Applied are used interchangeably, or as if they have the same meaning.

VA= V1+V2+V3

VA = 10 + 5 + 20

VA= 35V

Exercise 2. What is the algebraic sum of voltages in this circuit:

http://www.engineersblogsite.com/wp-content/uploads/2010/02/kvl.jpg

Let's assume

V1 = 5V

V2 = 3V

V3 = 2V

V4 = 5V

V5 = 5V

V6 = 10V

Answer: Moving from V1 through V6 on the circuit provided in the link, take on the sign that comes before the voltage as the voltage polarity.

-V1 + V2 + V3 - V4 - V5 + V6

= -5V + 3V + 2V -5V -5V +10V

= 0

1. that the sum of all voltages around the circuit is equal to total voltage applied,

VT = V1 + V2 + .......

2. and the algebraic sum of all voltage in the circuit and the applied total voltage is zero.

Exercise 1: The voltage at point one V1 is 10V, the voltage at point two V2 is 5V, and the voltage at point three V3 is 20V. Find the total applied voltage VA. During Chapter 3, Voltage Total and Voltage Applied are used interchangeably, or as if they have the same meaning.

VA= V1+V2+V3

VA = 10 + 5 + 20

VA= 35V

Exercise 2. What is the algebraic sum of voltages in this circuit:

http://www.engineersblogsite.com/wp-content/uploads/2010/02/kvl.jpg

Let's assume

V1 = 5V

V2 = 3V

V3 = 2V

V4 = 5V

V5 = 5V

V6 = 10V

Answer: Moving from V1 through V6 on the circuit provided in the link, take on the sign that comes before the voltage as the voltage polarity.

-V1 + V2 + V3 - V4 - V5 + V6

= -5V + 3V + 2V -5V -5V +10V

= 0

## Step 9: Power in Series Circuit

In this lecture you must familiarize yourself with the Power Formula in conjunction with Ohm's law.

P = V x I

P = I^2 R

P = V^2 / R

In order to find the amount of power dissipated by each component in a circuit we must use the power formula.

Exercis1: A circuit has 200Vdc , 2A current , and three resistors connected in series with the voltage source. The resistors are 40Ω, 50Ω, 10Ω respectively. How much power is dissipated by each resistor?

R1 = 40Ω

P = I^2 R

P = 2^2 x 40

P = 4 x 40

P = 160W

R2 = 50Ω

P = I^2 R

P = 2^2 x 50

P = 4 x 50

P = 200W

R3 = 10Ω

P = I^2 R

P = 2^2 x 10

P = 4 x 10

P = 40W

So in this circuit, 200W is dissipated through R2 which 50Ω, 160W is dissipated into R1 which is 40Ω, and 40W is dissipated into R3 which 10Ω. Notice that the higher the value of the resistor, the higher the power they can dissipate.

The total power PT can be calculated by adding the individual power dissipations.

PT = PR1 + PR2 + PR3

PT = 160 + 200 + 40

PT = 400W

To determine what percentage of power is dissipated in each component we must do the following:

PR1 = 160W

PT = 400W

160 / 400 = .4

The answer is .4 which mean that 40% of the total power is dissipated across resistor 1.

PR2 = 200W

PT = 400W

200 / 400 = .5

The answer is .5, which means that 50% of the total power is dissipated across resistor 2.

PR3 = 40W

PT = 400W

40 / 400 = .1

The answer is .1, which means that 10% of the total power is dissipated across resistor 3.

P = V x I

P = I^2 R

P = V^2 / R

In order to find the amount of power dissipated by each component in a circuit we must use the power formula.

Exercis1: A circuit has 200Vdc , 2A current , and three resistors connected in series with the voltage source. The resistors are 40Ω, 50Ω, 10Ω respectively. How much power is dissipated by each resistor?

R1 = 40Ω

P = I^2 R

P = 2^2 x 40

P = 4 x 40

P = 160W

R2 = 50Ω

P = I^2 R

P = 2^2 x 50

P = 4 x 50

P = 200W

R3 = 10Ω

P = I^2 R

P = 2^2 x 10

P = 4 x 10

P = 40W

So in this circuit, 200W is dissipated through R2 which 50Ω, 160W is dissipated into R1 which is 40Ω, and 40W is dissipated into R3 which 10Ω. Notice that the higher the value of the resistor, the higher the power they can dissipate.

The total power PT can be calculated by adding the individual power dissipations.

PT = PR1 + PR2 + PR3

PT = 160 + 200 + 40

PT = 400W

To determine what percentage of power is dissipated in each component we must do the following:

PR1 = 160W

PT = 400W

160 / 400 = .4

The answer is .4 which mean that 40% of the total power is dissipated across resistor 1.

PR2 = 200W

PT = 400W

200 / 400 = .5

The answer is .5, which means that 50% of the total power is dissipated across resistor 2.

PR3 = 40W

PT = 400W

40 / 400 = .1

The answer is .1, which means that 10% of the total power is dissipated across resistor 3.

## Step 10: Effects of a Short in Series Circuits

A short circuit is when an undesired low resistance is in or around a given circuit.

A short circuit can be created, when a piece of metal is dropped across a resistor. Remember, that current travels the path of least resistance. So instead of the current traveling across the resistor, it travels across the metal. When there is less resistance current becomes higher generating more heat in the circuit.

An undesired short in a circuit can cause damages to other components, and may cause resistor to blow out or the wire to burn out, among many things.

Circuit parameters without the short.

RT = 6kΩ...............R1 = 2kΩ................R2 = 4kΩ

VT = 10V................VR1 = 3.33V...........VR2 = 6.67V

IT = 1.67mA..........IR1 = 1.67mA.........IR2 = 1.67mA

PT = 16.7mW.......PR1 = 5.58mW......PR2 = 11.16mW

Let us find what happens to the values of this circuit when a short is placed on R1.

VT = 10V

RT = 4kΩ...............R1 = 0.......................R2 = 4kΩ

VT = 10V............... VR1 = 0.....................VR2 = 10V

IT = 2.5mA............ IR1 = 0......................IR2 = 2.5mA

PT = 25mW ..........PR1 = 0.....................PR2 = 25mW

Since there is just one resistor through which current flows, voltage will not be divided.

Since R1 is shorted there will be no voltage drop across it (VR1 = 0).

Since R2 is the only resistor not bypassed by current, it will maintain a 5V (VR2 = 5V)

PR1 = VR1 x IR1

PR1 = 0 X 0

PR1 = 0

PR2 = VR2 x IR2

PR2 = 10V x 2.5mA

PR2 = 25mW

PRT = PR1 + PR2

PRT = 0 + 25 mW

PRT = 25 mW

So in summary, this is what happened to the circuit when R1 was shorted with a piece of metal wire.

1. Since current travels the least path or resistance, it travelled across the metal instead of R1, making resistance across R1 = 0.

2. Since total resistance is lower, more current can pass through the circuit, making the current value higher.

3. R1 = 0, makes the voltage dropped across R1 also becomes 0.

4. The total voltage remains the same.

5. Power is higher since current is higher.

A short circuit can be created, when a piece of metal is dropped across a resistor. Remember, that current travels the path of least resistance. So instead of the current traveling across the resistor, it travels across the metal. When there is less resistance current becomes higher generating more heat in the circuit.

An undesired short in a circuit can cause damages to other components, and may cause resistor to blow out or the wire to burn out, among many things.

Circuit parameters without the short.

RT = 6kΩ...............R1 = 2kΩ................R2 = 4kΩ

VT = 10V................VR1 = 3.33V...........VR2 = 6.67V

IT = 1.67mA..........IR1 = 1.67mA.........IR2 = 1.67mA

PT = 16.7mW.......PR1 = 5.58mW......PR2 = 11.16mW

Let us find what happens to the values of this circuit when a short is placed on R1.

VT = 10V

RT = 4kΩ...............R1 = 0.......................R2 = 4kΩ

VT = 10V............... VR1 = 0.....................VR2 = 10V

IT = 2.5mA............ IR1 = 0......................IR2 = 2.5mA

PT = 25mW ..........PR1 = 0.....................PR2 = 25mW

Since there is just one resistor through which current flows, voltage will not be divided.

Since R1 is shorted there will be no voltage drop across it (VR1 = 0).

Since R2 is the only resistor not bypassed by current, it will maintain a 5V (VR2 = 5V)

PR1 = VR1 x IR1

PR1 = 0 X 0

PR1 = 0

PR2 = VR2 x IR2

PR2 = 10V x 2.5mA

PR2 = 25mW

PRT = PR1 + PR2

PRT = 0 + 25 mW

PRT = 25 mW

So in summary, this is what happened to the circuit when R1 was shorted with a piece of metal wire.

1. Since current travels the least path or resistance, it travelled across the metal instead of R1, making resistance across R1 = 0.

2. Since total resistance is lower, more current can pass through the circuit, making the current value higher.

3. R1 = 0, makes the voltage dropped across R1 also becomes 0.

4. The total voltage remains the same.

5. Power is higher since current is higher.

## Step 11: Effects of Open in Series Circuits

An open circuit is when there is a break in the path of current.

In a series circuit there is only one path for current to travel. If there is a break anywhere in the circuit then the circuit will have no current. Mainly because when current cannot flow, then the components in the circuit will not receive current. And when there is no current, the circuit will not work.

Circuit without a break in current path:

VA = 10V

VT = 10V..........................VR1 = 3.34V..............VR2 = 6.68V

RT = 6kΩ..........................R1 = 2kΩ...................R2 = 4kΩ

IT = 1.67mA.....................IR1 = 1.67mA...........IR2 = 1.67mA

PRT = 16.7mW............ .PR1 = 5.58mW........PR2 = 11.16mW

Circuit with a break in current path:

VA = 10V

VT = 0.................VR1 = 0...................VR2 = 0

RT = 6kΩ............R1 = 2KΩ................R2 = 4kΩ

IT = 0...................IR1 = 0.....................IR2 = 0

PRT = 0..............PR1 = 0...................PR2 = 0

In a series circuit there is only one path for current to travel. If there is a break anywhere in the circuit then the circuit will have no current. Mainly because when current cannot flow, then the components in the circuit will not receive current. And when there is no current, the circuit will not work.

Circuit without a break in current path:

VA = 10V

VT = 10V..........................VR1 = 3.34V..............VR2 = 6.68V

RT = 6kΩ..........................R1 = 2kΩ...................R2 = 4kΩ

IT = 1.67mA.....................IR1 = 1.67mA...........IR2 = 1.67mA

PRT = 16.7mW............ .PR1 = 5.58mW........PR2 = 11.16mW

Circuit with a break in current path:

VA = 10V

VT = 0.................VR1 = 0...................VR2 = 0

RT = 6kΩ............R1 = 2KΩ................R2 = 4kΩ

IT = 0...................IR1 = 0.....................IR2 = 0

PRT = 0..............PR1 = 0...................PR2 = 0

## Step 12: Designing a Series Circuit

As electronics technicians, we may be asked to construct a circuit to a certain specification. For this lecture we will learn how to design a series circuit with certain specifications. We will use Ohm's Law to aid us in doing so.

Exercise 1: Design a series circuit with three resistors, where two of the resistors are 4kΩ. The total circuit current is 1mA, and the applied voltage is 12V.

First we must list the known values;

VT = 12V ...............R1= 4kΩ.................VR1=.......................IR1 = ....................PR1=

IT = 1mA..................R2= 4kΩ.................VR2=.......................IR2 = ....................PR2=

RT =........................ R3= .........................VR3=...................... IR3=...................... PR3=

PT =

So far we only have the values of VT, IT, R1, and R2. By using our knowledge of Ohm's Law and the Power Formulwe can find the unknown values.

Exercise 1: Design a series circuit with three resistors, where two of the resistors are 4kΩ. The total circuit current is 1mA, and the applied voltage is 12V.

First we must list the known values;

VT = 12V ...............R1= 4kΩ.................VR1=.......................IR1 = ....................PR1=

IT = 1mA..................R2= 4kΩ.................VR2=.......................IR2 = ....................PR2=

RT =........................ R3= .........................VR3=...................... IR3=...................... PR3=

PT =

So far we only have the values of VT, IT, R1, and R2. By using our knowledge of Ohm's Law and the Power Formulwe can find the unknown values.

## Step 13: Design a Series Circuit 2

Since we know that in a series circuit, current is the same throughout the circuit, we now know the values of currents across Resistor 1, Resistor 2 and Resistor 3 (IR1, IR2, IR3), and that should be 1mA. So we incorporate this information below to reflect our finding.

VT = 12V ...............R1= 4kΩ.................VR1= .......................IR1 = 1mA.....................PR1=

IT = 1mA............... R2= 4kΩ.................VR2= ........................IR2 = 1mA.....................PR2=

RT =....................... R3= .........................VR3=...................... .IR3= 1mA.....................PR3=

PT =

VT = 12V ...............R1= 4kΩ.................VR1= .......................IR1 = 1mA.....................PR1=

IT = 1mA............... R2= 4kΩ.................VR2= ........................IR2 = 1mA.....................PR2=

RT =....................... R3= .........................VR3=...................... .IR3= 1mA.....................PR3=

PT =

## Step 14: Designing a Series Circuit 3

Since we know the current across R1 and R2, and we know the values of R1 and R2, we can find the voltage drops (VR1, VR2).

VR1 = IR1 x R1

VR1 = 1mA x 4kΩ

VR1 = 4V

VR2 = IR2 x R2

VR2 = 1mA x 4kΩ

VR2 = 4V

Ok, so now that we know VR1 and VR2, it will be easy to find the voltage drop across R3, since Total Voltage is all the voltage drops added together.

VT = VR1+VR2+VR3

12V = 4V+4V+VR3

12V = 8V+VR3

To isolate VR3 we must subtract 8V on both sides.

12V - 8V = 8V+VR3-8V

4V = VR3

VR3 = 4V

Then we can also find the Total Voltage (VT)

VT = VR1+VR2+VR3

VT = 4V+4V+4V

VT = 12V

VR1 = IR1 x R1

VR1 = 1mA x 4kΩ

VR1 = 4V

VR2 = IR2 x R2

VR2 = 1mA x 4kΩ

VR2 = 4V

Ok, so now that we know VR1 and VR2, it will be easy to find the voltage drop across R3, since Total Voltage is all the voltage drops added together.

VT = VR1+VR2+VR3

12V = 4V+4V+VR3

12V = 8V+VR3

To isolate VR3 we must subtract 8V on both sides.

12V - 8V = 8V+VR3-8V

4V = VR3

VR3 = 4V

Then we can also find the Total Voltage (VT)

VT = VR1+VR2+VR3

VT = 4V+4V+4V

VT = 12V

## Step 15: Design a Series Circuit 4

Since we now have VR3 and IR3, we can find the value of R3.

R3 = VR3 / IR3

R3 = 4V / 1mA

R3 = 4000Ω

R3 = 4kΩ

Since we know the values of all the resistors, we can now find total resistance (RT).

RT = R1+R2+R3

RT = 4kΩ + 4kΩ + 4kΩ

RT = 12kΩ

Let us reflect this information;

VT = 12V ...............R1= 4kΩ.................VR1= 4V.......................IR1 = 1mA.....................PR1=

IT = 1mA............... R2= 4kΩ.................VR2= 4V ......................IR2 = 1mA.....................PR2=

RT = 12kΩ............ R3= 4kΩ ................ VR3= 4V......................IR3= 1mA.....................PR3=

PT =

R3 = VR3 / IR3

R3 = 4V / 1mA

R3 = 4000Ω

R3 = 4kΩ

Since we know the values of all the resistors, we can now find total resistance (RT).

RT = R1+R2+R3

RT = 4kΩ + 4kΩ + 4kΩ

RT = 12kΩ

Let us reflect this information;

VT = 12V ...............R1= 4kΩ.................VR1= 4V.......................IR1 = 1mA.....................PR1=

IT = 1mA............... R2= 4kΩ.................VR2= 4V ......................IR2 = 1mA.....................PR2=

RT = 12kΩ............ R3= 4kΩ ................ VR3= 4V......................IR3= 1mA.....................PR3=

PT =

## Step 16: Designing a Series Circuit 5

According to the Power Formula, P = V x I. Let us solve.

P = VT x IT

P = 12V x 1mA

P = .012

P = 12mW

PR1 = VR1 x IR1

PR1 = 4V x 1mA

PR1 = .004

PR1 = 4mW

PR2 = VR2 x IR2

PR2 = 4V x 1mA

PR2 = .004

PR2 = 4mW

PR3 = VR3 x IR3

PR3 = 4V x 1mA

PR3 = .004

PR3 = 4mW

PT = PR1+PR2+PR3

PT = 4mW+4mW+4mW

PT = 12mW

Let us reflect this information;

VT = 12V ...............R1= 4kΩ.................VR1= 4V.......................IR1 = 1mA.....................PR1= 4mW

IT = 1mA............... R2= 4kΩ.................VR2= 4V ......................IR2 = 1mA.....................PR2= 4mW

RT = 12kΩ............ R3= 4kΩ ................ VR3= 4V......................IR3= 1mA.....................PR3= 4mW

PT = 12mW

Congratulations! you just designed your first circuit. Notice that in this exercise, our task was to basically find the value of the third resistor (R3) while keeping specifications given to us.

P = VT x IT

P = 12V x 1mA

P = .012

P = 12mW

PR1 = VR1 x IR1

PR1 = 4V x 1mA

PR1 = .004

PR1 = 4mW

PR2 = VR2 x IR2

PR2 = 4V x 1mA

PR2 = .004

PR2 = 4mW

PR3 = VR3 x IR3

PR3 = 4V x 1mA

PR3 = .004

PR3 = 4mW

PT = PR1+PR2+PR3

PT = 4mW+4mW+4mW

PT = 12mW

Let us reflect this information;

VT = 12V ...............R1= 4kΩ.................VR1= 4V.......................IR1 = 1mA.....................PR1= 4mW

IT = 1mA............... R2= 4kΩ.................VR2= 4V ......................IR2 = 1mA.....................PR2= 4mW

RT = 12kΩ............ R3= 4kΩ ................ VR3= 4V......................IR3= 1mA.....................PR3= 4mW

PT = 12mW

Congratulations! you just designed your first circuit. Notice that in this exercise, our task was to basically find the value of the third resistor (R3) while keeping specifications given to us.

## Step 17: Voltage Sources in Series

Voltage sources can be connected together in order to achieve a higher or lower Total Voltage (VT).

To achieve a higher total voltage, we use the Series Aiding technique.

To achieve a lower total voltage, we use the Series Opposing technique.

To achieve a higher total voltage, we use the Series Aiding technique.

To achieve a lower total voltage, we use the Series Opposing technique.

## Step 18: Series Aiding

To achieve a higher total voltage, we use the Series Aiding technique.

A Series Aiding technique can be achieved by

1. Connect the negative terminal of one voltage source to the positive terminal of the other voltage source.

AND/OR

2. Connect the positive terminal of one voltage source to the negative terminal of the other voltage source.

To determine the Total Voltage of a Series Aiding source, add all the voltages.

A Series Aiding technique can be achieved by

1. Connect the negative terminal of one voltage source to the positive terminal of the other voltage source.

AND/OR

2. Connect the positive terminal of one voltage source to the negative terminal of the other voltage source.

To determine the Total Voltage of a Series Aiding source, add all the voltages.

## Step 19: Series Opposing

To achieve a lower total voltage, we use the Series Opposing technique.

A Series Opposing technique can be achieved by

1. Connect the positive terminal of one voltage source to the positive terminal of the other voltage source.

AND/OR

2. Connect the negative terminal of one voltage source to the negative terminal of the other voltage source.

To determine the Total Voltage of a Series Opposing source, add all the smaller values and subtract it from the largest value.

A Series Opposing technique can be achieved by

1. Connect the positive terminal of one voltage source to the positive terminal of the other voltage source.

AND/OR

2. Connect the negative terminal of one voltage source to the negative terminal of the other voltage source.

To determine the Total Voltage of a Series Opposing source, add all the smaller values and subtract it from the largest value.

## Step 20: Series Opposing and Series Aiding in a Circuit

Click on the Image provided.

## Step 21: Chapter 3 Recap

Important things to remember:

1. A series circuit is a a circuit with just one path for current to flow.

In a series circuit:

a. Resistance is additive

b. Current is constant

c. Voltage drops are additive

2. An open circuit is when there is a break in the path of current.

3. A short circuit is when an undesired low resistance is in or around a given circuit.

4. To achieve a higher total voltage, we use the Series Aiding technique.

5. To achieve a lower total voltage, we use the Series Opposing technique.

6. VT = IT x RT

7. RT = VT / IT

8. IT = VT /RT

9. VT = VR1+VR2+...

10. RT = R1+R2+R3

11. VR1 = IR1 x R1

12. VR2 = IR2 x R2

13. VR3 = IR3 x R3

14. PT = VT x IT

15. PT = PR1+PR2+...

16. PR1 = VR1 x IR1

17. PR2 = VR2 x IR2

18. PR3 = VR3 x IR3

1. A series circuit is a a circuit with just one path for current to flow.

In a series circuit:

a. Resistance is additive

b. Current is constant

c. Voltage drops are additive

2. An open circuit is when there is a break in the path of current.

3. A short circuit is when an undesired low resistance is in or around a given circuit.

4. To achieve a higher total voltage, we use the Series Aiding technique.

5. To achieve a lower total voltage, we use the Series Opposing technique.

6. VT = IT x RT

7. RT = VT / IT

8. IT = VT /RT

9. VT = VR1+VR2+...

10. RT = R1+R2+R3

11. VR1 = IR1 x R1

12. VR2 = IR2 x R2

13. VR3 = IR3 x R3

14. PT = VT x IT

15. PT = PR1+PR2+...

16. PR1 = VR1 x IR1

17. PR2 = VR2 x IR2

18. PR3 = VR3 x IR3

## Step 22: Divide and Conquer

In troubleshooting, the divide and conquer technique saves the technician time when looking for a bad component.

In using the divide and conquer technique, the technician places one lead of the multi-meter on the first half of the circuit, and the other lead on the second half of the circuit. When the technician determines which half is defective, then he focuses on that half and divides it into two. He then places one lead of the multi-meter on the first half, and the other lead on the second half. The technician repeats this process until he isolates the component that is causing the problem.

Scenario 1: While conducting your routine tests, you noticed the TOTAL CURRENT has INCREASED. Normally, when resistance decreases, current increases. Current can increase when a circuit is shorted. So because of this, we will aim for finding where the short is located.

Scenario 2: While conducting your routine tests, you notice that the circuit is not working, and the Total Current is Zero while the TOTAL RESISTANCE is INFINITE. When the total current is ZERO, this means that there is no current flow and there is a break in current path (open circuit). A circuit with blown out resistors or broken wires can cause zero current to flow. So, we want to begin troubleshooting by finding where the break on the current path is.

In using the divide and conquer technique, the technician places one lead of the multi-meter on the first half of the circuit, and the other lead on the second half of the circuit. When the technician determines which half is defective, then he focuses on that half and divides it into two. He then places one lead of the multi-meter on the first half, and the other lead on the second half. The technician repeats this process until he isolates the component that is causing the problem.

Scenario 1: While conducting your routine tests, you noticed the TOTAL CURRENT has INCREASED. Normally, when resistance decreases, current increases. Current can increase when a circuit is shorted. So because of this, we will aim for finding where the short is located.

Scenario 2: While conducting your routine tests, you notice that the circuit is not working, and the Total Current is Zero while the TOTAL RESISTANCE is INFINITE. When the total current is ZERO, this means that there is no current flow and there is a break in current path (open circuit). A circuit with blown out resistors or broken wires can cause zero current to flow. So, we want to begin troubleshooting by finding where the break on the current path is.

## Step 23: Biasing Diodes

A diode is a device that allows current to pass on end and blocks it on the other end.

A simple example of a diode is the LED (Light emitting Diode).

A diode can be forward or reverse bias.

In a circuit where DC is the voltage source, a diode can only conduct when it's forward biased. This is because DC only travels one way, from positive to negative.

In a circuit where AC is the voltage source, we can do the following:

We can FORWARD BIAS a diode - if we want it to conduct with a positive voltage.

We can REVERSE BIAS a diode - if we want it to conduct with a negative voltage.

This is because in an AC voltage the polarity alternates from positive to negative and vice versa. So the current can travel two directions the polarities change.

A simple example of a diode is the LED (Light emitting Diode).

A diode can be forward or reverse bias.

In a circuit where DC is the voltage source, a diode can only conduct when it's forward biased. This is because DC only travels one way, from positive to negative.

In a circuit where AC is the voltage source, we can do the following:

We can FORWARD BIAS a diode - if we want it to conduct with a positive voltage.

We can REVERSE BIAS a diode - if we want it to conduct with a negative voltage.

This is because in an AC voltage the polarity alternates from positive to negative and vice versa. So the current can travel two directions the polarities change.