ee1 chapter3 resistors_inseries

Apr 10, 2023

Lecturer Name [email protected]

Contact Number

IT2001PAEngineering Essentials (1/2)

Chapter 3 - Resistors in Series Circuits


IT2001PA Engineering Essentials (1/2)

Number of slides for today

33Including this useless slide

2


IT2001PA Engineering Essentials (1/2) 3

Lesson Objectives

Upon completion of this topic, you should be able to: Apply Ohm’s Law to calculate voltages, currents and

resistances in a series circuit.



Specific Objectives

1. State the characteristics of series-connected resistors.

2. Calculate the total resistance for series-connected resistors

3. Calculate the current flow, voltage drops across the various resistors for series-connected resistors.

4. Use the voltage divider rule to calculate the voltage drops across series-connected resistors.



Introduction

Two resistors in series

Resistors in series are connected end to end or in a string as shown.

R1 R2

R1 R2 R3

Three resistors in series

Connecting Line



Two Resistors Connected in Series

Apply Ohm’s Law,

V1 = I x R1

V2 = I x R2

VT = I x RT

Where RT is the total resistance of the circuit.

VT = V1 + V2

= I x R1 + I x R2

= I x(R1+ R2)

= I x RT

R1 R2

V2V1

I

VT

RTI

VT



Two Resistors Connected in Series

Total resistance of the circuit,

RT = (R1+ R2)

R1 R2

V2V1

I

VT

RTI

VT



Characteristics of a Series Circuit

The current (I) is the same in all parts of a series circuit.

R1 R2 R3

V2V1 V3

I

V

Three resistors are connected in series.




VT = V1 + V2 + V3

R1 R2 R3

VT

I

V2V1 V3

The voltage applied to the circuit (VT) is equal the sum of the voltages across each individual parts.




Applying Ohm’s LawV1 = I R1

V2 = I R2

V3 = I R3

R1 R2 R3

VT

I

V2V1 V3

Individual Voltage drop = current x Individual resistance




GivenI = 2 AR1 = 2 R2 = 4 R3 = 6

R1 R2 R3

VT

I

V2V1 V3

Example

V1 = I x R1 = 2 x 2 = 4 VV2 = I x R2 = 2 x 4 = 8 VV3 = I x R3 = 2 x 6 = 12 VVT = 4 + 8 + 12 = 24 V




VT = V1 + V2 + V3

I RT = I R1 + I R2 + I R3

I RT = I ( R1 + R2 + R3)

RT = ( R1 + R2 + R3)

RT = R1 + R2 + R3

R1 R2 R3

VT

I

V2V1 V3

Total resistance = sum of Individual resistance




GivenI = 2 AR1 = 2 R2 = 4 R3 = 6

R1 R2 R3

VT

I

V2V1 V3

Example

RT = R1 + R2 + R3

= 2 + 4 + 6 = 12




RT = R1 + R2 + R3

R1 R2 R3

VT

I

V2V1 V3

Total resistance is greater than the larger individual resistance.




GivenI = 2 AR1 = 2 R2 = 4 R3 = 6

R1 R2 R3

VT

I

V2V1 V3

Example

RT = R1 + R2 + R3

= 2 + 4 + 6 = 12

Larger Individual Resistance= R3 = 6

RT > R3

12 > 6



Characteristics of a Series Circuit (Summary)

1. The current through all the resistors is the same.

2. The voltage applied to the circuit = the sum of the voltages across the individual parts.

3. Individual voltage drop = current x individual resistances.4. Total resistance = sum of individual

resistances.5. Total resistance is greater than the largest individual resistance.



Example 3-1

RT = R1 + R2 + R3

= 1 + 2 + 3

= 6

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

Determine the total resistance of the circuit?

RT

10 V

Combine three resistors into one single resistor



Example 3-1

I = V / RT

= 10 / 6 = 1.667 A

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

RT = 1 + 2 + 3 = 6

Determine the current flow of the circuit?



Example 3-1

V1 = I R1

= 1.667 X 1

= 1.667 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

Determine the voltage across resistor R1 of

the circuit?



Example 3-1

V2 = I R2 = 1.667 X 2 = 3.334 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3


the circuit?



Example 3-1

V3 = I R3

= 1.667 X 3 = 5 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3


the circuit?



Example 3-1 (Summary)

V3 = I R3

= 1.667 X 3 = 5 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

RT = 1 + 2 + 3 = 6

I = V / RT

= 10 / 6 = 1.667 A

V1 = I R1 = 1.667 X 1

= 1.667 V

V2 = I R2 = 1.667 X 2 = 3.334 V




5 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

RT = 1 + 2 + 3 = 6

1.667 A1.667 V

3.334 V




5 V

R1= 1 R2= 2 R3= 3

V= 10 V

I

V2V1 V3

1.667 A

1.667 V

3.334 VVoltage can be measured from a voltmeter connected in parallel

V V V

Acurrent can be measured from a ammeter connected in series



Example 3-2

Three resistors of 40 , 60 and X respectively are connected in series. The combination is connected across the 50 V supply. If the voltage drop across the 40 resistor is 16 V, determine the current in the circuit and the unknown resistor X.

V1=16V

R1= 40 R2= 60 R3= X

V= 50 V

I

V2 V3



Example 3-2 (Solution)

I = V1 / R1

V1=16V

R1= 40 R2= 60 R3= X

V= 50 V

I

V2 V3

I = 16 / 40

V2 = I R2

I = 0.4 A

V2 = 0.4 x 60

V2 = 24 V

V3 = 50 – 16 – 24

V3 = 10 V

V3 = I R3

10 = 0.4 x R3

R3 = 10 / 0.4

R3 = 25



Voltage Divider

s

T

xx V

RR

V WhereRT is the total or equivalent series resistanceVx is the voltage across any resistor, Rx

R1 R2

V2V1

I

Vs

General Voltage divider Formula is



Voltage Dividers

T

xx V

RR

V R1 R2

V2V1

I

Vs

s

T

11 V

RR

V

Rs

T

22 V

R V

RT = R1 + R2



Voltage Divider s

T

xx V

RR

V

Vs

I

R1

R2

R3

V1

V2

V3

RT = R1 + R2 + R3

s

T

11 V

RR

V

Rs

T

22 V

R V

Rs

T

33 V

R V



Example 3-3

R1= 82 R2= 64

V1V2

Vs = 10V

I

Determine the voltage across R1 and R2

RT = 82+ 64=146

Or

V2 = 10 – 5.617

= 4.383 V

V1 = Vs x R1 / RT

= 10 x 82 / (146) = 5.617 VV2 = Vs x R2 / RT

= 10 x 64 / (146) = 4.383 V



Potentiometer as an Adjustable Voltage Divider

Potentiometer is a variable resistor with three terminals A potentiometer connected to a voltage source is

shown:



Summary

1. The current through all the resistors is the same.

2. The voltage applied to the circuit = the sum of the voltages across the individual parts.

3. Individual voltage drop = current x individual resistances.4. Total resistance = sum of individual

resistances.5. Total resistance is greater than the largest individual resistance.



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