# In-02-02 Series _ Parallel Circuit

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```							Abu Dhabi National Oil Co. ADNOC Technical Institute

INSTRUMENTATION
INDUSTRIAL ELECTRONICS I

UNIT 2 SERIES AND PARALLEL CIRCUIT
INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09 PAGE 1 OF 19 IN-02-02

UNITS IN THIS COURSE
UNIT 1 UNIT 2 UNIT 3 UNIT 4 UNIT 5 UNIT 6 UNIT 7 UNIT 8 THE ELECTRICAL CIRCUIT SERIES AND PARALLEL CIRCUITS ELECTROMAGNETIC PRINCIPLES BASIC ELECTROSTATICS AND THE CAPACITOR THE INDUCTOR, CAPACITOR AND D.C. A.C. PRINCIPLES COMMON ELECTRICAL SYMBOLS PRACTICAL TASKS

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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Paragraph 2.0 COURSE OBJECTIVES 2.1 INTRODUCTION 2.2 THE SERIES CIRCUIT 2.3 SERIES CIRCUIT EXAMPLES 2.3.1 Example 1 2.3.2 Example 2 2.3.3 Example 3 2.4 THE PARALLEL CIRCUIT 2.5 PARALLEL CIRCUIT EXAMPLES 2.5.1 Example 1 2.5.2 Example 2 2.6 SERIES AND PARALLEL COMBINATION CIRCUITS 2.6.1 Example 1 2.6.2 Example 2 2.7 MEASUREMENT PROBLEMS 2.8 WHEATSTONE BRIDGE 2.8.1 The Bridge Circuit 2.9 SUMMARY OF IMPORTANT FORMULAS Page 4 5 5 6 6 7 8 9 10 10 11 12 12 14 16 18 18 19

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 2.0 COURSE OBJECTIVES The student will be able to:    State the formula for resistors in series. State the formula for resistors in parallel. Carry out simple calculations on series, parallel and series-parallel combination circuits.  Sketch a typical Wheatstone bridge and calculate the value of the unknown resistor.

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 2.1 INTRODUCTION The aim of this unit is to explain the three common circuits used in instrument electrical work. The series circuit, the parallel circuit and the Wheatstone bridge. 2.2 THE SERIES CIRCUIT

V1

V2

V3

V4

S
R1 SUPPLY VOLTAGE VS R2 R3 R4

SUPPLY CURRENT S

Figure 2-1 The Series Circuit Figure 2-1 shows an example of the basic series circuit with four loads (resistors) connected to a supply. The rules for this circuit are:a) Kirchhoff’s second law applies. The supply voltage equals the sum of the voltages around the circuit.

Vs = V1 + V2 + V3 +V4
b) The supply current is the same in all parts c) The TOTAL RESISTANCE of the circuit is the SUM OF THE INDIVIDUAL RESISTORS around the circuit.

R TOTAL = R1 + R2 + R3 + R4

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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NOTE:

The series circuit is not used much in industry because it has serious disadvantages.  If extra loads are added to the circuit the voltage across each load will fall.  If one load is broken (open circuit) the supply is cut off to all the loads on the circuit.

However the series circuit is used for the basic instrument electrical signal circuit (loop). This circuit will be shown during the instrument course.

2.3 SERIES CIRCUIT EXAMPLES The notes above give the rules for a series circuit. The example circuit has four loads. The following worked examples show the steps required to find unknown quantities in series circuits. 2.3.1 Example 1 Question Two 100 loads are connected in series to a 100V supply. Find a) b) Solution Circuit Diagram
S SUPPLY VOLTAGE 100V

The total circuit resistance The supply current.

100

100

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE RT = 100 + 100 = 200 Total Resistance Of Circuit = 200 From Ohm’s Law the supply current: S =

1 V S 100V = = A RT 2 200

Supply current = 0.5 AMP 2.3.2 Example 2
S

50

10

40

VS = 100V

Using the circuit diagram given above. Find :a) b) Solution: Total circuit resistance = 50 + 10 + 40 = 100 From Ohm’s Law S = The supply current (S) The power of the 40 load.

VS RT 100V 100
1A

=

=

Supply Current = 1A

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ADNOC TECHNICAL INSTITUTE The power of the 40 load is given by the formula P=V As the supply current in a series circuit is the same in all parts. P =  (R) = 2 R P = 1A x 1A x 40 P = 40W The power of the 40 load = 40W 2.3.3 Example 3 Question: 30 lights are connected in series across a 210V supply. Find the voltage across each bulb and the supply current if the resistance for each bulb = 2.93 Solution: The voltage across each bulb will be

210V = 7 VOLTS 30
The total resistance of the string of lights will be 30 x 2.93 = 87.9 The supply current will be

210V = 2.39 A 87.9

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 2.4 THE PARALLEL CIRCUIT
S 1 SUPPLY VOLTAGE (VS) 2 3

R1

R2

R3

Figure 2-2 The Parallel Circuit Figure 2-2 shows as an example the arrangement for three parallel electrical loads. The rules of this circuit are:a) In this case Kirchhoff’s first law applies, “Currents into a point on an electrical circuit equal the currents leaving that point “ so that

S = 1 + 2 + 3
The supply current increases as more loads are added in parallel. b) The supply voltage will be the same for each load. c) The total resistance of the circuit (RT) is given by the equation.

1 1 1 1 = + + R1 RT R2 R3
Note: - The parallel circuit is the standard industrial circuit because   The supply voltage is the same for each load. If one load is disconnected (open circuited) the other loads will still work.

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE 2.5 PARALLEL CIRCUIT EXAMPLES The notes above give the rules for a parallel circuit. The example circuit has three loads. The following worked examples show the steps required to find unknown quantities in a parallel circuit. 2.5.1 Example 1 Question: Two 100 loads are connected in parallel to a 100V supply. Find (a) (b) Solution: Circuit Diagram. The total circuit resistance. The supply current.

S

1 100 

2 100 

VS = 100V

(a) The total circuit resistance is found using the formula given.

1 1 1 = + RT 100 100 1 = 0.01 + 0.01 = 0.02 RT
RT =

1 = 50 0.02

Total circuit resistance = 50

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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b)

Supply current =

V S 100 V = = 2A RT 50 

Supply current = 2A 2.5.2 Example 2 Question
S =1A VS 20

10

40

An electrical circuit consists of three loads connected in parallel across a supply as shown in the circuit diagram above. Find :a) b) Solution: In a parallel circuit the supply voltage is the same for each load so with 1A flowing through the 10 resistor. VS = 1A x 10 = 10V Supply voltage = 10 V The current through the 20 resistor will be 20 = The supply voltage (VS) The supply current (S)

10V = 0.5A 20

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ADNOC TECHNICAL INSTITUTE The current through the 40 resistor will be 40 = from Kirchhoff’s Law

10V = 0.25A 40

S = 10 + 20 + 40
= 1A + 0.5A + 0.25A= 1.75A Supply current = 1.75A

2.6 SERIES AND PARALLEL COMBINATION CIRCUITS It can be difficult to calculate unknown quantities in a series and parallel combination circuit. answers. 2.6.1 Example 1 Question :
S 10

The following examples show ways of finding the

VS = 100V

20

30

Using the circuit diagram above calculate a) b) c) Total circuit resistance Supply current (S) The voltage across each resistor.

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE Solution: a) Work out the total resistance of the parallel part first.

20

30

1 1 = + RT 30

1 = 0.05 + 0.0333 = 0.0833 RT
RT =
b)

1 = 12.00 0.0833

Redraw the circuit using the parallel result.
10

100V

12

Total series resistance = 10 + 12 =22 Total circuit resistance = 22

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ADNOC TECHNICAL INSTITUTE c) Supply current

S =

V S 100V = = 4.55A RT 22

Supply current = 4.55A d) Voltage across the 10 resistor. V10 = S x 10 = 4.55A x 10 V10 = 45.5 V e) Voltage across 20 and 30 resistors will be the supply current times the parallel total resistance

V20 or V30 = S x 12
= 4.55A x 12

V20 or V30 = 54.6 V
Note:- To check your answer, add V20 to V10. The result should be VSUPPLY within the accuracy of the calculator. 2.6.2 Example 2
S = 2 A

5 VS 8

4

10

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ADNOC TECHNICAL INSTITUTE Using the above diagram calculate the supply voltage if S = 2A Solution: a) Work out the total resistance of each branch first.

5

4

1 1 1 = + 4 RT 5 1 = 0.2 + 0.25 = 0.45 RT

RT =

1 = 2.22 0.45

8

10

1 1 1 = + 8 RT 10
1 = 0.125 + 0.1 = 0.225 RT

RT =
b) Draw the new circuit

1 = 4.44 0.225

2.22 VS 4.44

RT = 2.22 + 4.44 RT = 6.66

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ADNOC TECHNICAL INSTITUTE c) The supply voltage VS = S x RT = 2A x 6.66 = 13.32V Supply voltage = 13.32 V 2.7 MEASUREMENT PROBLEMS

A R1 SUPPLY R2 (LOAD) V

Figure 2-3 Using an Ammeter and Voltmeter Figure 2-3 shows a circuit with an ammeter (in series) and voltmeter (in parallel) measuring the circuit current and potential difference (PD) across the load. Another circuit can be drawn which includes the resistance of the meters (see Figure 2-4)
RA R1 SUPPLY R2 RV

Figure 2-4 Equivalent Circuit with Ammeter and Voltmeter Added

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ADNOC TECHNICAL INSTITUTE For true readings the value of RA must be very small and RV very big. Otherwise, the current and voltage readings are all wrong because the meters change circuit values. Normally the ammeter causes no problems as it has a very low resistance. However, the voltmeter can cause problems, especially the old type of moving coil meter. An example is given below.
10 k R1 10V 10 k R2 V 10 k (Voltmeter)

The diagram shows a voltmeter (resistance 10k ) used to measure the PD across R2. Current with voltmeter not connected. =

10 = 0.5 mA 20,000

Voltage across R2 with voltmeter not connected. V=
10 x 10,000 = 5V 20,000

Current with voltmeter connected =
10 10 = = 0.66 mA 10,000  5,000 15,000

Voltage across R2 with voltmeter connected V=
10 x 5,000 = 3.33 V 15,000

From the example it can be seen that using a voltmeter of low resistance changes the readings a lot. Most measurements in instrumentation must be carried out with a digital voltmeter (DVM), with a high input resistance (> 10M ). If the voltmeter does not have a high input resistance the measurements will not be correct.

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ADNOC TECHNICAL INSTITUTE 2.8 WHEATSTONE BRIDGE

The Wheatstone bridge is widely used in instrumentation to measure resistance accurately. It is also used to show changes in the resistance of sensors used to measure pressure, level, temperature, etc. The following notes explain the basic principle of the Wheatstone bridge. applications will be shown later in the training . 2.8.1 The Bridge Circuit
A I1 R1 SUPPLY VOLTAGE D I3 R2 R3 ACCURATE VARIABLE RESISTOR (DECADE BOX) G B I2 Rx

Practical

C

Figure 2-5 The Wheatstone Bridge The Wheatstone bridge circuit (see Figure 2-5) consists of two very accurate (standard) resistors(R1 & R2) called the ratio arms. There is an accurate variable resistor (Decade Box R3). A very sensitive ammeter which will detect very small currents (called a Galvanometer (G)) is connected across points D and B. A supply voltage is connected across points A and C. There are also two terminals to connect an unknown resistor Rx across the points A and B. The value of the unknown resistor is given by the equation.

RX =

R3R1  R2

INDUSTRIAL ELECTRONICS I - SERIES AND PARALLEL CIRCUIT DATE OF ISSUE 8-DEC-09

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ADNOC TECHNICAL INSTITUTE Theory: The value of the unknown resistor is found by adjusting the value of the variable resistor until the galvanometer reads zero. balanced position so that at balance when I3 = 0 This is called the

1 R 1 = 2 R x 1 R 2 = 2 R 3
R1 = R2
and

}

DIVIDE

Rx R3

RX =

R3 R1  R2

This is the balance equation for a Wheatstone bridge and must be remembered. 2.9 SUMMARY OF IMPORTANT FORMULAS Resistors in Series RT = R1 + R2 +R3 Resistors in Parallel 1 / RT = 1 / R1 + 1/ R2 + 1/ R3 Wheatstone Bridge RX = Ratio Arms x Decade Box Value. etc. etc.

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