bbasic electric circuitsasic electric circuits

142

Basic Electric Circuits Basic Electric Circuits

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Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

143

Unit 6 Series Circuits Series Circuits

Why You Need to Know

I t is important to learn the rules governing the values of resistance, voltage, current, and power in a series circuit. Knowledge of series

circuits is essential in understanding more complex circuits that are encountered throughout an electrician’s career. There are some series circuits in everyday use, such as many street lighting systems, but knowl-edge of series circuits is most useful in understanding how components in combination circuits relate to each other. Combination circuits contain components or branches that are connected in both series and parallel. To apply the rules of a series circuit, this unit

■ provides a basic understanding of how voltage drop impacts devices that are connected in series.

■ explains why each series-connected device increases the total opposition to current fl ow.

■ provides an understanding of how current fl ows through a series circuit.

OUTLINE 6–1 Series Circuits6–2 Voltage Drops in a Series Circuit6–3 Resistance in a Series Circuit6–4 Calculating Series Circuit Values6–5 Solving Circuits6–6 Voltage Dividers6–7 The General Voltage Divider

Formula6–8 Voltage Polarity6–9 Using Ground as a Reference

KEY TERMS Chassis groundCircuit breakersEarth groundFusesGround pointSeries circuitVoltage dropVoltage polarity

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144 SECTION II Basic Electric Circuits

ObjectivesAfter studying this unit, you should be able to

■ discuss the properties of series circuits.

■ list three rules for solving electrical values of series circuits.

■ calculate values of voltage, current, resistance, and power for series circuits.

■ calculate the values of voltage drop in a series circuit using the voltage divider formula.

Preview

E lectric circuits can be divided into three major types: series, parallel, and combination. Combination circuits are circuits that contain both series and

parallel paths. The fi rst type discussed is the series circuit. ■

6–1 Series CircuitsA series circuit is a circuit that has only one path for current fl ow (Figure 6–1). Because there is only one path for current fl ow, the current is the same at any point in the circuit. Imagine that an electron leaves the negative terminal of the battery. This electron must fl ow through each resistor before it can complete the circuit to the positive battery terminal.

One of the best examples of a series-connected device is a fuse or circuit breaker (Figure 6–2). Because fuses and circuit breakers are connected in series with the rest of the circuit, all the circuit current must fl ow through them. If the current becomes excessive, the fuse or circuit breaker will open and disconnect the rest of the circuit from the power source.

Resistor 1

Battery

Resistor 2

Resistor 5 Resistor 4

Resistor 3

+

--

+ -- + --

+--+--

+

--

FIGURE 6–1 A series circuit has only one path for current fl ow.

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Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

UNIT 6 Series Circuits 145

6–2 Voltage Drops in a Series CircuitVoltage is the force that pushes the electrons through a resistance. The amount of voltage required is determined by the amount of current fl ow and resistance. If a voltmeter is connected across a resistor (Figure 6–3), the amount of voltage necessary to push the current through that resistor is indicated by the meter. This amount is known as voltage drop. It is similar to pressure drop in a water system. In a series circuit, the sum of all the voltage drops across all the resistors

+

--

Battery

-- + -- +

--+--+

Generator

Fuse

FIGURE 6–2 All the current must fl ow through the fuse.

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FIGURE 6–3 The voltage drops in a series circuit must equal the applied voltage.

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must equal the voltage applied to the circuit. The amount of voltage drop across each resistor is proportional to its resistance and the circuit current.

In the circuit shown in Figure 6–4, four resistors are connected in series. It is assumed that all four resistors have the same value. The circuit is connected to a 24-volt battery. Because all the resistors have the same value, the voltage drop across each will be 6 volts (24 V/4 resistors 5 6 V). Note that all four resistors will have the same voltage drop only if they all have the same value. The circuit shown in Figure 6–5 illustrates a series circuit comprising resistors

FIGURE 6–4 The voltage drop across each resistor is proportional to its resistance.

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+

--

24 VDC

-- + -- +

--+--+

6 VDC 6 VDC

6 VDC 6 VDC

FIGURE 6–5 Series circuit with four resistors having different voltage drops.

+

--

-- + -- +

--+--+

5 VDC 7 VDC

4 VDC 8 VDC

24 VDC

8 16

10 14

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of different values. Notice that the voltage drop across each resistor is propor-tional to its resistance. Also notice that the sum of the voltage drops is equal to the applied voltage of 24 volts.

6–3 Resistance in a Series CircuitBecause only one path exists for the current to fl ow through a series circuit, it must fl ow through each resistor in the circuit (Figure 6–1). Each resistor limits or impedes the fl ow of current in the circuit. Therefore, the total amount of resis tance to current fl ow in a series circuit is equal to the sum of the resis-tances in that circuit.

6–4 Calculating Series Circuit ValuesThree rules can be used with Ohm’s law for fi nding values of voltage, current, resistance, and power in any series circuit:

1. The current is the same at any point in the circuit.

2. The total resistance is the sum of the individual resistors.

3. The applied voltage is equal to the sum of the voltage drops across all the resistors.

The circuit shown in Figure 6–6 shows the values of current fl ow, volt-age drop, and resistance for each of the resistors. The total resistance (RT) of

FIGURE 6–6 Series circuit values.

E1 40 V I1 2 A R1 20

===

+

E3 60 V I3 2 A R3 30

===

ET 120 V IT 2 A RT 60

===

E2 20 V I2 2 A R2 10

===

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the circuit can be found by adding the values of the three resistors–resistance adds:

RT 5 R1 1 R2 1 R3

RT 5 20 V 1 10 V 1 30 V

RT 5 60 V

The amount of current fl ow in the circuit can be found by using Ohm’s law:

I 5 E __ R

I 5 120 V ______ 60 V

I 5 2 A

A current of 2 amperes fl ows through each resistor in the circuit:

IT 5 I1 5 I2 5 I3

Because the amount of current fl owing through resistor R1 is known, the voltage drop across the resistor can be found using Ohm’s law:

E1 5 I1 3 R1

E1 5 2 A 3 20 V

E1 5 40 V

In other words, it takes 40 volts to push 2 amperes of current through 20 ohms of resistance. If a voltmeter were connected across resistor R1, it would indicate a value of 40 volts (Figure 6–7). The voltage drop across resistors R2

and R3 can be found in the same way:

E2 5 I2 3 R2

E2 5 2 A 3 10 V

E2 5 20 V

E3 5 I3 3 R3

E3 5 2 A 3 30 V

E3 5 60 V

If the voltage drop across all the resistors is added, it equals the total applied voltage (ET):

ET 5 E1 1 E2 1 E3

ET 5 40 V 1 20 V 1 60 V

ET 5 120 V

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6–5 Solving CircuitsIn the following problems, circuits that have missing values are shown. The missing values can be found by using the rules for series circuits and Ohm’s law.

FIGURE 6–7 The voltmeter indicates a voltage drop of 40 volts.

E1 40 V I1 2 A R1 20

===

+

E3 60 V I3 2 A R3 30

===

ET 120 V IT 2 A RT 60

==

E2 20 V I2 2 A R2 10

===

40 V +

=

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■ EXAMPLE 6–1

The fi rst step in fi nding the missing values in the circuit shown in Figure 6–8 is to fi nd the total resistance (RT). This can be done using the second rule of series cir-cuits, which states that resistances add to equal the total resistance of the circuit:

RT 5 R1 1 R2 1 R3 1 R4

RT 5 100 V 1 250 V 1 150 V 1 300 V

RT 5 800 V

Now that the total voltage and the total resistance are known, the current fl ow through the circuit can be found using Ohm’s law:

I 5 E __ R

I 5 40 V ______ 800 V

I 5 0.050 A

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The fi rst rule of series circuits states that current remains the same at any point in the circuit. Therefore, 0.050 A fl ows through each resistor in the circuit (Figure 6–9). The voltage drop across each resistor can now be found using Ohm’s law (Figure 6–10):

E1 5 I1 3 R1

E1 5 0.050 A 3 100 V

E1 5 5 V

E2 5 I2 3 R2

E2 5 0.050 A 3 250 V

E2 5 12.5 V

E3 5 I3 3 R3

E3 5 0.050 A 3 150 V

E3 5 7.5 V

E4 5 I4 3 R4

E4 5 0.050 A 3 300 V

E4 5 15 V

FIGURE 6–8 Series circuit, Example 1.

E1 I1 R1 100

P1

====

E2 I2 R2 250

P2

====ET 40 V

IT RT

PT

===

+

E4 I4 R4 300

P4

====

E3 I3 R3 150

P3

====

=

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FIGURE 6–9 The current is the same at any point in a series circuit.

E1 I1 0.05 A R1 100

P1

====

E2 I2 0.05 A R2 250

P2

====ET 40 V

IT 0.05 A RT 800

PT

===

+

E4 I4 0.05 A R4 300

P4

====

E3 I3 0.05 A R3 150

P3

====

=

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FIGURE 6–10 The voltage drop across each resistor can be found using Ohm’s law.

E1 5 V I1 0.05 A R1 100

P1

====

E2 12.5 V I2 0.05 A R2 250

P2

====ET 40 V

IT 0.05 A RT 800

PT

===

+

E4 15 V I4 0.05 A R4 300

P4

====

E3 7.5 V I3 0.05 A R3 150

P3

====

=

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Several formulas can be used to determine the amount of power dissipated (converted into heat) by each resistor. The power dissipation of resistor R1 will be found using the formula

P1 5 E1 3 I1 P1 5 5 V 3 0.05 A

P1 5 0.25 W

The amount of power dissipation for resistor R2 will be calculated using the formula

P2 5 E2

2

___ R2

P2 5 156.25 V2 _________

250 V

P2 5 0.625 W

The amount of power dissipation for resistor R3 will be calculated using the formula

P3 5 I32 3 R3

P3 5 0.0025 A2 3 150 V

P3 5 0.375 W

The amount of power dissipation for resistor R4 will be found using the formula

P4 5 E4 3 I4 P4 5 15 V 3 0.05 A

P4 5 0.75 W

A good rule to remember when calculating values of electric circuits is that the total power used in a circuit is equal to the sum of the power used by all parts. That is, the total power can be found in any kind of a circuit—series, parallel, or combination—by adding the power dissipation of all the parts. The total power for this circuit can be found using the formula

PT 5 P1 1 P2 1 P3 1 P4

PT 5 0.25 W 1 0.625 W 1 0.375 W 1 0.75 W

PT 5 2 W

39154_06_ch06_p142-171.indd 152 11/8/10 8:54 AM



Now that all the missing values have been found (Figure 6–11), the circuit can be checked by using the third rule of series circuits, which states that volt-age drops add to equal the applied voltage:

ET 5 E1 1 E2 1 E3 1 E4

ET 5 5 V 1 12.5 V 1 7.5 V 1 15 V

ET 5 40 V

FIGURE 6–11 The fi nal values for the circuit in Example 1.

E1 5 V I1 0.05 A R1 100

P1 0.25 W

====

E2 12.5 V I2 0.05 A R2 250

P2 0.625 W

====ET 40 V

IT 0.05 A RT 800

PT 2 W

===

+

E4 15 V I4 0.05 A R4 300

P4 0.75 W

====

E3 7.5 V I3 0.05 A R3 150

P3 0.375 W

====

=

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■ EXAMPLE 6–2

The second circuit to be solved is shown in Figure 6–12. In this circuit, the total resistance is known, but the value of resistor R2 is not. The second rule of series circuits states that resistances add to equal the total resistance of the circuit. Because the total resistance is known, the missing resistance of R2 can be found by adding the values of the other resistors and subtracting their sum from the total resistance of the circuit (Figure 6–13):

R2 5 RT 2 (R1 1 R3 1 R4 )

R2 5 6000 V 2 (1000 V 1 2000 V 1 1200 V)

R2 5 6000 V 2 4200 V

R2 5 1800 V

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The amount of current fl ow in the circuit can be found using Ohm’s law:

I 5 E __ R

I 5 120 V _______ 6000 V

I 5 0.020 A


E1 I1 R1 1000

P1

====

E2 I2 R2

P2

====ET 120 V

IT RT 6000

PT

===

+

E4 I4 R4 1200

P4

====

E3 I3 R3 2000

P3

====

=

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E1 I1 R1 1000

P1

====

E2 I2 R2 1800

P2

====ET 120 V

IT RT 6000

PT

===

+

E4 I4 R4 1200

P4

====

E3 I3 R3 2000

P3

====

=

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FIGURE 6–13 The missing resistor value.

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Because the amount of current fl ow is the same through all elements of a series circuit (Figure 6–14), the voltage drop across each resistor can be found using Ohm’s law (Figure 6–15):

E1 5 I1 3 R1

E1 5 0.020 A 3 1000 V

E1 5 20 V

E2 5 I2 3 R2

E2 5 0.020 A 3 1800 V

E2 5 36 V

E3 5 I3 3 R3

E3 5 0.020 A 3 2000 V

E3 5 40 V

E4 5 I4 3 R4

E4 5 0.020 A 3 1200 V

E4 5 24 V

The third rule of series circuits can be used to check the answers:

ET 5 E1 1 E2 1 E3 1 E4

ET 5 20 V 1 36 V 1 40 V 1 24 V

ET 5 120 V

E1 I1 0.020 A R1 1000

P1

====

E2 I2 0.020 A R2 1800

P2

====ET 120 V

IT 0.020 A RT 6000

PT

===

+

E4 I4 0.020 A R4 1200

P4

====

E3 I3 0.020 A R3 2000

P3

====

=

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FIGURE 6–14 The current is the same through each circuit element.

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FIGURE 6–15 Voltage drops across each resistor.

E1 20 V I1 0.020 AR1 1000

P1

====

E2 36 V I2 0.020 AR2 1800

P2

====ET 120 V

IT 0.020 A RT 6000

PT

===

+

E4 24 V I4 0.020 A R4 1200

P4

====

E3 40 V I3 0.020 AR3 2000

P3

====

=

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The amount of power dissipation for each resistor in the circuit can be calcu-lated using the same method used to solve the circuit in Example 1. The power dissipated by resistor R1 is calculated using the formula

P1 5 E1 3 I1P1 5 20 V 3 0.02 A

P1 5 0.4 W

The amount of power dissipation for resistor R2 is found by using the formula

P2 5 E2

2

___ R2

P2 5 1296 V2 ________

1800 V

P2 5 0.72 W

The power dissipation of resistor R3 is found using the formula

P3 5 I32 3 R3

P3 5 0.0004 A2 3 2000 V

P3 5 0.8 W

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The power dissipation of resistor R4 is calculated using the formula

P4 5 E4 3 I4P4 5 24 V 3 0.02 A

P4 5 0.48 W

The total power is calculated using the formula

PT 5 ET 3 IT PT 5 120 V 3 0.02 A

PT 5 2.4 W

The circuit with all calculated values is shown in Figure 6–16.

FIGURE 6–16 The remaining unknown values for the circuit in Example 2.

E1 20 V I1 0.020 AR1 1000

P1 0.4 W

====

E2 36 V I2 0.020 AR2 1800

P2 0.72 W

====ET 120 V

IT 0.020 ART 6000

PT 2.4 W

===

+

E4 24 V I4 0.020 AR4 1200

P4 0.48 W

====

E3 40 V I3 0.020 AR3 2000

P3 0.8 W

====

=

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■ EXAMPLE 6–3

In the circuit shown in Figure 6–17, resistor R1 has a voltage drop of 6.4 V, resis-tor R2 has a power dissipation of 0.102 W, resistor R3 has a power dissipation of 0.154 W, resistor R4 has a power dissipation of 0.307 W, and the total power consumed by the circuit is 0.768 W.

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The only value that can be found with the given quantities is the amount of power dissipated by resistor R1. Because the total power is known and the power dissipated by the three other resistors is known, the power dissipated by resistor R1 can be found by subtracting the power dissipated by resistors R2, R3, and R4 from the total power used in the circuit:

P1 5 PT 2 (P2 1 P3 1 P4 )

or

P1 5 PT 2 P2 2 P3 2 P4

P1 5 0.768 W 2 0.102 W 2 0.154 W 2 0.307 W

P1 5 0.205 W

Now that the amount of power dissipated by resistor R1 and the voltage drop across R1 are known, the current fl ow through resistor R1 can be found using the formula

I 5 P __ E

I 5 0.205 W ________ 6.4 V

I 5 0.032 A

Because the current in a series circuit must be the same at any point in the circuit, it must be the same through all circuit components (Figure 6–18).


E1 6.4 V I1 R1

P1

====

E2 I2 R2

P2 0.102 W

====ET

IT RT

PT 0.768 W

===

+

E4 I4 R4

P4 0.307 W

====

E3 I3 R3

P3 0.154 W

====

=

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Now that the power dissipation of each resistor and the amount of current fl owing through each resistor are known, the voltage drop of each resistor can be calculated (Figure 6–19):

E2 5 P2 ___ I2

E2 5 0.102 W ________ 0.032 A

E2 5 3.188 V

E3 5 P3 ___ I3

E3 5 0.154 W ________ 0.032 A

E3 5 4.813 V

E4 5 P4 ___ I4

E4 5 0.307 W ________ 0.032 A

E4 5 9.594 V

FIGURE 6–18 The current fl ow in the circuit in Example 3.

E1 6.4 V I1 0.032 AR1

P1 0.205 W

====

E2 I2 0.032 AR2

P2 0.102 W

====ET

IT 0.032 A RT

PT 0.768 W

===

+

E4 I4 0.032 A R4

P4 0.307 W

====

E3 I3 0.032 AR3

P3 0.154 W

====

=

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Ohm’s law can now be used to fi nd the ohmic value of each resistor in the circuit (Figure 6–20):

R1 5 E1 ___ I1

R1 5 6.4 V ________ 0.032 A

R1 5 200 V

R2 5 E2 ___ I2

R2 5 3.188 V ________ 0.032 A

R2 5 99.625 V

R3 5 E3 ___ I3

R3 5 4.813 V ________ 0.032 A

R3 5 150.406 V

R4 5 E4 ___ I4

R4 5 9.594 V ________ 0.032 A

R4 5 299.813 V

FIGURE 6–19 Voltage drops across each resistor.

E1 6.4 V I1 R1

P1

====

E2 I2 R2

P2 0.102 W

====ET

IT RT

PT 0.768 W

===

+

E4 I4 R4

P4 0.307 W

====

E3 I3 R3

P3 0.154 W

====

=

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The voltage applied to the circuit can be found by adding the voltage drops across the resistor (Figure 6–21):

ET 5 E1 1 E2 1 E3 1 E4

ET 5 6.4 V 1 3.188 V 1 4.813 V 1 9.594 V

ET 5 23.995 V

FIGURE 6–20 The ohmic value of each resistor.

E1 6.4 V I1 0.032 AR1 200

P1 0.205 W

====

E2 3.188 V I2 0.032 AR2 99.625

P2 0.102 W

====ET

IT 0.032 ART

PT 0.768 W

===

+

E4 9.594 V I4 0.032 AR4 299.813

P4 0.307 W

====

E3 4.813 V I3 0.032 AR3 150.406

P3 0.154 W

====

=

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FIGURE 6–21 The applied voltage and the total resistance.

E1 6.4 V I1 0.032 AR1 200

P1 0.205 W

====

E2 3.2 V I2 0.032 AR2 100

P2 0.102 W

====ET 24 V

IT 0.032 ART 750

PT 0.768 W

===

+

E4 9.6 V I4 0.032 AR4 300

P4 0.307 W

====

E3 4.8 V I3 0.032 AR3 150

P3 0.154 W

====

=

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The total resistance of the circuit can be found in a similar manner (Figure 6–21). The total resistance is equal to the sum of all the resistive elements in the circuit:

RT 5 R1 1 R2 1 R3 1 R4

RT 5 200 V 1 99.625 V 1 150.406 V 1 299.813 V

RT 5 749.844 V

If Ohm’s law is used to determine total voltage and total resistance, slightly different answers are produced:

ET 5 PT ___ IT

ET 5 0.768 W ________ 0.032 A

ET 5 24 V

RT 5 ET ___ IT

RT 5 24 V ________ 0.032 A

RT 5 750 V

The slight difference in answers is caused by the rounding off of values. Al-though there is a small difference between the answers, they are within 1% of each other. This small difference has very little effect on the operation of the circuit, and most electric measuring instruments cannot measure values this accurately anyway.

6–6 Voltage DividersOne common use for series circuits is the construction of voltage dividers. A voltage divider works on the principle that the sum of the voltage drops across a series circuit must equal the applied voltage. Voltage dividers are used to provide different voltages between certain points (Figure 6–22). If a voltmeter is connected between Point A and Point B, a voltage of 20 volts will be seen. If the voltmeter is connected between Point B and Point D, a voltage of 80 volts will be seen.

Voltage dividers can be constructed to produce any voltage desired. For example, assume that a voltage divider is connected to a source of 120 volts

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and is to provide voltage drops of 36 volts, 18 volts, and 66 volts. Notice that the sum of the voltage drops equals the applied voltage. The next step is to decide how much current is to fl ow through the circuit. Because there is only one path for current fl ow, the current will be the same through all the resistors. In this circuit, a current fl ow of 15 milliamperes (0.015 A) will be used. The resistance value of each resistor can now be determined:

R 5 E __ I

R1 5 36 V ________ 0.015 A

R1 5 2.4 kV (2400 V)

R2 5 18 V ________ 0.015 A

R2 5 1.2 kV (1200 V)

R3 5 66 V ________ 0.015 A

R3 5 4.4 kV (4400 V)

FIGURE 6–22 Series circuit used as a voltage divider.

–

+

200 V

20 V

30 V

50 V

100 V

A

B

C

D

E

–

+

–

+

–

+

–

+

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6–7 The General Voltage Divider FormulaAnother method of determining the voltage drop across series elements is to use the general voltage divider formula. Because the current fl ow through a series circuit is the same at all points in the circuit, the voltage drop across any particular resistance is equal to the total circuit current times the value of that resistor:

EX 5 IT 3 RX

The total circuit current is proportional to the source voltage (ET) and the total resistance of the circuit:

IT 5 ET ___ RT

If the value of IT is substituted for ET/RT in the previous formula, the expression now becomes:

EX 5 ( ET ___ RT

) RX

If the formula is rearranged, it becomes what is known as the general voltage divider formula:

EX 5 ( RX ___ RT

) ET

The voltage drop across any series component (EX) can be calculated by substituting the value of RX for the resistance value of that component when the source voltage and total resistance are known.

■ EXAMPLE 6–4

Three resistors are connected in series to a 24-V source. Resistor R1 has a resistance of 200 V, resistor R2 has a value of 300 V, and resistor R3 has a value of 160 V. What is the voltage drop across each resistor?

Solution Find the total resistance of the circuit:

RT 5 R1 1 R2 1 R3

RT 5 200 V 1 300 V 1 160 V

RT 5 660 V

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6–8 Voltage PolarityIt is often necessary to know the polarity of the voltage dropped across a resistor. Voltage polarity can be determined by observing the direction of current fl ow through the circuit. In the circuit shown in Figure 6–22, it will be assumed that the current fl ows from the negative terminal of the battery to the positive terminal. Point A is connected to the negative battery terminal, and Point E is connected to the positive terminal. If a voltmeter is connected across Terminals A and B, Terminal B will be positive with respect to A. If a voltmeter is connected across Terminals B and C, however, Terminal B will be negative with respect to Terminal C. Notice that Terminal B is closer to the negative terminal of the battery than Terminal C is. Consequently, electrons fl ow through the resistor in a direction that makes Terminal B more negative than C. Terminal C would be negative with respect to Terminal D for the same reason.

Now use the voltage divider formula to calculate the voltage drop across each resistor:

E1 5 ( R1 ___ RT

) ET

E1 5 ( 200 V ______ 660 V

) 24 V

E1 5 7.273 V

E2 5 ( R2 ___ RT

) ET

E2 5 ( 300 V ______ 660 V

) 24 V

E2 5 10.909 V

E3 5 ( R3 ___ RT

) ET

E3 5 ( 160 V ______ 660 V

) 24 V

E3 5 5.818 V

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6–9 Using Ground as a ReferenceTwo symbols are used to represent ground (Figure 6–23). The symbol shown in Figure 6–23(A) is an earth ground symbol. It symbolizes a ground point that is made by physically driving an object such as a rod or a pipe into the ground. The symbol shown in Figure 6–23(B) symbolizes a chassis ground. This is a point that is used as a common connection for other parts of a circuit, but it is not actually driven into the ground. Although the symbol shown in Figure 6–23(B) is the accepted symbol for a chassis ground, the symbol shown in Figure 6–23(A) is often used to represent a chassis ground also.

An excellent example of using a chassis ground as a common connection can be found in the electric system of an automobile. The negative terminal of the battery is grounded to the frame or chassis of the vehicle. The frame of the automobile is not connected directly to earth ground; it is insulated from the ground by rubber tires. In the case of an automobile electric system, the chassis of the vehicle is the negative side of the circuit. An electric circuit using ground as a common connection point is shown in Figure 6–24. This circuit is an electronic burglar alarm. Notice the numerous ground points in the sche-matic. In practice, when the circuit is connected, all the ground points will be connected together.

In voltage divider circuits, ground is often used to provide a common refer-ence point to produce voltages that are above and below ground (Figure 6–25). An above-ground voltage is a voltage that is positive with respect to ground. A below-ground voltage is negative with respect to ground. In Figure 6–25, one terminal of a zero-center voltmeter is connected to ground. If the probe is connected to Point A, the pointer of the voltmeter gives a negative indication for voltage. If the probe is connected to Point B, the pointer indicates a posi-tive voltage.

FIGURE 6–23 Ground symbols.

(A)Earth ground

(B)Chassis ground

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Summary ■ Series circuits have only one path for current fl ow.

■ The individual voltage drops in a series circuit can be added to equal the applied voltage.

■ The current is the same at any point in a series circuit.

■ The individual resistors can be added to equal the total resistance of the circuit.

FIGURE 6–24 Burglar alarm with battery backup.

12 VDC

Arm switch

Key lock switch

LED

(Power)

120 VAC

POWER SW

Fuse1 Amp 1N4004

1N4004

LED(Arm)

1N4004

(Relay)

2.2 K

1 mF

12 VDCAlarm

1N4004

1000 mF50 VDC

2N1598

24 VCT

K1

K1

Door andwindow switches

F1

S1T1

LED

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+

24 V

12 V

12 V

A

B

+--

0

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FIGURE 6–25 A common ground used to produce above- and below-ground voltage.

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■ Fuses and circuit breakers are connected in series with the devices they are intended to protect.

■ The total power in any circuit is equal to the sum of the power dissipated by all parts of the circuit.

■ When the source voltage and the total resistance are known, the voltage drop across each element can be calculated using the general voltage divider formula.

Review Questions 1. A series circuit has individual resistor values of 200 V, 86 V, 91 V, 180 V,

and 150 V. What is the total resistance of the circuit?

2. A series circuit contains four resistors. The total resistance of the circuit is 360 V. Three of the resistors have values of 56 V, 110 V, and 75 V. What is the value of the fourth resistor?

3. A series circuit contains fi ve resistors. The total voltage applied to the circuit is 120 V. Four resistors have voltage drops of 35 V, 28 V, 22 V, and 15 V. What is the voltage drop of the fi fth resistor?

4. A circuit has three resistors connected in series. Resistor R2 has a re-sistance of 220 V and a voltage drop of 44 V. What is the current fl ow through resistor R3?

5. A circuit has four resistors connected in series. If each resistor has a volt-age drop of 60 V, what is the voltage applied to the circuit?

6. Defi ne a series circuit.

7. State the three rules for series circuits.

8. A series circuit has resistance values of 160 V, 100 V, 82 V, and 120 V. What is the total resistance of this circuit?

9. If a voltage of 24 V is applied to the circuit in Question 8, what will be the total amount of current fl ow in the circuit?

10. Referring to the circuit described in Questions 8 and 9, determine the voltage drop across each of the resistors.

160 V, ____________ V

100 V, ____________ V

82 V, ____________ V

120 V, ____________ V

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11. A series circuit contains the following values of resistors:

R1 5 510 V R2 5 680 V R3 5 390 V R4 5 750 V

Assume a source voltage of 48 V. Use the general voltage divider formula to calculate the voltage drop across each of the resistors.

E1 5 _______ V E2 5 _______ V E3 5 _______ V E4 5 _______ V

Practical Applications

A 12-V DC automobile head lamp is to be used on a fi shing boat with a 24-V power system. The head lamp is rated at 50 W. A resistor is to be connected in

series with the lamp to permit it to operate on 24 V. What should be the resistance and power rating of the resistor? ■


T hree wire-wound resistors have the following values: 30 V, 80 V, and 100 V. Each resistor has a voltage rating of 100 V. If these three resistors are connected

in series, can they be connected to a 240-V circuit without damage to the resistors? Explain your answer. ■


Y ou are an electrician working in an industrial plant. A circuit contains eight incandescent lamps connected in series across 480 volts. One lamp has burned

out, and you must determine which one is defective. You have available a voltmeter, ammeter, and ohmmeter. Which meter would you use to determine which lamp is defective in the shortest possible time? Explain how you would use this meter and why. ■

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Practice Problems Series Circuits

1. Using the three rules for series circuits and Ohm’s law, solve for the miss-ing values.


A direct current motor is connected to a 250-volt DC supply. The armature has a current draw of 165 amperes when operating at full load. You have been

assigned the task of connecting two resistors in the armature circuit to provide speed control for the motor. When both resistors are connected in the circuit, the armature current is to be limited to 50% of the full-load current draw. When only one resistor is connected in the circuit, the armature current is to be limited to 85% of full-load current. Determine the ohmic value and minimum power rating of each resistor. Refer to Figure 6–26. When both switches S1 and S2 are open (off), both resistors are connected in the armature circuit, limiting current to 50% of its normal value. When switch S1 is closed, it causes the current to bypass resistor 1. Resistor 2 now limits the current to 85% of the full-load current. When both switches S1 and S2 are closed, all resistance is bypassed, and the armature is connected to full power. ■

FIGURE 6–26 Determine the resistance and power rating of the two series resistors.

Resistor 1 Resistor 2Armature

Series field

Shunt field

250 VDC

S1 S2

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ET 120 V E1 E2 E3 E4 E5

IT I1 I2 I3 I4 I5

RT R1 430 V R2 360 V R3 750 V R4 1000 V R5 620 V

PT P1 P2 P3 P4 P5

ET E1 E2 E3 11 V E4 E5

IT I1 I2 I3 I4 I5

RT R1 R2 R3 R4 R5

PT 0.25 W P1 0.03 W P2 0.0825 W P3 P4 0.045 W P5 0.0375 W

ET 340 V E1 44 V E2 94 V E3 60 V E4 40 V E5

IT I1 I2 I3 I4 I5

RT R1 R2 R3 R4 R5

PT P1 P2 P3 P4 P5 0.204 W

2. Use the general voltage divider formula to calculate the values of voltage drop for the following series-connected resistors. Assume a source volt-age of 120 V.

R1 5 1K V R2 5 2.2K V R3 5 1.8K V R4 5 1.5K V

E1 5 _____ V E2 5 _____ V E3 5 _____ V E4 5 _____ V

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172

Unit 7 Parallel Circuits Parallel Circuits


T here are probably more devices connected in parallel than any other way. Learning the rules concerning voltage, current, resis-

tance, and power in parallel circuits is paramount to understanding all electric circuits. Without this knowledge, it is impossible for an electri-cian to ever know how or why things work the way they do. In this unit, you will

■ use Ohm’s law and three rules governing parallel circuits to de-termine different electrical values.

■ use Ohm’s law to understand how voltage remains constant while current changes in a parallel circuit.

■ learn why total resistance decreases when resistors are con-nected in parallel.

Although working Ohm’s law problems often becomes tedious, it is the only way to gain an understanding of the relationship of voltage, current, resistance, and power in a parallel circuit.

OUTLINE7–1 Parallel Circuit Values7–2 Parallel Resistance Formulas

KEY TERMSCircuit branchCurrent dividersLoadParallel circuits

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UNIT 7 Parallel Circuits 173


■ discuss the characteristics of parallel circuits.

■ state three rules for solving electrical values of parallel circuits.

■ solve the missing values in a parallel circuit using the three rules and Ohm’s law.

■ discuss the operation of a current divider circuit.

■ calculate current values using the current divider formula.

Preview

Parallel circuits are probably the type of circuit with which most people are familiar. Most devices such as lights and receptacles in homes and offi ce

buildings are connected in parallel. Imagine if the lights in your home were wired in series. All the lights in the home would have to be turned on in order for any light to operate, and, if one were to burn out, all the lights would go out. The same is true for receptacles. If receptacles were connected in series, some device would have to be connected into each receptacle before power could be supplied to any other device. ■

7–1 Parallel Circuit ValuesTotal CurrentParallel circuits are circuits that have more than one path for current fl ow (Figure 7–1). If it is assumed that current leaves Terminal A and returns to Terminal B, it can be seen that the electrons can take three separate paths. In Figure 7–1, 3 amperes of current leave Terminal A. One ampere fl ows through Resistor R1, and 2 amperes fl ow to Resistors R2 and R3. At the junction of Resis-tors R2 and R3, 1 ampere fl ows through Resistor R2, and 1 ampere fl ows to Re-sistor R3. Notice that the power supply, Terminals A and B, must furnish all the current that fl ows through each individual resistor, or circuit branch. One of the rules for parallel circuits states that the total current flow in the circuit is equal to the sum of the currents through all the branches. This rule is known as current adds. Notice that the amount of current leaving the source must return to the source.

Voltage DropFigure 7–2 shows another parallel circuit and gives the values of voltage, current, and resistance for each individual resistor or branch. Notice that the voltage drop across all three resistors is the same. If the circuit is traced, it can be

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seen that each resistor is connected directly to the power source. A second rule for parallel circuits states that the voltage drop across any branch of a parallel circuit is the same as the applied voltage. For this reason, most electric circuits in homes are connected in parallel. Each lamp and receptacle is supplied with 120 volts (Figure 7–3).

Total ResistanceIn the circuit shown in Figure 7–4, three separate resistors have values of15 ohms, 10 ohms, and 30 ohms. The total resistance of the circuit, however, is 5 ohms. The total resistance of a parallel circuit is always less than the resistance of the lowest value resistor, or branch, in the circuit. Each

FIGURE 7–1 Parallel circuits provide more than one path for current fl ow.

+

R1

1 A

B

A

3 A

1 A

2 A

R2

1 A

R3

1 A

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FIGURE 7–2 Parallel circuit values.

+

ET 120 V IT 24 A RT 5

E1 120 V I1 8 A R1 15

E2 120 V I2 12 A R2 10

E3 120 V I3 4 A R3 30

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time another element is connected in parallel, there is less opposition to the fl ow of current through the entire circuit. Imagine a water system consisting of a holding tank, a pump, and return lines to the tank (Figure 7–5). Although large return pipes have less resistance to the fl ow of water than small pipes, the small pipes do provide a return path to the holding tank. Each time another return path is added, regardless of size, there is less overall resistance to fl ow and the rate of fl ow increases.

That concept often causes confusion concerning the defi nition of load among students of electricity. Students often think that an increase of resistance constitutes an increase of load. An increase of current, not resistance, results in an increase of load. In laboratory exercises, students often see the circuit current increase each time a resistive element is connected to the circuit, and they conclude that an increase of resistance must therefore cause an increase of current. That conclusion is, of course, completely contrary to Ohm’s law,

FIGURE 7–3 Lights and receptacles are connected in parallel.

Panelbox

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FIGURE 7–4 Total resistance is always less than the resistance of any single branch.

+ RT 5

R1 15

R2 10

R3 30

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which states that an increase of resistance must cause a proportional decrease of current. The false concept that an increase of resistance causes an increase of current can be overcome once the student understands that if the resistive elements are being connected in parallel, the circuit resistance is actually being decreased and not increased.

7–2 Parallel Resistance FormulasResistors of Equal ValueThree formulas can be used to determine the total resistance of a parallel circuit. The fi rst formula can be used only when all the resistors in the circuit are of equal value. This formula states that when all resistors are of equal value, the total resistance is equal to the value of one individual resistor, or branch, divided by the number (N) of resistors or branches.

RT 5 R __ N

For example, assume that three resistors, each having a value of 24 ohms, are connected in parallel (Figure 7–6). The total resistance of this circuit can

FIGURE 7–5 Each new path reduces the total resistance to the fl ow of water.

Pump

Holding tank

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be found by dividing the resistance of one single resistor by the total number of resistors:

RT 5 R __ N

RT 5 24 V _____ 3

RT 5 8 V

Product over SumThe second formula used to determine the total resistance in a parallel circuit divides the product of pairs of resistors by their sum sequentially until only one pair is left. This is commonly referred to as the product-over-sum method for fi nding total resistance.

RT 5 R1 3 R2 _______ R1 1 R2

In the circuit shown in Figure 7–7, three branches having resistors with values of 20 ohms, 30 ohms, and 60 ohms are connected in parallel. To fi nd the total resistance of the circuit using the product-over-sum method, fi nd the total resistance of any two branches in the circuit (Figure 7–8):

RT 5 R2 3 R3 _______ R2 1 R3

RT 5 30 V 3 60 V ____________ 30 V 1 60 V

RT 5 1800 V ________ 90 V

RT 5 20 V

FIGURE 7–6 Finding the total resistance when all resistors have the same value.

+ RT 8

R1 24

R2 24

R3 24

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FIGURE 7–7 Finding the total resistance of a parallel circuit by dividing the product of two resistorsby their sum.

+ RT 10

R1 20

R2 30

R3 60

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FIGURE 7–8 The total resistance of the last two branches.

+ RT 10

R1 20

R2 30

R3 60

RT 20

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The total resistance of the last two resistors in the circuit is 20 ohms. This20 ohms, however, is connected in parallel with a 20-ohms resistor. The total resis-tance of the last two resistors is now substituted for the value of R1 in the formula, and the value of the fi rst resistor is substituted for the value of R2 (Figure 7–9):

RT 5 R1 3 R2 _______ R1 1 R2

RT 5 20 V 3 20 V ____________ 20 V 1 20 V

RT 5 400 V _______ 40 V

RT 5 10 V

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Reciprocal FormulaThe third formula used to fi nd the total resistance of a parallel circuit is

1 ___ RT

5 1 ___ R1

1 1 ___ R2

1 1 ___ R3

1 1 ___ RN

Notice that this formula actually fi nds the reciprocal of the total resistance, instead of the total resistance. To make the formula equal to the total resistance, it can be rewritten as follows:

RT 5 1 __________________

1 ___ R1

1 1 ___ R2

1 1 ___ R3

1 1 ___ RN

The value RN stands for the number of resistors in the circuit. If the circuit has 25 resistors connected in parallel, for example, the last resistor in the formula would be R25.

This formula is known as the reciprocal formula. The reciprocal of any number is that number divided into 1. The reciprocal of 4, for example, is 0.25 because 1⁄4 5 0.25. Another rule of parallel circuits is that the total resis-tance of a parallel circuit is the reciprocal of the sum of the reciprocalsof the individual branches. A modifi ed version of this formula is used in several different applications to fi nd values other than resistance. Some of those other formulas are covered later.

Before the invention of handheld calculators, the slide rule was often em-ployed to help with the mathematical calculations in electrical work. At that time, the product-over-sum method of fi nding total resistance was the most popular. Since the invention of calculators, however, the reciprocal formula has become the most popular because scientifi c calculators have a reciprocal key (1/X), which makes calculating total resistance using the reciprocal method very easy.

FIGURE 7–9 The total value of the fi rst two resistors is used as Resistor 2.

+ RT 10

R1 20

R2 20

Total resistanceof the last tworesistors

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In Figure 7–10, three resistors having values of 150 ohms, 300 ohms, and 100 ohms are connected in parallel. The total resistance can be found using the reciprocal formula:

RT 5 1 _____________

1 ___ R1

1 1 ___ R2

1 1 ___ R3

RT 5 1 _______________________ 1 ______ 150 V

1 1 ______ 300 V

1 1 ______ 100 V

RT 5 1 ______________________________ (0.006667 1 0.003333 1 0.01) 1 __

V

RT 5 1 ________ (0.02) 1 __

V

RT 5 50 V

To fi nd the total resistance of the previous example using a scientifi c calculator, press the following keys. Note that the calculator automatically carries each answer to the maximum number of decimal places. This increases the accuracy of the answer.

1 5 0 1⁄3 1 3 0 0 1⁄3 1 1 0 0 1⁄3 5 1⁄3

Note that this is intended to illustrate how total parallel resistance can be determined using many scientifi c calculators. Some calculators may require a different key entry or pressing the equal key at the end.

FIGURE 7–10 Finding the total resistance using the reciprocal method.

+ RT 50

R1 150

R2 300

R3 100

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■ EXAMPLE 7–1

In the circuit shown in Figure 7–11, three resistors having values of 300 V,200 V, and 600 V are connected in parallel. The total current fl ow through the circuit is 0.6 A. Find all the missing values in the circuit.

SolutionThe fi rst step is to fi nd the total resistance of the circuit. The reciprocal formula is used:

RT 5 1 _____________

1 ___ R1

1 1 ___ R2

1 1 ___ R3

RT 5 1 _______________________ 1 ______ 300 V

1 1 ______ 200 V

1 1 ______ 600 V

RT 5 1 ______________________________

(0.00333 1 0.0050 1 0.00167) 1 __ V

RT 5 1 _______

(0.01) 1 __ V

RT 5 100 V

Now that the total resistance of the circuit is known, the voltage applied to the circuit can be found by using the total current value and Ohm’s law:

ET 5 lT 3 RT

ET 5 0.6 A 3 100 V

ET 5 60 V

FIGURE 7–11 Parallel circuit, Example 1.

+ETIT 0.6 ARTPT

E1I1

R1 300P1

E2I2

R2 200P2

E3I3

R3 600P3

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One of the rules for parallel circuits states that the voltage drops across all the parts of a parallel circuit are the same as the total voltage. Therefore, the voltage drop across each resistor is 60 V (Figure 7–12):

ET 5 E1 5 E2 5 E3

Because the voltage drop and the resistance of each resistor are known, Ohm’s law can be used to determine the amount of current fl ow through each resistor (Figure 7–13):

I1 5 E1 ___ R1

I1 5 60 V ______ 300 V

I1 5 0.2 A

I2 5 E2 ___ R2

I2 5 60 V ______ 200 V

I2 5 0.3 A

I3 5 E3 ___ R3

I3 5 60 V ______ 600 V

I3 5 0.1 A

FIGURE 7–12 The voltage is the same across all branches of a parallel circuit.

+ET 60 VIT 0.6 ART 100PT

E1 60 VI1

R1 300P1

E2 60 VI2

R2 200P2

E3 60 VI3

R3 600P3

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FIGURE 7–13 Ohm’s law is used to calculate the amount of current through each branch.


E1 60 VI1 0.2 AR1 300P1

E2 60 VI2 0.3 A

R2 200P2

E3 60 VI3 0.1 A

R3 600P3

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The amount of power (W) used by each resistor can be found by using Ohm’s law. A different formula is used to fi nd the amount of electrical energy converted into heat by each of the resistors:

P1 5 E1

2

___ R1

P1 5 60 V 3 60 V ___________ 300 V

P1 5 3600 V2 ________

300 V

P1 5 12 W

P2 5 I22 3 R2

P2 5 0.3 A 3 0.3 A 3 200 V

P2 5 0.09 A2 3 200 V

P2 5 18 W

P3 5 E3 3 I3 P3 5 60 V 3 0.1 A

P3 5 6 W

In Unit 6, it was stated that the total amount of power in a circuit is equal to the sum of the power used by all the parts. This is true for any type of circuit. Therefore, the total amount of power used by this circuit can be found by taking the sum of the power used by all the resistors (Figure 7–14):

PT 5 P1 1 P2 1 P3

PT 5 12 W 1 18 W 1 6 W

PT 5 36 W

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FIGURE 7–14 The amount of power used by the circuit.

+ET 60 VIT 0.6 ART 100PT 36 W

E1 60 VI1 0.2 AR1 300P1 12 W

E2 60 VI2 0.3 A

R2 200P2 18 W

E3 60 VI3 0.1 A

R3 600P3 6 W

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■ EXAMPLE 7–2

In the circuit shown in Figure 7–15, three resistors are connected in parallel. Two of the resistors have a value of 900 V and 1800 V. The value of Resistor R2 is unknown. The total resistance of the circuit is 300 V. Resistor R2 has a current fl ow of 0.2 A. Find the missing circuit values.

SolutionThe fi rst step in solving this problem is to fi nd the missing resistor value. This can be done by changing the reciprocal formula as shown:

1 ___ R2

5 1 ___ RT

2 1 ___ R1

2 1 ___ R3

or

R2 5 1 _____________ 1 ___ RT

2 1 ___ R1

2 1 ___ R3

One of the rules for parallel circuits states that the total resistance is equal to the reciprocal of the sum of the reciprocals of the individual resistors. Therefore, the reciprocal of any individual resistor is equal to the reciprocal of the difference

39154_07_ch07_p172-197.indd 184 11/8/10 8:57 AM



between the reciprocal of the total resistance and the sum of the reciprocals of the other resistors in the circuit:

R2 5 1 _____________ 1 ___ RT

2 1 ___ R1

2 1 ___ R3

R2 5 1 ________________________ 1 ______ 300 V

2 1 ______ 900 V

2 1 _______ 1800 V

R2 5 1 ____________________________________ (0.003333 2 0.001111 2 0.0005556) 1 __

V

R2 5 1 ____________ (0.001666) 1 __

V

R2 5 600 V

Now that the resistance of Resistor R2 has been found, the voltage drop across Resistor R2 can be determined using the current fl ow through the re-sistor and Ohm’s law (Figure 7–16):

E2 5 I2 3 R2

E2 5 0.2 A 3 600 V

E2 5 120 V


+ET

IT

RT 300

PT

E1

I1

R1 900P1

E2

I2 0.2 A

R2

P2

E3

I3

R3 1800P3

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If 120 V is dropped across Resistor R2, the same voltage is dropped across each component of the circuit:

E2 5 ET 5 E1 5 E3

Now that the voltage drop across each part of the circuit is known and the resistance is known, the current fl ow through each branch can be determined using Ohm’s law (Figure 7–17):

IT 5 ET ___ RT

IT 5 120 V ______ 300 V

IT 5 0.4 A

I1 5 E1 ___ R1

I1 5 120 V _______ 1800 V

I1 5 0.1333 A

I3 5 E3 ___ R3

I3 5 120 V _______ 1800 V

I3 5 0.0667 A

FIGURE 7–16 The missing resistor and voltage values.

+ET 120 VIT

RT 300PT

E1 120 VI1

R1 900P1

E2 120 VI2 0.2 A

R2 600P2

E3 120 VI3

R3 1800P3

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FIGURE 7–17 Determining the current using Ohm’s law.


E1 120 VI1 0.133 AR1 900P1

E2 120 VI2 0.2 A

R2 600P2

E3 120 VI3 0.0667 A

R3 1800P3

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The amount of power used by each resistor can be found using Ohm’s law (Figure 7–18):

P1 5 E1

2

___ R1

P1 5 120 V 3 120 V ______________ 900 V

P1 5 14,400 V2

_________ 900 V

P1 5 16 W

P2 5 I22 3 R2

P2 5 0.2 A 3 0.2 A 3 600 V

P2 5 0.04 A2 3 600 V

P2 5 24 W

P3 5 E3 3 I3

P3 5 120 V 3 0.0667 A

P3 5 8.004 W

PT 5 ET 3 IT

PT 5 120 V 3 0.4 A

PT 5 48 W

If the wattage values of the three resistors are added to calculate total power for the circuit, it will be seen that their total is 48.004 W instead of the calculated 48 W.The small difference in answers is caused by the rounding off of other values. In this instance, the current of Resistor R3 was rounded from 0.066666666 to 0.0667.

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FIGURE 7–18 The values of power for the circuit in Example 2.

+ET 120 V

IT 0.4 ART 300

PT 48 W

E1 120 V

I1 0.133 A

R1 900P1 16 W

E2 120 V

I2 0.2 A

R2 600

P2 24 W

E3 120 V

I3 0.0667 A

R3 1800P3 8.004 W

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■ EXAMPLE 7–3

In the circuit shown in Figure 7–19, three resistors are connected in parallel. Re-sistor R1 is producing 0.075 W of heat, R2 is producing 0.45 W of heat, and R3 is producing 0.225 W of heat. The circuit has a total current of 0.05 A.

SolutionBecause the amount of power dissipated by each resistor is known, the total power for the circuit can be found by fi nding the sum of the power used by the components:


+ET

IT 0.05 A

RT

PT

E1

I1

R1

P1 0.075 W

E2

I2

R2

P2 0.45 W

E3

I3

R3

P3 0.225 W

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PT 5 P1 1 P2 1 P3

PT 5 0.075 W 1 0.45 W 1 0.225 W

PT 5 0.75 W

Now that the amount of total current and total power for the circuit are known, the applied voltage can be found using Ohm’s law (Figure 7–20):

ET 5 PT ___ IT

ET 5 0.75 W _______ 0.05 A

ET 5 15 VThe amount of current fl ow through each resistor can now be found using

Ohm’s law (Figure 7–21):

FIGURE 7–20 The applied voltage for the circuit.

+ET 15 VIT 0.05 ART

PT 0.75 W

E1 15 VI1

R1

P1 0.075 W

E2 15 VI2

R2

P2 0.45 W

E3 15 VI3

R3

P3 0.225 W

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FIGURE 7–21 The current through each branch.

+ET 15 V

IT 0.05 ART

PT 0.75 W

E1 15 V

I1 0.005 A

R1

P1 0.075 W

E2 15 V

I2 0.03 A

R2

P2 0.45 W

E3 15 V

I3 0.015 A

R3

P3 0.225 W

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I1 5 P1 ___ E1

I1 5 0.075 W ________ 15 V

I1 5 0.005 A

I2 5 P2 ___ E2

I2 5 0.45 W _______ 15 V

I2 5 0.03 A

I3 5 P3 ___ E3

I3 5 0.225 W ________ 15 V

I3 5 0.015 A

All resistance values for the circuit can now be found using Ohm’s law(Figure 7–22):

R1 5 E1 ___ I1

R1 5 15 V ________ 0.005 A

R1 5 3000 V

R2 5 E2 ___ I2

R2 5 15 V ______ 0.03 A

R2 5 500 V

R3 5 E3 ___ I3

R3 5 15 V ________ 0.015 A

R3 5 1000 V

RT 5 ET ___ IT

RT 5 15 V ______ 0.05 A

RT 5 300 V

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FIGURE 7–22 The remaining values for the circuit.

+ET 15 V

IT 0.05 ART 300

PT 0.75 W

E1 15 V

I1 0.005 A

R1 3000P1 0.075 W

E2 15 V

I2 0.03 A

R2 500

P2 0.45 W

E3 15 V

I3 0.015 A

R3 1000P3 0.225 W

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Current DividersAll parallel circuits are current dividers (Figure 7–23). As previously discussed in this unit, the sum of the currents in a parallel circuit must equal the total current. Assume that a current of 1 ampere enters the circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2, and then recom-bine at Point B. The amount of current that fl ows through each resistor is inversely proportional to the resistance value. A greater amount of current will

FIGURE 7–23 Parallel circuits are current dividers.

I1R1

I2R2

A

B

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fl ow through a low-value resistor, and less current will fl ow through a high-value resistor. In other words, the amount of current fl owing through each resistor is inversely proportional to its resistance.

In a parallel circuit, the voltage across each branch must be equal(Figure 7–24). Therefore, the current fl ow through any branch can be calcu-lated by dividing the source voltage (ET) by the resistance of that branch. The current fl ow through Branch 1 can be calculated using the formula

I1 5 ET ___ R1

It is also true that the total circuit voltage is equal to the product of the total circuit current and the total circuit resistance:

ET 5 lT 3 RT

If the value of ET is substituted for (IT 3 RT) in the previous formula, it becomes

I1 5 IT 3 RT _______ R1

If the formula is rearranged and the values of I1 and R1 are substituted for IX and RX, it becomes what is generally known as the current divider formula:

IX 5 ( RT ___ RX

) ITThis formula can be used to calculate the current fl ow through any branch by substituting the values of IX and RX for the branch values when the total circuit current and the resistance are known. In the circuit shown in Figure 7–25,

FIGURE 7–24 The voltage is the same across all branches of a parallel circuit.

ET

24 V 24 V 24 V24 V

R1 R2 R3

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Resistor R1 has a value of 1200 ohms, Resistor R2 has a value of 300 ohms, and Resistor R3 has a value of 120 ohms, producing a total resistance of 80 ohms for the circuit. It is assumed that a total current of 2 amperes fl ows in the cir-cuit. The amount of current fl ow through Resistor R1 can be found using the formula

I1 5 ( RT ___ R1

) ITI1 5 ( 80 V _______

1200 V ) 2 A

I1 5 0.133 A

The current fl ow through each of the other resistors can be found by substitut-ing in the same formula:

I2 5 ( RT ___ R2

) IT I2 5 ( 80 V ______

300 V ) 2 A

I2 5 0.533 A

I3 5 ( RT ___ R3

) IT I3 5 ( 80 V ______

120 V ) 2 A

I3 5 1.333 A

FIGURE 7–25 The current divides through each branch of a parallel circuit.

IT = 2ART = 80

R3 = 120

R2 = 300

R1 = 1200

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Summary ■ A Parallel circuit is characterized by the fact that it has more than one path

for current fl ow.

■ Three rules for solving parallel circuits are as follows:

a. The total current is the sum of the currents through all of the branches of the circuit.

b. The voltage across any part of the circuit is the same as the total voltage.

c. The total resistance is the reciprocal of the sum of the reciprocals of each individual branch.

■ Circuits in homes are connected in parallel.

■ The total power in a parallel circuit is equal to the sum of the power dissipa-tion of all the components.

■ Parallel circuits are current dividers.

■ The current fl owing through each branch of a parallel circuit can be calcu-lated when the total resistance and the total current are known.

■ The amount of current fl ow through each branch of a parallel circuit is in-versely proportional to its resistance.

Review Questions 1. What characterizes a parallel circuit?

2. Why are circuits in homes connected in parallel?

3. State three rules concerning parallel circuits.

4. A parallel circuit contains four branches. One branch has a current fl ow of 0.8 A, another has a current fl ow of 1.2 A, the third has a current fl ow of 0.25 A, and the fourth has a current fl ow of 1.5 A. What is the total current fl ow in the circuit?

5. Four resistors having a value of 100 V each are connected in parallel. What is the total resistance of the circuit?

6. A parallel circuit has three branches. An ammeter is connected inseries with the output of the power supply and indicates a total cur-rent fl ow of 2.8 A. If Branch 1 has a current fl ow of 0.9 A and Branch 2 has a current fl ow of 1.05 A, what is the current fl ow through Branch 3?

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7. Four resistors having values of 270 V, 330 V, 510 V, and 430 V are connected in parallel. What is the total resistance in the circuit?

8. A parallel circuit contains four resistors. The total resistance of the circuit is 120 V. Three of the resistors have values of 820 V, 750 V, and 470 V. What is the value of the fourth resistor?

9. A circuit contains a 1200-V, a 2200-V, and a 3300-V resistor connected in parallel. The circuit has a total current fl ow of 0.25 A. How much current fl ows through each of the resistors?


Y ou have been hired by a homeowner to install a ceiling fan and light kit in a liv-ing room. The living room luminaire (light fi xture) being used at the present time

contains two 60-W lamps. After locating the circuit in the panel box, you fi nd that the circuit is protected by a 15-A circuit breaker and is run with 14 AWG copper wire. After turning on all lights connected to this circuit, you have a current draw of 8.6 A and a voltage of 120 V. The ceiling fan light kit contains four 60-W lamps, and the fan has a maximum current draw of 1.6 A. Can this fan be connected to the existing living room circuit? Recall that a continuous-use circuit should not be loaded to more than 80% of its rated capacity. ■


Y ou are employed in a large industrial plant. A 480-V, 5000-W heater is used to melt lead in a large tank. It has been decided that the heater is not suffi cient to

raise the temperature of the lead to the desired level. A second 5000-W heater is to be installed on the same circuit. What will be the circuit current after installation of the second heater, and what is the minimum size circuit breaker that can be used if this is a continuous-duty circuit? ■

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Y ou are an electrician. You have been asked by a homeowner to install a lighted mirror in a bathroom. The mirror contains eight 40-watt lamps. Upon check-

ing the service panel you discover that the bathroom circuit is connected to a single 120-volt, 20-ampere circuit breaker. At the present time, the circuit supplies power to an electric wall heater rated at 1000 watts, a ceiling fan with a light kit, and a light fi xture over the mirror. The fan motor has a full-load current draw of 3.2 amperes and the light kit contains three 60-watt lamps. The light fi xture presently installed over the mirror contains four 60-watt lamps. The homeowner asked whether the present light fi xture over the mirror can be replaced by the lighted mirror. Assuming all loads are continuous, can the present circuit supply the power needed to operate all the loads without overloading the circuit? ■


A car lot uses incandescent lamps to supply outside lighting during the night.There are three strings of lamps connected to a single 20-ampere circuit. Each

string contains eight lamps. What is the largest standard lamp that can be used with-out overloading the circuit? Standard size lamps are 25 watt, 40 watt, 60 watt,75 watt, and 100 watt. ■

Practice Problems Parallel Circuits

Using the rules for parallel circuits and Ohm’s law, solve for the missing values.

1.

ET E1 E2 E3 E4

IT 0.942 A I1 I2 I3 I4

RT R1 680 V R2 820 V R3 470 V R4 330 V

PT P1 P2 P3 P4

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2.

ET E1 E2 E3 E4

IT 0.00639 A I1 I2 0.00139 A I3 0.00154 A I4 0.00115 A

RT R1 R2 R3 R4

PT P1 0.640 W P2 P3 P4

3.

ET E1 E2 E3 E4

IT I1 I2 I3 3.2 A I4

RT 3.582 V R1 16 V R2 10 V R3 R4 20 V

PT P1 P2 P3 P4

4.

ET E1 E2 E3 E4

IT I1 I2 I3 I4

RT R1 82 kV R2 75 kV R3 56 kV R4 62 kV

PT 3.436 W P1 P2 P3 P4

5. A parallel circuit contains the following resistor values:

R1 5 360 V R2 5 470 V R3 5 300 V

R4 5 270 V IT 5 0.05 A

Find the following missing values:

RT 5 V I1 5 A I2 5 A

I3 5 A I4 5 A

6. A parallel circuit contains the following resistor values:

R1 5 270K V R2 5 360K V R3 5 430K V

R4 5 100K V IT 5 0.006 A

Find the following missing values:

RT 5 V I1 5 A I2 5 A

I3 5 A I4 5 A

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198

Unit 8 Combination Circuits Combination Circuits


U nderstanding how values of voltage, current, and resistance relate to each other in combination circuits is essential to understanding

how all electric circuits work. Although it is not a common practice for electricians on the job to use a calculator to calculate values of voltage and current, they must have an understanding of how these electrical values relate to each other in different kinds of circuits. An understand-ing of Ohm’s law and basic circuits is also essential in being able to troubleshoot problems that occur in an electric circuit. In this unit

■ the rules for series circuits are employed for those components connected in series, and parallel rules are used for components connected in parallel.

■ you are shown how to reduce a complex combination circuit to a simple series or parallel circuit and then apply circuit rules and Ohm’s law to determine the values of voltage, current, and power for the different components.

OUTLINE 8–1 Combination Circuits8–2 Solving Combination Circuits8–3 Simplifying the Circuit

KEY TERMS Combination circuitNodeParallel blockRedrawReduceSimple parallel circuitTrace the current path

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UNIT 8 Combination Circuits 199


■ defi ne a combination circuit.

■ list the rules for parallel circuits.

■ list the rules for series circuits.

■ solve combination circuits using the rules for parallel circuits, the rules for series circuits, and Ohm’s law.

Preview

Combination circuits contain a combination of both series and parallel elements. To determine which components are in parallel and which are

in series, trace the fl ow of current through the circuit. Remember that a series circuit has only one path for current fl ow and a parallel circuit has more than one path for current fl ow. ■

8–1 Combination CircuitsA simple combination circuit is shown in Figure 8–1. It is assumed that the current in Figure 8–1 fl ows from Point A to Point B. To identify the series and parallel elements, trace the current path. All the current in the circuit must fl ow through the fi rst resistor, R1. R1 is therefore in series with the rest of the circuit. When the current reaches the junction point of R2 and R3, however, it splits. A junction point such as this is often referred to as a node. Part of the current fl ows through R2, and part fl ows through R3. These two resistors are in parallel. Because this circuit contains both series and parallel elements, it is a combination circuit.

FIGURE 8–1 A simple combination circuit.

R2

R1

R3A

B

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8–2 Solving Combination CircuitsThe circuit shown in Figure 8–2 contains four resistors with values of 325 ohms, 275 ohms, 150 ohms, and 250 ohms. The circuit has a total current fl ow of 1 ampere. In order to determine which resistors are in series and which are in parallel, trace the path for current fl ow through the circuit. When the path of current fl ow is traced, it can be seen that current can fl ow by two separate paths from the negative terminal to the positive terminal. One path is through R1 and R2, and the other path is through R3 and R4. These two paths are there-fore in parallel. However, the same current must fl ow through R1 and R2. So these two resistors are in series. The same is true for R3 and R4.

To solve the unknown values in a combination circuit, use series circuit rules for those sections of the circuit that are connected in series and parallel circuit rules for those sections connected in parallel. The circuit rules are as follows:

Series Circuits

1. The current is the same at any point in the circuit.

2. The total resistance is the sum of the individual resistances.

3. The applied voltage is equal to the sum of the voltage drops across all the resistors.

Parallel Circuits

1. The voltage drop across any branch of a parallel circuit is the same as the applied voltage.

2. The total current fl ow is equal to the sum of the currents through all of the circuit branches.

FIGURE 8–2 Tracing the current paths through a combination circuit.

E3

I3

R3 150 ===

ET

IT 1 A

RT ===

E4

I4

R4 250 ===

E1

I1

R1 325 ===

E2

I2

R2 275 ===+

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UNIT 8 Combination Circuits 201

3. The total resistance is equal to the reciprocal of the sum of the recipro-cals of the branch resistances.

8–3 Simplifying the CircuitThe circuit shown in Figure 8–2 can be reduced or simplifi ed to a simple parallel circuit (Figure 8–3). Because R1 and R2 are connected in series, their values can be added to form one equivalent resistor, Rc(1&2), which stands for a combination of Resistors 1 and 2. The same is true for R3 and R4. Their values are added to form Rc(3&4). Now that the circuit has been reduced to a simple parallel circuit, the total resistance can be found:

RT 5 1 _____________

1 _____ Rc(1& 2)

1 1 _____ Rc(3&4)

RT 5 1 _______________

1 ______ 600 V

1 1 ______ 400 V

RT 5 1 ______________________

( 0.0016667 1 0.0025 ) 1 __ V

RT 5 1 _____________

(0.0041667) 1 __ V

RT 5 240 V

FIGURE 8–3 Simplifying the combination circuit.

Ec(3&4) 240 V

Ic(3&4) 0.6 A

Rc(3&4) 400 ===Ec(1&2)

240 VIc(1&2)

0.4 ARc(1&2)

600 ===ET

240 VIT

1 ART

240 ===

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bbasic electric circuitsasic electric circuits

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