Modelo Cognitivo de John Hayes y Linda Flower

Resistances in series produce ratios of voltages, and these ratios can be tailored to meet certain needs by means of voltage divider networks.

When a voltage divider network is designed and assembled, the resistance values should be as small as possible without causing too much current drain on the battery or power supply. (In prac- tice, the optimum values depend on the nature of the circuit being designed. This is a matter for en- gineers, and specific details are beyond the scope of this course.) The reason for choosing the smallest possible resistances is that, when the divider is used with a circuit, you do not want that circuit to upset the operation of the divider. The voltage divider “fixes” the intermediate voltages best when the resistance values are as small as the current-delivering capability of the power supply will allow.

Figure 5-7 illustrates the principle of voltage division. The individual resistances are R1,R2,

R3, . . . , Rn. The total resistance is R=R1+R2+R3+. . .+Rn. The supply voltage is E, and the cur-

rent in the circuit is therefore I=E/R.At the various points P1,P2,P3, . . . , Pn, the potential differ-

ences relative to the negative battery terminal are E1,E2,E3, . . . , En, respectively. The last voltage,

En, is the same as the battery voltage, E.All the other voltages are less than E, and ascend in succes-

sion, so that E1<E2<E3<. . .<En. (The mathematical symbol <means “is less than.”)

The voltages at the various points increase according to the sum total of the resistances up to each point, in proportion to the total resistance, multiplied by the supply voltage. Thus, the voltage

E1is equal to ER1/R.The voltage E2is equal to E(R1+R2)/R.The voltage E3is equal to E(R1+R2+

R3)/R. This process goes on for each of the voltages at points all the way up to En=E(R1+R2+

Problem 5-15

Suppose you are building an electronic circuit, and the battery supplies 9.0 V. The minus terminal is at common (chassis) ground. You need to provide a circuit point where the dc voltage is +2.5 V. Give an example of a pair of resistors that can be connected in a voltage divider configuration, such that+2.5 V appears at some point.

Examine the schematic diagram of Fig. 5-8. There are infinitely many different combinations of resistances that will work here! Pick some total value, say R=R1+R2=1000Ω. Keep in mind

that the ratio R1:R will always be the same as the ratio E1:E. In this case, E1 =2.5 V, so E1:E =

2.5/9.0=0.28. This means that you want the ratio R1:R to be equal to 0.28. You have chosen to

Voltage Divider Networks 79

5-7 General arrangement for a voltage-divider circuit. Illustration for Quiz Questions 19 and 20.

5-8 A voltage divider network in which 2.5 V dc is derived from a 9.0-V dc source.

makeRequal to 1000 Ω. This means R1must be 280 Ωin order to get the ratio R1:R=0.28. The

value of R2is the difference between RandR1. That is 1000 −280=720Ω.

In a practical circuit, you would want to choose the smallest possible value for R.This might be less than 1000 Ω, or it might be more, depending on the nature of the circuit and the current- delivering capability of the battery. It’s not the actual values of R1andR2that determine the voltage

you get at the intermediate point, but their ratio.

Problem 5-16

What is the current I, in milliamperes, drawn by the entire network of series resistances in the situ- ation described in Problem 5-15 and its solution?

Use Ohm’s Law to get I=E/R=9.0/1000=0.0090 A =9.0 mA.

Problem 5-17

Suppose that it is all right for the voltage divider network to draw up to 100 mA of current in the situation shown by Fig. 5-8 and posed by Problem 5-15. You want to design the network to draw this amount of current, because that will offer the best voltage regulation for the circuit to be oper- ated from the network. What values of resistances R1andR2should you use?

Calculate the total resistance first, using Ohm’s Law. Remember to convert 100 mA to amperes! That means you use the figure I = 0.100 A in your calculations. Then R = E/I = 9.0/0.100 = 90 Ω. The ratio of resistances that you need is R1:R2 = 2.5/9.0 = 0.28. You should use R1 =

0.28 × 90 = 25 Ω. The value of R2 is the difference between R and R1. That is, R2 = R −

R1=90−25=65Ω.

Quiz

Refer to the text in this chapter if necessary. A good score is at least 18 correct answers. The answers are in the back of the book.

1. In a series-connected string of ornament bulbs, if one bulb gets shorted out, which of the following will occur?

(a) All the other bulbs will go out. (b) The current in the string will go up. (c) The current in the string will go down. (d) The current in the string will stay the same.

2. Imagine that four resistors are connected in series across a 6.0-V battery, and the ohmic values are R1=10Ω,R2=20Ω,R3=50Ω, and R4=100Ω, as shown in Fig. 5-9. What is the voltage

across the resistance R2?

(a) 0.18 V (b) 33 mV (c) 5.6 mV (d) 0.67 V

3. In the scenario shown by Fig. 5-9, what is the voltage across the combination of R3andR4?

(a) 0.22 V (b) 0.22 mV (c) 5.0 V (d) 3.3 V

4. Suppose three resistors are connected in parallel across a battery that delivers 15 V, and the ohmic values are R1=470Ω,R2=2.2 kΩ, and R3=3.3 kΩ, as shown in Fig. 5-10. The voltage

across the resistance R2is

(a) 4.4 V. (b) 5.0 V. (c) 15 V.

(d) not determinable from the data given.

Quiz 81

5-9 Illustration for Quiz Questions 2, 3, 8, and 9. Resistance values are in ohms.

5-10 Illustration for Quiz Questions 4, 5, 6, 7, 10, and 11. Resistances are in ohms, where k indicates

multiplication by 1000.

5. In the situation shown by Fig. 5-10, what is the current through R2?

(a) 6.8 mA (b) 43 mA (c) 0.15 A (d) 6.8 A

6. In the situation shown by Fig. 5-10, what is the total current drawn from the source? (a) 6.8 mA

(b) 43 mA (c) 0.15 A (d) 6.8 A

7. In the situation shown by Fig. 5-10, suppose that resistor R2opens up. The current through

the other two resistors will (a) increase.

(b) decrease. (c) drop to zero. (d) not change.

8. Suppose that four resistors are connected in series with a 6.0-V supply, with values shown in Fig. 5-9. What is the power dissipated by the whole combination?

(a) 0.2 W (b) 6.5 mW (c) 200 W (d) 6.5 W

9. In the situation shown by Fig. 5-9, what is the power dissipated by R4?

(a) 11 mW (b) 0.11 W (c) 0.2 W (d) 6.5 mW

10. Suppose that three resistors are in parallel as shown in Fig. 5-10. What is the power dissipated by the whole set of resistors?

(a) 5.4 W (b) 5.4 µW (c) 650 W (d) 0.65 W

11. In the situation shown by Fig. 5-10, what is the power dissipated in resistance R1?

(a) 32 mW (b) 0.48 W (c) 2.1 W (d) 31 W

12. Fill in the blank in the following sentence to make it true: “In a series or parallel dc circuit, the sum of the s in each component is equal to the total provided by the power supply.”

(a) current (b) voltage (c) wattage (d) resistance

13. Look at Fig. 5-5A. Suppose the resistors each have values of 33 Ωand the battery supplies 24 V. What is the current I1?

(a) 1.1 A (b) 0.73 A

(c) 0.36 A

(d) Not determinable from the information given

14. Look at Fig. 5-5B. Let each resistor have a value of 820 Ω. Suppose the top three resistors all lead to identical light bulbs. If I1=50 mA and I2=70 mA, what is the power dissipated in the

resistor carrying current I4?

(a) 33 W (b) 40 mW (c) 1.3 W

(d) It can’t be found using the information given.

15. Refer to Fig. 5-6. Suppose the resistances R1,R2,R3, and R4are in exactly the ratio 1:2:4:8

from left to right, and the battery supplies 30 V. What is the voltage E2?

(a) 4.0 V (b) 8.0 V (c) 16 V

(d) It is not determinable from the data given.

16. Refer to Fig. 5-6. Suppose the resistances are each 3.3 kΩ, and the battery supplies 12 V. If the plus terminal of a dc voltmeter is placed between resistances R1andR2(with voltages E1andE2

across them, respectively), and the minus terminal of the voltmeter is placed between resistances R3

andR4(with voltages E3andE4across them, respectively), what will the meter register?

(a) 0.0 V (b) 3.0 V (c) 6.0 V (d) 12 V

17. In a voltage divider network, the total resistance (a) should be large to minimize current drain. (b) should be as small as the power supply will allow. (c) is not important.

(d) should be such that the current is kept to 100 mA. 18. The maximum voltage output from a voltage divider

(a) is a fraction of the power supply voltage. (b) depends on the total resistance.

(c) is equal to the supply voltage. (d) depends on the ratio of resistances.

19. Refer to Fig. 5-7. Suppose the battery voltage Eis 18.0 V, and there are four resistances in the network such that R1=100Ω,R2=22.0Ω,R3=33.0Ω, and R4=47.0Ω. What is the voltage

E3atP3?

(a) 4.19 V (b) 13.8 V

(c) 1.61 V (d) 2.94 V

20. Refer to Fig. 5-7. Suppose the battery voltage is 12 V, and you want to obtain intermediate voltages of 3.0 V, 6.0 V, and 9.0 V. Suppose that a maximum of 200 mA is allowed to be drawn from the battery. What should the resistances, R1,R2,R3, and R4be, respectively?

(a) 15 Ω, 30 Ω, 45 Ω, and 60 Ω (b) 60 Ω, 45 Ω, 30 Ω, and 15 Ω (c) 15 Ω, 15 Ω, 15 Ω, and 15 Ω

ALL ELECTRICAL COMPONENTS,DEVICES,AND SYSTEMS HAVE SOME RESISTANCE.IN EVERYDAY PRACTICE,

there is no such thing as a perfect electrical conductor. You’ve seen some examples of circuits con- taining components that are deliberately designed to oppose the flow of current. These components are resistors.In this chapter, you’ll learn all about them.