After completing this chapter, the student should be able to:

DC Circuits

OBJECTIVES

After completing this chapter, the student should be able to:

• Solve for all unknown values (current, voltage, resistance, and power) in a series, parallel, or series-parallel circuit.

• Understand the importance of voltage dividers.

• Design and solve for all unknown values in a voltage-divider circuit.

See accompanying CD for interactive presentations, tutorials,and DC Circuit examples in MultiSim,Chapter 8.

Inthe study of electronics, certain circuits appear again and again. The most commonly used cir-cuits are the series circuit, the parallel circuit, and the series-parallel circuit.

This chapter applies information from the last few chapters to the solving of all unknowns in these three basic types of circuits.Voltagedividers can make available several voltages from asingle voltage source. Voltage dividers are essentially se-ries circuits with parallel loads.Thischapter helps the student understand how significant voltage dividers are as an application of series circuits and how to design one for a specific application.

B

Ill

SERIES CIRCUITS

A series circuit (Figure 8-1) provides only one path for current flow. The factors governing the operation ofa series circuit are:

80

FIGURE 8-1 Series circuit.

1. The same current flows through each com-ponent in a series circuit.

4

=IR,= IR2= IR,... = IRN

2. Thetotal resistance inaseries circuit is equal to the sum of the individual resistances.

~ =R] +R2 +R3 ... +~

3. The total voltage across a series circuit is equal to the sum of the individual voltage drops.

CHAPTER 8 DC CIRCUITS

4. The voltage drop across a resistor in aseries circuit is proportional to the size of the re

-sistor. (I =E/R)

5. The total power dissipated in a series circuit is equal to the sum of the individual power dissipations.

PT= PR,+ PR,+PR,... +PR•

EXAMPLE: Three resistors, 47 ohms, 100 ohms,

and 150ohms, areconnected inseries with abattery rated at 12volts. Solve for allvalues in the circuit.

The first step is to draw a schematic ofthe cir-cuit and list all known variables. See Figure 8-2.

Given: ~ =? ET= 12volts ~=?

PT =?

R[ =47 ohms R2 = 100 ohms R3

=

150 ohms

ER,= ? ER,=? ER,=? PR,

=

? PR, = ? PR, = ? In solving for all values in a circuit, the total resistance must be found first. Then the total cir-cuit current can be determined. Once the current is known, the voltage drops and power dissipation can be determined.

Given: ~=?

R[ =47 ohms R2 = 100 ohms R3 = 150 ohms

Solution:

~

=

R[ +R2 +R3 RT

=

47 + 100 + 150 ~ =297 ohms

FIGURE 8-2

-Using Ohm's law, the current is:

Given: Solution:

ET IT=

-RT

12 I

=-T 297

ET= 12volts

R, = 297 ohms ~ = 0.040 amp

Since I, =IR,=IR,=IR"the voltage drop (ER) across resistor R[is:

Given: Solution:

ER,

IR = -, R[

ER

0.040 =-' 47 IR[=0.040 amp

R[=47 ohms ER,= 1.88 volts The voltage drop (ER,l across resistor R2 is:

Given: Solution:

ER,

IR,=~

ER 0.040 =-'

100 IR,= 0.040 amp

R2

=

100 ohms ER,

=

4volts The voltage drop (ER,) across resistor R3is:

Given: Solution:

ER,

IR,

=

"

R

3

ER 0.040 =-'

150 IR,= 0.040 amp

R3

=

150 ohms ER,

=

6 volts Verify that the sum ofthe individual voltages is equal to the total voltage.

Given:

ET= 12volts ER,= 1.88 volts ER,= 4 volts ER,= 6volts

Solution:

ET=ER,+ ER,+ER,

ET

=

1.88 +4 + 6 ET= 11.88 volts

CHAPTER 8 DC CIRCUITS

4. The voltage drop across a resistor in aseries circuit is proportional to the size of the re

-sistor. (I =E/R)

5. The total power dissipated in a series circuit is equal to the sum of the individual power dissipations.

PT= PR,+ PR,+PR,... +PR•

EXAMPLE: Three resistors, 47 ohms, 100 ohms,

and 150ohms, areconnected inseries with abattery rated at 12volts. Solve for allvalues in the circuit.

The first step is to draw a schematic ofthe cir-cuit and list all known variables. See Figure 8-2.

Given: ~ =? ET= 12volts ~=?

PT =?

R[ =47 ohms R2 = 100 ohms R3

=

150 ohms

ER,= ? ER,=? ER,=? PR,

=

? PR, = ? PR, = ? In solving for all values in a circuit, the total resistance must be found first. Then the total cir-cuit current can be determined. Once the current is known, the voltage drops and power dissipation can be determined.

Given: ~=?

R[ =47 ohms R2 = 100 ohms R3 = 150 ohms

Solution:

~

=

R[ +R2 +R3 RT

=

47 + 100 + 150 ~ =297 ohms

FIGURE 8-2

-Using Ohm's law, the current is:

Given: Solution:

ET IT=

-RT

12 I

=-T 297

ET= 12volts

R, = 297 ohms ~ = 0.040 amp

Since I, =IR,=IR,=IR"the voltage drop (ER) across resistor R[is:

Given: Solution:

ER,

IR = -, R[

ER

0.040 =-' 47 IR[=0.040 amp

R[=47 ohms ER,= 1.88 volts The voltage drop (ER,l across resistor R2 is:

Given: Solution:

ER,

IR,=~

ER 0.040 =-'

100 IR,= 0.040 amp

R2

=

100 ohms ER,

=

4volts The voltage drop (ER,) across resistor R3is:

Given: Solution:

ER,

IR,

=

"

R

3

ER 0.040 =-'

150 IR,= 0.040 amp

R3

=

150 ohms ER,

=

6 volts Verify that the sum ofthe individual voltages is equal to the total voltage.

Given:

ET= 12volts ER,= 1.88 volts ER,= 4 volts ER,= 6volts

Solution:

ET=ER,+ ER,+ER,

ET

=

1.88 +4 + 6 ET= 11.88 volts

SECTION 1 DC CIRCUITS

There is a difference between the calculated andthe totalgiven voltage due to the rounding of the total current to three decimal places.

The power dissipated across resistorR] is:

Given: Solution:

PR, = ? PR, =IR,ER,

IR, = 0.040 amp PR, = (0.040)(1.88)

ER, = 1.88 volts PR, = 0.075 watt

The power dissipated acrossresistorR2is: Given:

PR, =?

Solution: PR,

=

IR,ER, IR, =0.040 amp PR, = (0.040)(4)

ER,

=

4volts PR,

=

0.16 watt

The power dissipatedacross resistorR3is:

Given: Solution:

PR) = ? PR)= IR)ER) IR)

=

0.040 amp PR,

=

(0.040)(6)

ER,

=

6volts PR,

=

0.24 watt

Thetotal power dissipated is: Given:

P_T =?_•

PR, = 0.075 watt PR,

=

0.16 watt PR, = 0.24 watt

Solution:

Pr = PR, + PR, +PR) PT= 0.075 +0.16 +0.24 PT= 0.475 watt or475 mW

8-

1

QUESTION

1. Four resistors, 270 ohms, 560 ohms, 1200

ohms, and 1500 ohms, are connected in series with a battery rated at28 volts. Solve for all values of the circuit,

r

m

PARALLEL CIRCUITS

Aparallel circuit (Figure8-3)is defined as a circuit having more than one current path. The factors governing the operations of a parallel circuit are:

FIGURE 8-3

Parallel circuit.

1. The same voltage exists acrosseach branch of the parallel circuit and is equal to that of the voltage source.

Er =ER,=ER,=ER)· . . =ERn

2. The current through each branch of a paral-lel circuit is inversely proportional to the amount of resistance ofthebranch, (I=E/R). 3. The total current in a parallel circuit is the

sum of the individual branch currents. IT =IR, +IR, +IR)... +IRn

4. The reciprocal of the total resistance in a parallel circuitisequal to the sum ofthe re-ciprocals of the individual resistances.

11111

-=-+-+- ... +

-Rr R] R2 R3 Rn

5. Thetotal power consumed in aparallel cir-cuit is equal to the sum of the power con-sumed bythe individual resistors.

Pr =PR,+PR,+PR,... +PRn

EXAMPLE: Three resistors, 100 ohms, 220

ohms, and 470 ohms, are connected in parallel with abattery rated at 48 volts. Solve forall val-ues of the circuit.

First draw a schematic of the circuit andlistall known variables (Figure8-4).

Given: Iy=? Er= 48 volts R, =? Pr =?

R]= 100 ohms

R2 = 220 ohms R3

=

470 ohms

IR, = ? IR, = ? IR, = ? PR, =?

PR, = ?

CHAPTER 8 DC CIRCUITS

In solving the circuit for all values, the total resistance must be found first. Then the individ

-ual currents flowing through each branch of the circuit can be found. Once the current is known, the power dissipation across each resistor can be

determined.

Given:

~=7

RJ = 100 ohms R2=220 ohms R3=470 ohms

Solution:

III 1

-=-+-+-RT RJ R2 R3

1 1 1 1

-=--+--+--RT 100 220 470

~ = 59.98 ohms

The current (IR,) through resistor RJ is: Given: Solution: ER

IR =-'

, R[

48 1-

-R, - 100

RJ

=

100 ohms IR,

=

0.48 amp

The current (IR,) through resistor R2is: ER,=48 volts

Given: Solution: ER,

IR,=~

ER,

=

48 volts 1-- 48

R, - 220

IR,

=

0.218 amp R2

=

220 ohms

FIGURE 8-4

-R3= 470 Q R, = R2

=

100Q 220Q

The current (IR,) through resistor R3is:

Given: Solution:

ER

IR,= R3'

ER,

=

48 volts _{IR,= 470}48

IR,

=

0.102 amp R3=470 ohms

The total current is:

Given: Solution: IT

=

IR,+IR,+ IR,

4

=0.48 +0.218 +0.102

4

= 0.800amp

4=7

IR,= 0048 amp IR,= 0.218 amp IR,= 0.102 amp

The total current can also be found using

Ohm's law:

Given: Solution:

ET

I =

-T RT

48

1=

--T 59.98

4=7

ET=48volts

~ = 59.98ohms

4

= 0.800 amps Again, a difference occurs because of r ound-ing. The power dissipated by resistor R[ is:

Given: Solution:

PR,=7 PR,= IR,ER,

IR,

=

0.48 amp PR,= (0.48)(48) ER,= 48 volts PR,= 23.04 watts

The power dissipated by resistor R2is:

Given: Solution:

PR,

=

7 PR,

=

IR,ER,

IR,

=

0.218 amp PR,

=

(0.218)(48) ER,= 48 volts PR,= 10.46 watts

The power dissipated by resistor R3is:

Given: Solution: PR,= 7 PR,

=

IR,ER,

IR,= 0.102 amp PR,= (0.102)(48)

SECTION 1 DC CIRCUITS

Thetotal power is: Given:

P_T=?_.

PR, =23.04 watts PR, = 10.46watts PR, = 4.90 watts

Thetotal power canalsobe determined using

Ohm's law:

Solution:

PT = PR, +PR, +PR,

PT = 23.04 + 10.46 +4.90

PT =38.40watts

Given:

P_T=?_.

~ =0.80amp ET =48volts

Solution:

PT

=

ITET

Py = (0.80)(48) PT=38.4watts

8-

2

QUESTION

1. Four resistors, 2200 ohms, 2700 ohms,

3300 ohms, and 5600 ohms, are

connected in parallel with a battery rated at 9volts. Solve for all values of the circuit.

BIll

SERIES

-

PARALLEL

CIRCUITS

Mostcircuits consist of both series andparallel cir

-cuits. Circuits of this type are referred to asseries -parallel circuits (Figure 8-5). The solution of most series-parallel circuits is simply amatter ofapply

-ing the laws and rules discussed earlier. Series

formulas are applied to the series part of the

FIGURE 8-5

Series-parallel circuit

'tt.t'f

~ ---

-circuit, and parallel formulas are applied to the

parallel parts ofthe circuit.

EXAMPLE: Solve for all unknown quantities in

Figure 8-6. Given:

~ = 7, ET= 48 volts, ~ = 7, PT= 7 R] = 820ohms IR, = 7 ERJ = 7 PRJ = 7 R2= 330ohms IR, = 7 ER, = 7 PR, = 7 R3= 680ohms IR, = 7 ER, = 7 PR, = 7 R4= 470 ohms IR, = 7 ER, = 7 PR, = 7 Rs= 120ohms IR, = 7 ER, = 7 PR, = 7 R6= 560ohms IR6 = 7 ER6 = 7 PR6= 7 To solve for total resistance (~), first find the equivalent resistance (RA)for parallel resistors R2

and R3.Then solve forthe equivalent resistance of resistors RAand R4(identified as Rs) and Rsand R6(identified asRs,). Then the equivalent resist

-ance (RBTSca)n be determined for Rs,and Rs,. F

i-nally,find the total series resistance for R]and RB. Given:

R_A =?_.

R2= 330ohms

R3=680ohms

Solution:

1 1 1

--=--+-

-RA R2 R3

1 1 1

--

=

---

+-- -RA 330 680

RA=222.22 ohms

Redraw the circuit, substituting resistor RAfor

resistors R2and R3. See Figure 8-7.

Now determine series resistance Rs,for resis -tors RAand R4·

Given:

Rs,

=

7

RA=222.22 ohms

R4=470 ohms

Solution:

Rs,= RA

+

R4

Rs,= 222.22 +470

Rs,= 692.22 ohms

Determine series resistance Rs,for resistors

CHAPTER 8 DC CIRCUITS

FIGURE 8-6

R2= 330Q

-R4=470 Q

R3=680Q

R6= 560Q _{Rs= 120Q}

J

1 ~

~---~I~---~~ Er= 48 V

R1 = 820Q

FIGURE 8-7

R1=820Q

.

----

--~w~

----

~

RA=222.2 Q R4=470 Q_~----,

R6= 560Q Rs =120Q

L---li 1I---=---=---.J Er= 48V

FIGURE 8-8

..•

RS1 =692.2 Q

~

''I'

R1= 820Q _.u..

T'

:11

RS2= 680Q

Given: RS2 = ?

R5 = 120ohms R6

=

560 ohms

Solution:

RS2 = R5

+

R6 RS2

=

120 + 560 RS2

=

680 ohms

Given:

R_B =?_.

Rs, = 692.22 ohms

RS2 = 680 ohms

Solution:

1 1 1

-

=

-

+

-RB Rs, Rs,

1 1 1

---

+--RB 692.22 680 RB =343.64 ohms

Redraw the circuit using resistor RB.See

Fig-ure 8-9.

Now determine the total resistance in the circuit.

Given:

Ry =?

R] = 820 ohms

RB

=

343.64 ohms

Solution:

Ry =R[ +RB R,

=

820

+

343.64 Rr= 1163.64 ohms

Redraw the circuit with resistors Rs,and RS2' See Figure 8-8.

Now determine the parallel resistance (RB) for resistors Rs, and Rs2·

SECTION 1 DC CIRCUITS

FIGURE 8-9

.

""

...

,.

.,.

RB=343.64 Q

J

ET= 48 V IT

'---;: 111-------'----'

R1 = 820 Q

The totalcurrent in the circuit can now be de

-termined using Ohm's law.

Given: Solution:

ET

IT =

-RT

48

1=--

-T 1163.64

ly=0.0412amp or41.2 mA

ly=?

ET=48volts

R,

=

1163.64ohms

Thevoltage drop across resistorRl can now be determined.

Given:

IR, = 0.0412 amp

ER, =?

R. = 820 ohms

Solution:

ER,

IR, =~

ER 0.0412 =_1

820 ER, = 33.78 volts

The voltage drop acrossequivalent resistance

RB is:

Given:

IR•

=

0.0412 amp

ER•

=

?

RB = 343.64 ohms

I

Solution:

ER•

IR =

-• RB

ERB 0.0412

=

_--0.._

343.64 ERB = 14.158 volts

The current through each branch in a parallel circuit has to be calculated separately.

The current through branch resistance Rs,is:

Given: Solution:

ER

I =_s.

Rs, R

51

ERs, = 14.158 volts

14.158

I ---

-Rs, - 692.22

R51

=

692.22 ohms IRs,

=

0.0205 amps

The current through branch resistance Rs,is:

Given: Solution:

ER

I =_s,

Rs, R

5,

ERs, = 14.158 volts 14.158 680

R5, = 680 ohms IRs, = 0.0208 amp The voltage drop across resistors RA and R4

can now be determined.

Given: Solution:

IRA = 0.0205 amp IR =-ERA

A RA

ERA

0.0205 = ---.::...

-222.22

ERA =4.56 volts RA = 222.22 ohms

Given: Solution:

ER4

IR. =

-R4 ER

0.0205 =_4 470

ER4

=

9.64 volts

IR4

=

0.0205 amp

CHAPTER 8 DC CIRCUITS

The voltagedrop across resistorsR5andR6 is:

Given: Solution:

ER I =-'

R, R5 ER 0.0208 =-'

120 ER,

=

2.50volts

IR, = 0.0208 amp

R5 = 120ohms

Given: Solution:

ER

IR,

=

R6'

ER

0.0208 =-' 560 IR•

=

0.0208 amp

R6 = 560ohms ER• = 11.65 volts The current through equivalent resistanceRA

splitsthrough parallel branches with resistors R2

and Ry Current through each resistor has to be calculated separately.

Thecurrent through resistorR2is:

Given: Solution:

ER I =-'

R, R2

ER, = 4.56 volts _{IR, =} 4.56₃₃₀

IR, = 0.0138 amp R2 = 330ohms

Given: Solution:

ER,

IR =

-, R3 4.56 I

--R, - 680

R3 =680ohms IR, = 0.00671 amp

Now the power dissipation through each

re-sistorcan bedetermined. The power consumed by

resistor Rl is:

ER, = 4.56volts

Given:

PRl =?

IRl = 0.0412 amp ERl = 33.78 volts

Solution: PR, = IRIERl

PRl = (0.0412)(33.78) PRl= 1.39 watts

The power consumed byresistorR2 is: Given:

PR,

=

?

Solution: PR,= IR,ER,

IR,

=

0.0138 amp PR, = (0.0138)(4.56)

ER, = 4.56 volts PR, = 0.063watt or 63 mW

The power consumed by resistorR3is:

Given: Solution:

PR,

=

? PR,

=

IR,ER,

ER,

=

4.56 volts PR,

=

(0.00670)(4.56)

IR, = 0.00670 amp PR, = 0.031 watt or 31 mW The power consumed by resistorR4is: Given:

P_R4 =?_.

ER4 = 9.64 volts IR• = 0.0205 amp

Solution:

PR• =Ia.ER4

PR, = (0.0205)(9.64)

PR, = 0.20watt or200mW The power consumed byresistorR5is:

Given: Solution: PR, = ? PR, =IR,ER,

ER, = 2.50volts PR,

=

(0.0208)(2.50) IR,

=

0.0208 amp PR,

=

0.052 watt or 52mW

The power consumed byresistorR6is:

Given: Solution:

PR, = ? PR, = IR,ER,

ER,

=

11.65 volts PR,

=

(0.0208)(11.65)

IR, = 0.0208 amp Pa. = 0.24watt or 240mW Thetotal power consumption ofthe circuit is: Given:

P_T=?_.

PRl = 1.39 watts PR, = 0.063watt PR, = 0.031watt PR, = 0.20watt

PR,

=

0.052 watt PR, = 0.24watt

SECTION 1 DC CIRCUITS

FIGURE 8-10

,

•

.

,

•.•.

1

R2=330

n

R3=150o

1

I

,

•.

1

T"

R4=470

n

R7=270

n

Rl =1k.{2 _.•.,.•.

..

_,

•

.

'T R5=560

n

Rs =680

n

:11

Solution:

PT = PR, + PR, + PR, + PR,+ PR, + PR,

PT = 1.39 + 0.063+ 0.031+ 0.20+ 0.052+ 0.24 PT= 1.98 watts

The total power consumption could also be determined by the use of the power formula.

Given:

P_T=?_.

ET =48 volts I,

=

0.0413 amp

Solution:

PT =ETLr

PT

=

(48)(0.0413)

PT= 1.98 watts

8-3

QUESTION

1.Solvefor all unknown values in Figure 8-10.

11II

VOLTAGE DIVIDERS

One of the more common uses of resistors is voltage division, which is used extensively in electronics. Voltage dividers are often used to set a bias or operating point of various active electronic components such as transistors or in -tegrated circuits. A voltage divider is also used to convert a higher voltage to a lower one so that an instrument can read or measure a volt-age above its normal range. This is often re -ferred to asscaling.

In Chapters 4-6, the concept of connecting resistorsin serieswaspresented. Ohm's law (pre -se~ted inChapter 5)is one of the mostimportant concepts tounderstand because power formulas, voltage dividers, current dividers, and so on can all be traced back to and solved by it.Ohm's law statesthat the current through a circuitisdirectly proportional to the voltage acrossthe circuit and inverselyproportional to the resistance.

voltage current =

resistance

E

I=

-R

That the current is directly proportional to the voltage across the circuit isavery simple yet pro -found statement. Thefirst part of the statement simply means that as the voltagechanges, so does the current, and the change is in the same dir ec-tion (i.e.,if the voltage increases, then the cur -rent increases; similarly,ifthe voltage decreases, then the current decreases). The second part of Ohm'slaw states that current is inverselypropor -tional to resistance. Inversely proportional simply means that when one thing occurs, the opposite happens to the other quantity. In this case,if re -sistanceincreases, then current decreases.

Figure 8-11 depicts the simplest of circuits.It hasa singlevoltage source of10volts,whichis con -nected to a single 10-kn resistor.By Ohm's law, 1milliampere of current is flowingin this circuit.

CHAPTER 8 DC CIRCUITS

FIGURE 8-11

Simple circuit.

1=1 mAJ

R= 10 k.Q

Er=

-..=

10V

-FIGURE 8-12

-Simple voltage divider.

Er=-.. =-

10V-~ R1

=

-4 5kQ

J'

~

R2=

~-4 5k.Q 1= 1mA

E

1=-R

1=10110000

1=0.001 A or 1 mA

In Figure 8-12, the original io-kn resistor in

Figure 8-11 is shown divided into two equal

re-sistors, R, and R2.This circuit still has the same to

-tal resistance as Figure 8-11 and therefore has the

same amount of current flowing through it (i.e.. 1

mA). Ohm's law can be used to determine the

voltage across each of the resistors.

E] Ij

=-Rj 0.001

=

E/5000

5V= Ej

Therefore

E2

12=

-R2 0.001

=

E2/5000

5 V

=

E2

The voltage drop across R. and R2 has effectively

divided the source voltage of 10volts in half.

Now, let ustake Figure 8-12 another step and

divide resistor R2 into two resistors, R2and Ry

which are 4 kfl and 1 kG, as shown in Figure

8-13. As far as the voltage source is concerned,

nothing has really changed in the circuit because

the total resistance is still 10kG. There is still 1mA

of current flowing in the circuit. The important

difference is that now there are multiple voltage

drops. The voltage drop at the junction ofRj and

R2 is still 5 volts. However, at the junction of R2

and R3to the ground reference point, the voltage

drop is 1volt.

The product of this discussion is depicted in

Figure 8-14. It represents a 10-volt voltage source

connected to a 10-kfl potentiometer, orpot.Asthe

wiper moves up and down, the overall value of

the pot does not change asin the earlier examples.

However, the value of the resistance between the

wiper and the top of the pot does, thus effectively

creating a two-resistor divider of any value where

the resistors can be between zero and the value of

the pot. Thisisacommon control used to set vol

t-ages, audio levels (volume control), and so on. A

word of warning: This voltage divider action is

predictable only when these simple circuits are

connected to resistances greater than they are. For

example, if the I-volt drop in Figure 8-13 were

supplying a load less than approximately 10-kfl,

then, because it is in parallel with Ry the load

would reduce its value, thus changing the voltage

divider action (Figure 8-15).

What may not be obvious isthat voltage

di-viders work on a ratio principle. Look at Figure

8-12. Resistor R2is 5kfl, which just happens to

be half (or 50%) of the total network. It alsohap

-pens to have half (or 50%) of the total supplied

voltage dropped across it. Inspection of Figure

8-13 depicts a similar event occurring. Resistor R3

is 1kfl, which just happens to be one-tenth (or

10%) of the total network resistance. In this case,

the voltage dropped across it is 10% of the sup

SECTION 1 DC CIRCUITS

FIGURE 8-13

-Voltagedivider.

ET= _

10V

-1=1mAJ

FIGURE 8-14 Voltage divider application.

ET= _

1

0V-R= 10kQ

1=1mAJ

unchanged because the sum of the resistances R2 and R3make up R2 voltage, and their combined

value isstill5 kfl.

Ageneralization canbe stated for voltage di

-viders ofthis nature. Thevoltage drop is equal to thepercentage of thedropping resistortothe sum of the droppingnetwork.

Esource X Rorop

Eorop = R

Total

The next example (Figure 8-16) is a more

practical example of a real-world voltage divider.

The voltagesource isthe standard automotive ac -cessory power source of 14.8 volts. What is de

-R,=

5kQ

At---,

B t-------,

FIGURE 8-15 w _

Voltage divider with load.

R,=

5kQ

Er=

-

..

=

10V

-siredis avoltage-divider network that willdivide this 14.8-voltsource into a lower-voltage source specified by the manufacturer of a tape or CD player.Thefirst voltagedesiredis3volts,and the loadforthis voltage willdraw 150 mA.The sec

-ond voltage required is6volts,and this voltage's

loadwill draw 400 mA.The finalvoltage required

is 9 volts, and itsload draws 600 mA. Reviewing Figure 8-16 indicatesthat both series andparallel

networks are involved.Itisnecessaryto know the load currents for each voltage inadvance.

CHAPTER 8 DC CIRCUITS

FIGURE 8-16 Real-world voltage divider.

14.8

-.=

R4"

9V@600mA ,

6V @400mA

3V @150mA

'

"

-.=

At first glance, the design of this circuit may

seem very complex. In reality itis a simple process.

Refer to Figure 8-17. Selecting the amount of cur

-rent that is desired to flow through resistor R4

starts the process. In this example, an arbitrary value of 10 mA was chosen.

Next, inspect junction A between resistors R3

and R4.At this point, there is current summing or

addition. There is 10 mA of current flowing

through R4' and it is joined or summed with 150 mA of current through the 3-volt load, RL3.This

means that atotal of 160 mA ofcurrent is flowing

through resistor R3.

At junction B between resistors R2 and Ry

there is another summation process. The 400 mA

of current flowing through the 6-volt load now

joins the 160 mA of current through R3.The total amount of current flowing through resistor R2is

this summation, or 560 mA.

Junction Chas 560 mA of current entering it

through the divider chain, and itissummed with

the current from the 9-volt load, which is600 mA.

This summation becomes the total current flow

-ing through the final resistor Rl to be a value of

1160 mA, or 1.16 A.

The final step in the process is calculating the values of Rl' R2' R3' and R4.These calculations

in-volve nothing more than applying Ohm's law.

Again, start at the bottom ofthe network with re

-sistor R4.Atthis point in the circuit, it isdetermined

that there is only the previously selected 10 mA of

current flowmg. The desired voltage at junction A is

3volts. Therefore, the following relationships exist:

E4

I4=

-R4

3 0.01 =

-R4

300

n

= R4

The calculation ofR3is only slightly more dif

-ficult. From the previous calculations, the total

current flawing through the resistor was det

er-mined to be 160 mA. Remember that voltage

di-viders are ratios, or the relation of one potential point to another. The desired voltage junction Bis

SECTION 1 DC CIRCUITS

FIGURE 8-17

Analyzing the circuit.

Total1160mA

l

or U6A

10mA +150mA +400mA + _R1 600mA

I

0

9V@600mA ,

10mA +150mA +400mA R2

,

._~..~

--

=-0

-I

6V@400mA ,

10mA +150mA _R3

~

0

3V @150mA

10mA

~

R4

,

.

-

'-14.8

6 volts. However, this voltage is not used in the

calculation. Junction B is above junction A,which

has been established at 3 volts. Therefore, junc

-tion B is 6 V- 3 V,or 3 V greater than junction A.

This value is used in the calculation of resistor Rr

E3

13=

-R3

0.160 =~

R3

18.75 D= R3

The process is repeated in the calculation of

R2· The current flowing through R2is560mA.The

desired voltage atjunction A is 9 volts. Junction A

is 9 V- 6 V,or 3 volts more than at junction B.

E2

12 =

-R2

3 0.560 =

-R2

5.36 D =R2

The final calculation to determine the value of

resistor Rj is somewhat different. In this case, the

total current flowing through the circuits is1.16A.

In this example, the junction A desired voltage is 9

volts. The source voltage is 14.8 volts; therefore,

resistor RJ must drop 14.8V- 9 V, or 5.8volts.

EI

II=

-RI

5.8 1.16

=-Rj

5D=RI

As in the real world, the resistor values re

-quired in this circuit are not exact values but are

rounded (Figure 8-18).

NOTE: THE VALUES OFRESISTORS INDICATED BY THE COMPUTER SIMULATIONS WILL NOT PRECISELY MATCH THE CALCULATED VALUES. AGAIN, THIS ISA REAL-WORLD SIMULATION, AND THE RESISTORS' VALUES ARE NOT CARRIED PAST1OHM.

A REAL-WORLD AMMETER HAS SOME INTERNAL RESISTANCE ASSOCIATED WITH IT,AND A REAL-WORLD VOLTMETER HAS SOME CURRENT FLOWING THROUGH IT.

CHAPTER 8 DC CIRCUITS

FIGURE 8-18

Determining the resistor values.

.

-10mA +150mA +4oom;j

+600mA =1160mA

1OmA+150mA +400mA ,

=560mA

10mA +150mA =160mA ,

~ 14.8VDC-=

-10mA

BOTH OFTHESE DEVICES WILL CAUSE THE DISPLAYED VALUES TO BE SLIGHTLY DIFFERENT FROM THOSE CALCULATED.

COMPUTER SIMULATIONS USECALCULATED VALUES,

AND THIS PRESENTS A COMMON PROBLEM IN ELECTRONICS. COMPONENTS ARE MANUFACTURED IN STANDARD VALUES, AND USUALLY ONLY THESE VALUES ARE AVAILABLE WHEN A CIRCUIT NEEDSTOBECONSTRUCTED.

THE ONLY WAY TO OBTAIN A RESISTOR SUCH AS RESISTOR R2(5.36 D,)OR R3(18.75il) IS TO SPEND A LARGE AMOUNT OFMONEY FOR ACUSTOM-MADE RESISTOR. THESTUDENT SHOULD REALIZE THE GOAL IS ALMOST ALWAYS TO DESIGN AND BUILD PRACTICAL AND ECONOMICAL EQUIPMENT. THE USE OFPRECISION COMPONENTS OFTEN INCREASES COSTS WITH LlTILE OR NO BENEFIT.

8-4

QUESTIONS

1.What function does a voltage divider perform?

2.What is a common function for voltage dividers?

R1

5Q

0.

....---

,

\J

9V @ 600mA --,

R2

5.36Q

0

6V @ 400mA

--,

R3

18.75 Q

0

3V @ 150mA

R4 ~

300Q

3. What principle do voltage dividers work on?

4. What generalization can be stated for voltage dividers?

5. Design a voltage divider to use a 14.8-V

voltage source to control a 3 V load at

300mA.

_

WHEATSTONE BRIDGE

AWheatstone bridge isa measuring instrument

that wasinvented by SamuelHunter Christie in

1833. In 1843, a talented and versatile scientist

named SirCharles Wheatstone improved on the

original designof the Wheatstone bridge.He did

not claimtohave invented the circuit named af

-ter him,but he was the first touse the circuit ef

-fectively for making resistance measurements. TheWheatstone bridge consists essentially of

two voltage dividers (Figure8-19). It isused to measure anunknown electricalresistance byba

SECTION 1 DC CIRCUITS

FIGURE 8-19

AWheatstone bridge consists oftwo voltage dividers.

R •

x

-

=

-Voltmeter or

Ammeter

RA

~

.

component against another leg of known compo-nents with one being variable.

Inthe circuit,Rxisthe unknown resistor value.

Reis a potentiometer and is adjusteduntilthe vol t-age fromthe secondvoltagedivider is equalto volt -agefromthe voltagedividercontainingRx' When the voltagevalues areequal,thebridge is said to be balanced. The balance point can be detected by connecting either a voltmeter or an ammeter across the output terminals. Both meters willgive a zero reading when a balance isachieved.

NOTE: ADIGITAL METER WORKS WELL FORTHIS

APPLICATION.

Inabalanced circuit, the ratioR / R is equal_{x A} tothe ratioRB/Rc

Rx RB

RA Re

Therefore:

RA X RB

Rx=----=..:-~

Re

Inother words, ifthe values ofR_A' Rand_B R_e are known, it iseasy to calculate Rx' Re would

FIGURE 8-20

Wheatstone bridge schematic.

have to be measured with a meter to determine its value after adjusting for a zero reading. In an actual Wheatstone bridge test instruments r RA

andRBare fixed andRe is adjusted and R canbe easilyread off a sliding scale connected t~Rc

The circuitforaWheatstone bridge asdesigned by Wheatstone is shown in Figure 8-20. This is how the circuitisdrawnfor applicationsusing it.

Variations on the Wheatstone bridge can be used to measure capacitance, inductance, and im

-pedance. Wheatstone bridge circuits are rarely used todayto measure resistance values. Theyare now used for designing sensing circuits such as: strain gauges for transforming a strain applied,to aproportional change of resistance and are widely usedinindustry; variometers todetect changesin air pressure to alerting glider pilots to updrafts or thermals so height can be gained allowing for longer flights; and explosimeters for samplingthe amount of combustible gases ina space.

8-5

QUESTIONS

1.Who invented the Wheatstone bridge? 2. What is the Wheatstone bridge circuit

CHAPTER 8 DC CIRCUITS

3. When the meter reads zero, what does this mean?

4. What is the value ofRx when RAand RB are 10kfl each and Rcis 96,432 fl? 5.What are other quantities that can be

measured using the Wheatstone bridge concept?

6.What applications areused today for the Wheatstone bridge?

SUMMARY

• Aseries circuit provides only one path for

current flow.

• Formulas governing the operation ofa

series circuit include:

IT

=

IR•

=

IR,

=

IRJ

=

IR" RT

=

RI

+

R2

+

R3

+

Rn

ET

=

ER•

+

ER,

+

ERJ

+

ERn

1= E/R

PT

=

PR•

+

PR,

+

PRJ ...

+

PRo

• A parallel circuitprovides more than one

path for current flow.

• Formulas governing the operation of a

parallel circuit include:

IT

=

IR•

+

IR,

+

IRJ •.•

+

IR"

1 1 1 1 1

-

=

-

+-

+

-

.

.

.

+-RT Rl R2 R3 Rn

ET = ER• = ER, = ERJ ... = ERn 1= E/ R

PT

=

PR•

+

PR2

+

PRJ.•.

+

PR"

• Series-parallel circuits aresolved by using

seriesformulas for the series parts of the

circuit and parallel formulas for the

parallel parts ofthe circuit.

• Voltage dividers are used to set the bias or

operating point of activeelectronic

components.

• A Wheatstone bridge is used to measure

anunknown electrical resistance.

CHAPTER

8

SELF-TEST.

"

R,=150 0

1.Solve for all unknown quantities in the circuits shown.

(A)

R2=3000

(C)

(B)

R,=1500 R2=3000

R,=1000