4

CHAPTER

Direct-Current Circuit Basics

IN THIS CHAPTER, YOU’LL LEARN MORE ABOUT CIRCUIT DIAGRAMS, WHICH ENGINEERS AND TECHnicians call schematic diagrams, or simply schematics, because they detail the schemes for designing and assembling circuits. You’ll also learn more about how current, voltage, resistance, and power interact in simple DC circuits.

Schematic Symbols

When we want to denote an electrical conductor such as a wire, we draw a straight, solid line either horizontally across, or vertically up and down, the page. We can “turn a corner” when we draw a conductor line, but we should strive to minimize the total number of “corners” in a diagram. By following this convention, we can keep schematics neat and ensure that they’re easy for others to read.

When two conductor lines cross, assume that they do not connect at the crossing point unless you see a heavy, solid dot where they meet. Whenever you draw a “connecting dot,” make it clearly visible, no matter how many conductors meet at the junction.

We portray a resistor by drawing a “zig-zag,” as shown in Fig. 4-1A. We portray a two-terminal variable resistor or potentiometer by drawing a “zig-zag” with an arrow through it (Fig. 4-1B). We portray a three-terminal potentiometer by drawing a “zig-zag” with an arrow pointing sideways at it (Fig. 4-1C).

4-1   Schematic symbols for a fixed resistor (A), a two-terminal variable resistor (B), and a three-terminal potentiometer (C).

We symbolize an electrochemical cell by drawing two parallel lines, one longer than the other (Fig. 4-2A). The longer line represents the positive (+) terminal, while the shorter line represents the negative (−) terminal. We symbolize a battery, which is a combination of two or more cells in series, by drawing several parallel lines, alternately long and short (Fig. 4-2B). As with the cell, the longer end line represents the positive terminal and the shorter end line represents the negative terminal.

4-2   Schematic symbols for a single electrochemical cell (A) and a multiple-cell electrochemical battery (B).

We can portray meters as circles. Sometimes the circle has an arrow inside it, and the meter type, such as mA (milliammeter) or V (voltmeter) is written alongside the circle, as shown in Fig. 4-3A. Sometimes the meter type is denoted inside the circle, and no arrow appears (Fig. 4-3B). It doesn’t matter which way you draw meters in your schematics, as long as you keep the style consistent throughout your work.

4-3   Meter symbols can have the designator either outside the circle (A) or inside (B). In this case, both symbols represent a milliammeter (mA).

Some other common symbols include the incandescent lamp, the capacitor, the air-core coil, the iron-core coil, the chassis ground, the earth ground, the AC source, the set of terminals, and the black box (specialized component or device), a rectangle with the designator written inside. These symbols appear in Fig. 4-4.

4-4   Schematic symbols for an incandescent lamp (A), a fixed capacitor (B), a fixed inductor with air core (C), fixed inductor with laminated-iron core (D), chassis ground (E), earth ground (F), a signal generator or source of alternating current (G), a pair of terminals (H), and a specialized component or device (I).

Schematic and Wiring Diagrams

Schematics illustrate the interconnections among components in a circuit or system, but the actual values of the components are not necessarily indicated. You might see a diagram of a radio-frequency (RF) power amplifier with resistors, capacitors, coils, and transistors, but without any data concerning the values or ratings of the components. That’s a schematic diagram, but it’s not a wiring diagram. The schematic tells you the scheme for the circuit, but you can’t wire up the circuit and make it work because you don’t have enough information.

Suppose you want to build a certain amplifier circuit. You go to an electronics store to get the parts. What values of resistors should you buy? How about capacitors? What type of transistor will work best? Must you wind the coils, or can you get them ready-made? Should you install test points for the benefit of technicians who might have to repair the amplifier someday? How many watts should the potentiometers be able to handle? A wiring diagram tells you all of these things.

Circuit Simplification

We can simplify most DC circuits to three major components: a voltage source, a set of conductors, and a resistance, as shown in Fig. 4-5. We call the source voltage E (or sometimes V), the current I, and the resistance R. The standard units for these components are the volt (V), the ampere (A), and the ohm, respectively. Italicized letters represent mathematical variables (voltage, current, and resistance in this case). Non-italicized characters represent abbreviations for physical units.

4-5   Basic elements of a DC circuit with voltage E, current I, and resistance R.

We already know that a relationship exists between the voltage, current, and resistance in a DC circuit. If one of these parameters changes, then one or both of the others will also change. If we make the resistance smaller, the current will get larger. If we increase the EMF, the current will also increase. If the current in the circuit increases, the voltage across the resistor will increase. Ohm’s law comprises a simple set of formulas defining the relationship between these three quantities.

Ohm’s Law

Scientists gave Ohm’s Law its name in honor of Georg Simon Ohm, a German physicist who (according to some historians) first expressed it in the 1800s. To calculate the voltage when we know the current and the resistance, use the formula

E = IR

To calculate the current when we know the voltage and the resistance, use

I = E/R

To calculate the resistance when we know the voltage and the current, use

R = E/I

You need only remember one of these formulas to derive the other two. You can arrange the three variables geometrically into an Ohm’s Law triangle, as shown in Fig. 4-6. When you want to find the formula for a particular parameter, cover up its symbol and read the positions of the others.

4-6   The Ohm’s Law triangle showing voltage E, current I, and resistance R, expressed in volts, amperes, and ohms respectively.

If you want Ohm’s Law to produce the correct results, you must use the proper units. Under most circumstances, you’ll want to use the standard units of volts, amperes, and ohms. If you use volts, milliamperes (mA), and ohms, or if you use kilovolts (kV), microamperes (μA), and megohms (M), you can’t expect to get the right answers. If you see initial quantities in units other than volts, amperes, and ohms, you should convert to these standard units before you begin your calculations. After you’ve done all the arithmetic, you can convert the individual units to whatever you like. For example, if you get 13,500,000 ohms as a calculated resistance, you might prefer to call it 13.5 megohms. But in the calculation, you should use the number 13,500,000 (or 1.35 × 10⁷) and stay with units of ohms.

Current Calculations

In order to determine the current in a circuit, we must know the voltage and the resistance, or be able to deduce them. Figure 4-7 illustrates a generic circuit with a variable DC generator, a voltmeter, some wire, an ammeter, and a potentiometer.

4-7   A circuit for doing calculations with Ohm’s Law.

Problem 4-1

Suppose that the DC generator in Fig. 4-7 produces 36 V and we set the potentiometer to a resistance of 18 ohms. What’s the current?

Solution

Use the formula I = E/R. Plug in the values for E and R in volts and ohms, getting

I = E/R = 36/18 = 2.0 A

Problem 4-2

Imagine that the DC generator in Fig. 4-7 produces 72 V and the potentiometer is set to 12 k. What’s the current?

Solution

First, convert the resistance to ohms, getting 12 k = 12,000 ohms. Then input the values in volts and ohms to get

I = E/R = 72/12,000 = 0.0060 A = 6.0 mA

Problem 4-3

Suppose that we adjust the DC generator in Fig. 4-7 so that it produces 26 kV, and we adjust the potentiometer so that it has a resistance of 13 M. What’s the current?

Solution

First, change the resistance value from 13 M to 13,000,000 ohms. Then change the voltage value from 26 kV to 26,000 V. Finally, plug the voltage and resistance into the Ohm’s Law formula, getting

I = E/R = 26,000/13,000,000 = 0.0020 A = 2.0 mA

Voltage Calculations

We can use Ohm’s Law to calculate the DC voltage between two points when we know the current and the resistance.

Problem 4-4

Suppose we set the potentiometer in Fig. 4-7 to 500 ohms, and we measure the current as 20 mA. What’s the DC voltage?

Solution

Use the formula E = IR. First, convert the current to amperes: 20 mA = 0.020 A. Then multiply the current by the resistance to obtain

E = IR = 0. 020 × 500 = 10 V

Problem 4-5

We set the potentiometer in Fig. 4-7 to 2.33 k, and we get 250 mA of current. What’s the voltage?

Solution

Before doing any arithmetic, we convert the resistance and current to ohms and amperes. A resistance of 2.33 k equals 2,330 ohms, and a current of 250 mA equals 0.250 A. Now we can calculate the voltage as

E = IR = 0.250 × 2,330 = 582.5 V

We can round this result up to 583 V.

Problem 4-6

We set the potentiometer in Fig. 4-7 to get 1.25 A of current, and we measure the resistance as 203 ohms. What’s the voltage?

Solution

These values are both in standard units. Input them directly to get

E = IR = 1.25 × 203 = 253.75 V = 254 V

We can (and should) round off because we can’t justify a result that claims more precision than the data that we start with.

The Rule of Significant Figures

Competent engineers and scientists go by the rule of significant figures, also called the rule of significant digits. After completing a calculation, we always round the answer off to the least number of digits given in the input data numbers.

If we follow this rule in Problem 4-6, we must round off the answer to three significant digits, getting 254 V. That’s because the resistance (203 ohms) is specified only to that level of accuracy. If the resistance were given as 203.0 ohms, then we would again round off the answer to 254 V. If the resistance were given as 203.00 ohms, then we could still state the answer to only three significant digits, getting 254 V because we know the current only to three significant digits.

This rule takes some “getting-used-to” if you haven’t known about it or practiced it before. But after a while, you’ll use it “automatically” without giving it much thought.

Resistance Calculations

We can use Ohm’s Law to calculate the resistance between two points when we know the voltage and the current.

Problem 4-7

If the voltmeter in Fig. 4-7 reads 12 V and the ammeter shows 2.0 A, what’s the resistance of the potentiometer?

Solution

Use the formula R = E/I. We can plug in the values directly because they’re expressed in volts and amperes. It works out as

R = E/I = 12/2.0 = 6.0 ohms

Problem 4-8

What’s the value of the resistance in Fig. 4-7 if the current equals 24 mA and the voltage equals 360 mV?

Solution

First, convert to amperes and volts, obtaining I = 0.024 A and E = 0.360 V. Then plug the numbers into the Ohm’s Law equation to get

R = E/I = 0.360/0.024 = 15 ohms

Problem 4-9

Suppose that the ammeter in Fig. 4-7 reads 175 μA and the voltmeter indicates 1.11 kV. What’s the resistance?

Solution

Convert to amperes and volts, getting I = 0.000175 A and E = 1110 V. Then input these numbers, rounding off to get

R = E/I = 1110/0.000175 = 6,342,857 ohms = 6.34 M

Power Calculations

We can calculate the power P in a DC circuit, such as the one in Fig. 4-7, using the formula

P = EI

If we aren’t given the voltage directly, we can calculate it if we know the current and the resistance. Recall the Ohm’s Law formula for obtaining voltage:

E = IR

If we know I and R but we don’t know E, we can get the power P as

P = EI = (IR)I = I²R

If we know only E and R but don’t know I, we can restate I as

I = E/R

Then we can substitute into the voltage-current power formula to obtain

P = EI = E(E/R) = E²/R

Problem 4-10

Suppose that the voltmeter in Fig. 4-7 reads 15 V and the ammeter shows 70 mA. How much power does the potentiometer dissipate?

Solution

Use the formula P = EI. First, convert the current to amperes, getting I = 0.070 A. (The last 0 counts as a significant digit.) Then multiply by 15 V, getting

P = EI = 15 × 0.070 = 1.05 W

The input data only has two significant digits, while this answer, as it stands, has three. Rounding up gives 1.1 A. That’s the number we should use.

Problem 4-11

If the resistance in the circuit of Fig. 4-7 equals 470 ohms and the voltage source delivers 6.30 V, what’s the power dissipated by the potentiometer?

Solution

We don’t have to do any unit conversions. Plug in the values directly and then do the arithmetic to get

P = E²/R = 6.30 × 6.30 / 470 = 0.0844 W = 84.4 mW

Problem 4-12

Suppose that the resistance in Fig. 4-7 is 33 k and the current is 756 mA. What’s the power dissipated by the potentiometer?

Solution

We can use the formula P = I²R after converting to ohms and amperes: R = 33,000 and I = 0.756. Then calculate and round off to get

P = 0.756 × 0.756 × 33,000 = 18,861 W = 18.9 kW

Obviously, a common potentiometer can’t dissipate that much power! Most potentiometers are rated at 1 W or so.

Problem 4-13

How much voltage would we need to drive 60.0 μA through 33.0 k?

Solution

These input numbers both have three significant figures because the zeros on the far right are important. (Without them, you’d only have two significant figures in your values.) Use Ohm’s Law to find the voltage after converting to amperes and ohms, obtaining

E = IR = 0.0000600 × 33,000 = 1.98 V

Resistances in Series

When we connect two or more resistances in series, their ohmic values add up to get the total (or net) resistance.

Problem 4-14

We connect three resistors in series with individual resistances of 220 ohms, 330 ohms, and 470 ohms, as shown in Fig. 4-8. What’s the net resistance of the combination?

4-8   Three resistors in series. Illustration for Problem 4-14. All resistances are expressed in ohms.

Solution

Because we know all the values in ohms, we can add without doing any unit conversions to get

R = 220 + 330 + 470 = 1020 ohms = 1.02 k

That’s the pure theory. But when we build a real-life circuit, the exact resistances depend on the component tolerances: how much we should expect the actual values to vary, as a result of manufacturing quirks, from the values specified by the vendor.

Resistances in Parallel

We can evaluate resistances in parallel by considering them as conductances instead. Engineers express conductance in units called siemens, symbolized S. (The word “siemens” serves both in the singular and the plural sense). Some older physics or engineering documents use the mho (“ohm” spelled backwards) as the fundamental unit of conductance; the mho and the siemens represent the same thing. When we connect conductances in parallel, their values add up, just as resistances add up in series. If we change all the ohmic values to siemens, we can add these figures up and convert the result back to ohms.

Engineers use the uppercase, italic letter G to symbolize conductance as a parameter or mathematical variable. The conductance in siemens equals the reciprocal of the resistance in ohms. We can express this fact using two formulas, assuming that neither R nor G ever equals zero:

G = 1/R

and

R = 1/G

Problem 4-15

Consider five resistors in parallel. Call the resistors R₁ through R₅, and call the total resistance R, as shown in Fig. 4-9. Suppose that the individual resistors have values of R₁ = 10 ohms, R₂ = 20 ohms, R₃ = 40 ohms, R₄ = 50 ohms, and R₅ = 100 ohms. What’s the total resistance R of this parallel combination? (Note that we should not italicize R when it means “resistor” as a physical object, but we should italicize R when it means “resistance” as in a mathematical variable.)

4-9   Five resistors R₁ through R₅, connected in parallel, produce a net resistance R. Illustration for Problems 4-15 and 4-16.

Solution

To solve this problem, start by converting the resistances to conductances by taking their reciprocals. We’ll get

When we add these numbers, we obtain

G = 0.10 + 0.050 + 0.025 + 0.020 + 0.0100 = 0.205 S

The total resistance, rounded to two significant figures, turns out as

R = 1/G = 1/0.205 = 4.9 ohms

We can calculate the net resistance of a parallel combination directly, but the arithmetic can get messy. Refer again to Fig. 4-9. The resistances combine according to the formula

R = 1/(1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅)

Once in a while, you’ll encounter a situation where you have multiple resistances in parallel and their values are all equal. In a case of that sort, the total resistance equals the resistance of any one component divided by the number of components. For example, two 80-ohm resistors combine in parallel to yield a net resistance of 80/2 = 40 ohms; four of the same resistors combine in parallel to produce 80/4 = 20 ohms; five of them combine in parallel to give you 80/5 = 16 ohms.

Problem 4-16

We have five resistors R₁ through R₅ connected in parallel, as shown in Fig. 4-9. Suppose that each one of the resistances is 1.800 k. What’s the total resistance, R, of this combination?

Solution

Here, we can convert the resistances to 1800 ohms and then divide by 5 to get

R = 1800/5 = 360.0 ohms

We’re entitled to four significant figures here because we know the input value as stated, 1.800 K, to that many digits. We can treat the divisor 5 as exact, accurate to however many significant digits we want because the arrangement contains exactly five resistors.

Division of Power

When we connect sets of resistors to a source of voltage, each resistor draws some current. If we know the voltage, we can figure out how much current the entire set demands by calculating the net resistance of the combination, and then considering the combination as a single resistor.

If the resistors in the network all have the same ohmic value, the power from the source divides up equally among them, whether we connect the resistors in series or in parallel. For example, if we have eight identical resistors in series with a battery, the network consumes a certain amount of power, each resistor bearing 1/8 of the load. If we rearrange the circuit to connect the resistors in parallel with the same battery, the network as a whole dissipates more power than it does when the resistors are in series, but each individual resistor handles 1/8 of the total power, just as when they’re in series.

If the resistances in a network do not all have identical ohmic values, then some resistors dissipate more power than others.

Resistances in Series-Parallel

We can connect sets of resistors, all having identical ohmic values, in parallel sets of series networks, or in series sets of parallel networks. In either case, we get a series-parallel network that can greatly increase the total power-handling capacity of the network over the power-handling capacity of a single resistor.

Sometimes, the total resistance of a series-parallel network equals the value of any one of the resistors. This happens if the components are all identical, and are arranged in a network called an n-by-n (or n × n) matrix. That means when n is a whole number, we have n series-connected sets of n resistors with all the sets connected in parallel (Fig. 4-10A), or else n parallel-connected sets of n resistors, all connected in series (Fig. 4-10B). In practice, these arrangements yield identical results.

4-10   Two examples of series-parallel resistance matrices. At A, sets of series resistors join in parallel. At B, sets of parallel resistances join in series. These examples show symmetrical n-by-n matrices, where n = 3.

A series-parallel array of n × n resistors, all having identical ohmic values and identical power ratings, has n² times the power-handling capability of any resistor by itself. For example, a 3 × 3 series-parallel matrix of 2 W resistors can handle up to 3² × 2 = 9 × 2 = 18 W. If we have a 10 × 10 array of 1/2 W resistors, then it can dissipate up to 10² × 1/2 = 50 W. Simply multiply the power-handling capacity of each individual resistor by the total number of resistors in the matrix.

The above-described scheme works if, but only if, all of the resistors have identical ohmic values and identical power-dissipation ratings. If the resistors have values and/or ratings that differ (even slightly), one of the components might draw more current than it can withstand, so it will burn out. Then the current distribution in the network will change, increasing the likelihood that a second resistor will fail. We can end up with a chain reaction of component destruction!

If we need a resistor that can handle 50 W and a certain series-parallel network will handle 75 W, that’s fine. But we shouldn’t “push our luck” and expect to get away with a network that will handle only 48 W in the same application. We should allow some extra tolerance, say 10 percent over the minimum rating. If we expect the network to dissipate 50 W, we should build it to handle 55 W, or a bit more. We don’t have to engage in “overkill,” however. We’ll waste resources if we build a network that can handle 500 W when we only expect it to cope with 50 W—unless that’s the only convenient combination we can cobble together with resistors we have on hand.

Non-symmetrical series-parallel networks, made up from identical resistors, can increase the power-handling capability over that of a single resistor. But in these cases, the total resistance differs from the value of any individual resistor. To obtain the overall power-handling capacity, we can always multiply the power-handling capacity of any individual resistor by the total number of resistors, whether the network is symmetrical or not—again, if and only if, all the resistors have identical ohmic values and identical power-dissipation ratings.

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.  We have an unlimited supply of 33-ohm resistors, each one capable of dissipating 0.50 W. We want a 33-ohm resistor that can dissipate 18 W (a figure that includes a 2-W safety margin). We can get that component by wiring up

(a)  a 6 × 6 series-parallel matrix of individual resistors.

(b)  a 9 × 4 series-parallel matrix of individual resistors.

(c)  a 3 × 12 series-parallel matrix of individual resistors.

(d)  Any of the above

2.  We connect a 6.30-V lantern battery across a 330-ohm resistor. The resistor dissipates

(a)  19.0 mW of power.

(b)  8.31 mW of power.

(c)  120 mW of power.

(d)  We need more information to calculate it.

3.  If we connect 10 components in parallel, each one with a DC conductance of 0.15 S, what’s the net DC conductance of the combination?

(a)  0.015 S

(b)  0.15 S

(c)  1.5 S

(d)  15 S

4.  We have an unlimited supply of 100-ohm resistors, each one capable of dissipating 1.00 W. We want a resistance of 100 ohms capable of dissipating up to 12 W (a figure that includes a 2.5-W safety margin). Which of the following circuits is the smallest n × n matrix that will work here?

(a)  A 5 × 5 matrix

(b)  A 4 × 4 matrix

(c)  A 3 × 3 matrix

(d)  A 2 × 2 matrix

5.  If we connect a 6.3-V battery across a 330-ohm resistor, the current is

(a)  72 mA.

(b)  36 mA.

(c)  12 mA.

(d)  19 mA.

6.  The voltage across a resistor is 2.2 V. The resistor dissipates 400 mW. What’s its resistance?

(a)  12 ohms

(b)  24 ohms

(c)  48 ohms

(d)  96 ohms

7.  If we connect eight resistors in parallel, all identical and each with a value of 1.100 k, we get a component with a resistance of

(a)  8800 ohms.

(b)  4840 ohms.

(c)  1100 ohms.

(d)  137.5 ohms.

8.  We wire up three resistors in parallel: 600 ohms, 300 ohms, and 200 ohms. Then we connect a 12-V battery across the combination. How much current does the 300-ohm resistor draw all by itself?

(a)  80 mA

(b)  40 mA

(c)  33 mA

(d)  11 mA

9.  If we decrease the conductance of a resistor by a factor of 16 while leaving it connected to a source of constant DC voltage, then the power that the resistor dissipates will

(a)  decrease by a factor of 16.

(b)  decrease by a factor of 4.

(c)  increase by a factor of 4.

(d)  increase by a factor of 16.

10.  If we double the DC voltage across a resistor and double its resistance as well, then the power that the resistor dissipates will

(a)  get cut in half.

(b)  stay the same.

(c)  double.

(d)  quadruple.

11.  If we double the DC voltage across a resistor and double its resistance as well, then the current that the resistor draws will

(a)  get cut in half.

(b)  stay the same.

(c)  double.

(d)  quadruple.

12.  If we know the current through a component (in amperes) and its resistance (in ohms), how can we calculate the energy (in joules) that the component consumes?

(a)  Square the current and then multiply by the resistance.

(b)  Multiply the current by the resistance.

(c)  Divide the resistance by the current.

(d)  We need more information to do it.

13.  Suppose that 33.300 mA DC flows through a resistance of 3.333333 k. How can we best express the voltage across this resistance, taking significant figures into account?

(a)  111 V

(b)  111.0 V

(c)  111.00 V

(d)  110.999 V

14.  If a potentiometer carries 18.5 mA DC and we set its resistance to 1.12 k, how much power does it dissipate?

(a)  383 mW

(b)  20.7 mW

(c)  60.5 mW

(d)  67.8 mW

15.  We wire up seven 70.0-ohm resistors in parallel, and then connect a 12.6-V battery across the whole combination. How much current gets drawn from the battery?

(a)  25.7 mA

(b)  1.26 A

(c)  794 mA

(d)  180 mA

16.  We remove three of the resistors from the circuit in Question 15. What will happen to the current drawn by any one of the remaining four resistors?

(a)  It will go down to zero.

(b)  It will become 4/7 of its previous value.

(c)  It will stay the same.

(d)  It will become 7/4 of its previous value.

17.  We connect resistors with values of 180, 270, and 680 ohms in series with a 12.6-V battery. How much power does the set of resistors dissipate as a whole?

(a)  7.12 W

(b)  89.7 W

(c)  11.2 mW

(d)  140 mW

18.  The three primary units that engineers use when working with DC systems are the

(a)  ampere, volt, and ohm.

(b)  watt, joule, and volt.

(c)  siemens, ampere, and joule.

(d)  erg, joule, and ohm.

19.  A direct current of 3.00 A flows through a component whose conductance is 0.250 S. What’s the voltage across the component?

(a)  0.750 V

(b)  12.0 V

(c)  36.0 V

(d)  We need more information to calculate it.

20.  A direct current of 3.00 A flows through a component whose conductance is 0.250 S. How much power does the component dissipate?

(a)  750 mW

(b)  2.25 W

(c)  36.0 W

(d)  We need more information to calculate it.