To make numbers easier to read, some international scientific committees have recommended the practice of separating digits into groups of three to the right and to the left of decimal points, as in 64 325.473 53. No separation is necessary, however, for just four digits, and they are preferably not separated. For example, either 4138 or 4 138 is acceptable, as is 0.1278 or 0.127 8, with 4138 and 0.1278 preferred. The international committees did not approve of the use of the comma to separate digits because in some countries the comma is used in place of the decimal point. This digit grouping is used throughout this book.
The International System of Units (SI) is the international measurement language. SI has nine base units, which are shown in Table 1-1 along with the unit symbols. Units of all other physical quantities are derived from these.
Table 1-1
There is a decimal relation, indicated by prefixes, among multiples and submultiples of each base unit. An SI prefix is a term attached to the beginning of an SI unit name to form either a decimal multiple or submultiple. For example, since “kilo” is the prefix for one thousand, a kilometer equals 1000 m. And because “micro” is the SI prefix for one-millionth, one microsecond equals 0.000 001 s.
The SI prefixes have symbols as shown in Table 1-2, which also shows the corresponding powers of 10. For most circuit analyses, only mega, kilo, milli, micro, nano, and pico are important. The proper location for a prefix symbol is in front of a unit symbol, as in km for kilometer and cm for centimeter.
Table 1-2
Scientists have discovered two kinds of electric charge: positive and negative. Positive charge is carried by subatomic particles called protons, and negative charge by subatomic particles called electrons. All amounts of charge are integer multiples of these elemental charges. Scientists have also found that charges produce forces on each other: Charges of the same sign repel each other, but charges of opposite sign attract each other. Moreover, in an electric circuit there is conservation of charge, which means that the net electric charge remains constant—charge is neither created nor destroyed. (Electric components interconnected to form at least one closed path comprise an electric circuit or network.)
The charge of an electron or proton is much too small to be the basic charge unit. Instead, the SI unit of charge is the coulomb with unit symbol C. The quantity symbol is Q for a constant charge and q for a charge that varies with time. The charge of an electron is –1.602 × 10–19 C and that of a proton is 1.602 × 10–19 C. Put another way, the combined charge of 6.241 × 1018 electrons equals —1 C, and that of 6.241 × 1018 protons equals 1 C.
Each atom of matter has a positively charged nucleus consisting of protons and uncharged particles called neutrons. Electrons orbit around the nucleus under the attraction of the protons. For an undisturbed atom the number of electrons equals the number of protons, making the atom electrically neutral. But if an outer electron receives energy from, say, heat, it can gain enough energy to overcome the force of attraction of the protons and become a free electron. The atom then has more positive than negative charge and is a positive ion. Some atoms can also “capture” free electrons to gain a surplus of negative charge and become negative ions.
Electric current results from the movement of electric charge. The SI unit of current is the ampere with unit symbol A. The quantity symbol is I for a constant current and i for a time-varying current. If a steady flow of 1 C of charge passes a given point in a conductor in 1 s, the resulting current is 1 A. In general,
in which t is the quantity symbol for time.
Current has an associated direction. By convention the direction of current flow is in the direction of positive charge movement and opposite the direction of negative charge movement. In solids only free electrons move to produce current flow —the ions cannot move. But in gases and liquids, both positive and negative ions can move to produce current flow. Since electric circuits consist almost entirely of solids, only electrons produce current flow in almost all circuits. But this fact is seldom important in circuit analyses because the analyses are almost always at the current level and not the charge level.
In a circuit diagram each I (or i) usually has an associated arrow to indicate the current reference direction, as shown in Fig. 1-1. This arrow specifies the direction of positive current flow, but not necessarily the direction of actual flow. If, after calculations, I is found to be positive, then actual current flow is in the direction of the arrow. But if I is negative, current flow is in the opposite direction.
Fig. 1-1
Fig. 1-2
A current that flows in only one direction all the time is a direct current (dc), while a current that alternates in direction of flow is an alternating current (ac). Usually, though, direct current refers only to a constant current, and alternating current refers only to a current that varies sinusoidally with time.
A current source is a circuit element that provides a specified current. Figure 1-2 shows the circuit diagram symbol for a current source. This source provides a current of 6 A in the direction of the arrow irrespective of the voltage (discussed next) across the source.
The concept of voltage involves work, which in turn involves force and distance. The SI unit of work is the joule with unit symbol J, the SI unit of force is the newton with unit symbol N, and of course the SI unit for distance is the meter with unit symbol m.
Work is required for moving an object against a force that opposes the motion. For example, lifting something against the force of gravity requires work. In general the work required in joules is the product of the force in newtons and the distance moved in meters:
where W, F, and s are the quantity symbols for work, force, and distance, respectively.
Energy is the capacity to do work. One of its forms is potential energy, which is the energy a body has because of its position.
The voltage difference (also called the potential difference) between two points is the work in joules required to move 1 C of charge from one point to the other. The SI unit of voltage is the volt with unit symbol V. The quantity symbol is For v, although E and e are also popular. In general,
The voltage quantity symbol V sometimes has subscripts to designate the two points to which the voltage corresponds. If the letter a designates one point and b the other, and if W joules of work are required to move Q coulombs from point b to a, then Vab = W/Q. Note that the first subscript is the point to which the charge is moved. The work quantity symbol sometimes also has subscripts as in Vab = Wab/Q.
If moving a positive charge from b to a (or a negative charge from a to b) actually requires work, the point a is positive with respect to point b. This is the voltage polarity. In a circuit diagram this voltage polarity is indicated by a positive sign (+) at point a and a negative sign (–) at point b, as shown in Fig. 1-3a for 6 V. Terms used to designate this voltage are a 6-V voltage or potential rise from b to a or, equivalently, a 6-V voltage or potential drop from a to b.
Fig. 1-3
If the voltage is designated by a quantity symbol as in Fig. 1-3b, the positive and negative signs are reference polarities and not necessarily actual polarities. Also, if subscripts are used, the positive polarity sign is at the point corresponding to the first subscript (a here) and the negative polarity sign is at the point corresponding to the second subscript (b here). If after calculations, Vab is found to be positive, then point a is actually positive with respect to point b, in agreement with the reference polarity signs. But if Vab is negative, the actual polarities are opposite those shown.
A constant voltage is called a dc voltage. And a voltage that varies sinusoidally with time is called an ac voltage.
A voltage source, such as a battery or generator, provides a voltage that, ideally, does not depend on the current flow through the source. Figure l-4a shows the circuit symbol for a battery. This source provides a dc voltage of 12 V. This symbol is also often used for a dc voltage source that may not be a battery. Often, the + and – signs are not shown because, by convention, the long end-line designates the positive terminal and the short end-line the negative terminal. Another circuit symbol for a dc voltage source is shown in Fig. 1-4b. A battery uses chemical energy to move negative charges from the attracting positive terminal, where there is a surplus of protons, to the repulsing negative terminal, where there is a surplus of electrons. A voltage generator supplies this energy from mechanical energy that rotates a magnet past coils of wire.
Fig. 1-4
The sources of Figs. 1-2 and 1-4 are independent sources. An independent current source provides a certain current, and an independent voltage source provides a certain voltage, both independently of any other voltage or current. In contrast, a dependent source (also called a controlled source) provides a voltage or current that depends on a voltage or current elsewhere in a circuit. In a circuit diagram, a dependent source is designated by a diamond-shaped symbol. For an illustration, the circuit of Fig. 1-5 contains a dependent voltage source that provides a voltage of 5V1, which is five times the voltage V1 that appears across a resistor elsewhere in the circuit. (The resistors shown are discussed in the next chapter.) There are four types of dependent sources: a voltage-controlled voltage source as shown in Fig. 1-5, a current-controlled voltage source, a voltage-controlled current source, and a current-controlled current source. Dependent sources are rarely separate physical components. But they are important because they occur in models of electronic components such as operational amplifiers and transistors.
Fig. 1-5
The rate at which something either absorbs or produces energy is the power absorbed or produced. A source of energy produces or delivers power and a load absorbs it. The SI unit of power is the watt with unit symbol W. The quantity symbol is P for constant power and p for time-varying power. If 1 J of work is either absorbed or delivered at a constant rate in 1 s, the corresponding power is 1 W. In general,
The power absorbed by an electric component is the product of voltage and current if the current reference arrow is into the positively referenced terminal, as shown in Fig. 1-6:
Fig. 1-6
Such references are called associated references. (The term passive sign convention is often used instead of “associated references.”) If the references are not associated (the current arrow is into the negatively referenced terminal), the power absorbed is P = – VI.
If the calculated P is positive with either formula, the component actually absorbs power. But if P is negative, the component produces power--it is a source of electric energy.
The power output rating of motors is usually expressed in a power unit called the horsepower (hp) even though this is not an SI unit. The relation between horsepower and watts is 1 hp = 745.7 W.
Electric motors and other systems have an efficiency (η) of operation defined by
Efficiency can also be based on work output divided by work input. In calculations, efficiency is usually expressed as a decimal fraction that is the percentage divided by 100.
The overall efficiency of a cascaded system as shown in Fig. 1-7 is the product of the individual efficiencies:
Fig. 1-7
Electric energy used or produced is the product of the electric power input or output and the time over which this input or output occurs:
Electric energy is what customers purchase from electric utility companies. These companies do not use the joule as an energy unit but instead use the much larger and more convenient kilowatthour (kWh) even though it is not an SI unit. The number of kilowatthours consumed equals the product of the power absorbed in kilowatts and the time in hours over which it is absorbed:
1.1 Find the charge in coulombs of (a) 5.31 × 1020 electrons, and (b) 2.9 × 1022 protons.
(a) Since the charge of an electron is –1.602 × 10–19 C, the total charge is
(b) Similarly, the total charge is
1.2 How many protons have a combined charge of 6.8 pC?
Because the combined charge of 6.241 × 1018 protons is 1 C, the number of protons is
1.3 Find the current flow through a light bulb from a steady movement of (a) 60 C in 4 s, (b) 15C in 2 min, and (c) 1022 electrons in 1 h.
Current is the rate of charge movement in coulombs per second. So,
(a) 
(b) 
(c) 
The negative sign in the answer indicates that the current flows in a direction opposite that of electron movement. But this sign is unimportant here and can be omitted because the problem statement does not specify the direction of electron movement.
1.4 Electrons pass to the right through a wire cross section at the rate of 6.4 × 1021 electrons per minute. What is the current in the wire?
Because current is the rate of charge movement in coulombs per second,
The negative sign in the answer indicates that the current is to the left, opposite the direction of electron movement.
1.5 In a liquid, negative ions, each with a single surplus electron, move to the left at a steady rate of 2.1 × 1020 ions per minute and positive ions, each with two surplus protons, move to the right at a steady rate of 4.8 × 1019 ions per minute. Find the current to the right.
The negative ions moving to the left and the positive ions moving to the right both produce a current to the right because current flow is in a direction opposite that of negative charge movement and the same as that of positive charge movement. For a current to the right, the movement of electrons to the left is a negative movement. Also, each positive ion, being doubly ionized, has double the charge of a proton. So,
1.6 Will a 10-A fuse blow for a steady rate of charge flow through it of 45 000 C/h?
The current is
which is more than the 10-A rating. So the fuse will blow.
1.7 Assuming a steady current flow through a switch, find the time required for (a) 20 C to flow if the current is 15 mA, (b) 12 μC to flow if the current is 30 pA, and (c) 2.58 × 1015 electrons to flow if the current is –64.2 nA.
Since I = Q/t solved for t is t = Q/I,
(a) 
(b) 
(c) 
1.8 The total charge that a battery can deliver is usually specified in ampere-hours (Ah). An ampere-hour is the quantity of charge corresponding to a current flow of 1 A for 1 h. Find the number of coulombs corresponding to 1 Ah.
Since from Q = It, 1 C is equal to one ampere second (As),
1.9 A certain car battery is rated at 700 Ah at 3.5 A, which means that the battery can deliver 3.5 A for approximately 700/3.5 = 200 h. However, the larger the current, the less the charge that can be drawn. How long can this battery deliver 2 A?
The time that the current can flow is approximately equal to the ampere-hour rating divided by the current:
Actually, the battery can deliver 2 A for longer than 350 h because the ampere-hour rating for this smaller current is greater than that for 3.5 A.
1.10 Find the average drift velocity of electrons in a No. 14 AWG copper wire carrying a 10-A current, given that copper has 1.38 × 1024 free electrons per cubic inch and that the cross-sectional area of No. 14 AWG wire is 3.23 × 10—3 in2.
The average drift velocity (v) equals the current divided by the product of the cross-sectional area and the electron density:
The negative sign in the answer indicates that the electrons move in a direction opposite that of current flow. Notice the low velocity. An electron travels only 1.28 m in 1 h, on the average, even though the electric impulses produced by the electron movement travel at near the speed of light (2.998 × 108 m/s).
1.11 Find the work required to lift a 4500-kg elevator a vertical distance of 50 m.
The work required is the product of the distance moved and the force needed to overcome the weight of the elevator. Since this weight in newtons is 9.8 times the mass in kilograms.
1.12 Find the potential energy in joules gained by a 180-lb man in climbing a 6-ft ladder.
The potential energy gained by the man equals the work he had to do to climb the ladder. The force involved is his weight, and the distance is the height of the ladder. The conversion factor from weight in pounds to a force in newtons is 1 N = 0.225 lb. Thus,
1.13 How much chemical energy must a 12-V car battery expend in moving 8.93 × 1020 electrons from its positive terminal to its negative terminal?
The appropriate formula is W= QV. Although the signs of Q and V are important, obviously here the product of these quantities must be positive because energy is required to move the electrons. So, the easiest approach is to ignore the signs of Q and V. Or, if signs are used, V is negative because the charge moves to a more negative terminal, and of course Q is negative because electrons have a negative charge. Thus,
1.14 If moving 16 C of positive charge from point b to point a requires 0.8 J, find Vab, the voltage drop from point a to point b.
1.15 In moving from point a to point b, 2 × 1019 electrons do 4 J of work. Find Vab, the voltage drop from point a to point b.
Work done by the electrons is equivalent to negative work done on the electrons, and voltage depends on work done on charge. So, Wba = – 4 J, but Wab = – Wba = 4 J. Thus,
The negative sign indicates that there is a voltage rise from a to b instead of a voltage drop. In other words, point b is more positive than point a.
1.16 Find Vab, the voltage drop from point a to point b, if 24 J are required to move charges of (a) 3 C, (b) – 4 C, and (c) 20 × 1019 electrons from point a to point b.
If 24 J are required to move the charges from point a to point b, then — 24 J are required to move them from point b to point a. In other words, Wab = — 24 J. So,
(a) 
The negative sign in the answer indicates that point a is more negative than point b —there is a voltage rise from a to b.
(b) 
(c) 
1.17 Find the energy stored in a 12-V car battery rated at 650 Ah.
From W = QV and the fact that 1 As = 1 C,
1.18 Find the voltage drop across a light bulb if a 0.5-A current flowing through it for 4 s causes the light bulb to give off 240 J of light and heat energy.
Since the charge that flows is Q = It = 0.5 × 4 = 2 C,
1.19 Find the average input power to a radio that consumes 3600 J in 2 min.
1.20 How many joules does a 60-W light bulb consume in 1 h?
From rearranging P = W/t and from the fact that 1 Ws = 1 J,
1.21 How long does a 100-W light bulb take to consume 13 kJ?
From rearranging P = W/t,
1.22 How much power does a stove element absorb if it draws 10 A when connected to a 115-V line?
1.23 What current does a 1200-W toaster draw from a 120-V line?
From rearranging P = VI,
1.24 Figure 1-8 shows a circuit diagram of a voltage source of V volts connected to a current source of I amperes. Find the power absorbed by the voltage source for
(a) V = 2 V, I = 4 A
(b) V = 3 V, I = –2 A
(c) V = –6 V, I = –8 A
Fig. 1-8
Because the reference arrow for I is into the positively referenced terminal for V, the current and voltage references for the voltage source are associated. This means that there is a positive sign (or the absence of a negative sign) in the relation between power absorbed and the product of voltage and current: P = VI. With the given values inserted,
(a) P = VI = 2 × 4 = 8 W
(b) P = VI = 3 × (–2)= –6W
The negative sign for the power indicates that the voltage source delivers rather than absorbs power.
(c) P = VI = –6 × (–8) = 48 W
1.25 Figure 1-9 shows a circuit diagram of a current source of I amperes connected to an independent voltage source of 8 V and a current-controlled dependent voltage source that provides a voltage that in volts is equal to two times the current flow in amperes through it. Determine the power P1 absorbed by the independent voltage source and the power P2 absorbed by the dependent voltage source for (a) I = 4 A, (b) I = 5 mA, and (c) I = —3 A.
Fig. 1-9
Because the reference arrow for I is directed into the negative terminal of the 8-V source, the power-absorbed formula has a negative sign: P1 = —8I. For the dependent source, though, the voltage and current references are associated, and so the power absorbed is P2 = 2I (I) = 2I2. With the given current values inserted,
(a) P1 = –8(4)= –32W and P2 = 2(4)2 = 32 W. The negative power for the independent source indicates that it is producing power instead of absorbing it.
(b) P1 = –8(5 × 10–3)= –40 × 10–3 W = –40 mW
P2 = 2(5 × 10–3)2 = 50 × 10–6 W = 50 μW
(c) P1 = –8(–3) = 24 W and P2 = 2(–3)2 = 18 W. The power absorbed by the dependent source remains positive because although the current reversed direction, the polarity of the voltage did also, and so the actual current flow is still into the actual positive terminal.
1.26 Calculate the power absorbed by each component in the circuit of Fig. 1-10.
Fig. 1-10
Since for the 10-A current source the current flows out of the positive terminal, the power it absorbs is P1 = — 16(10) = — 160 W. The negative sign indicates that this source is not absorbing power but rather is delivering power to other components in the circuit. For the 6-V source, the 10-A current flows into the negative terminal, and so P2 = –6(10) = –60 W. For the 22-V source, P3 = 22(6) = 132W. Finally, the dependent source provides a current of 0.4(10) = 4 A. This current flows into the positive terminal since this source also has 22 V, positive at the top, across it. Consequently, P4 = 22(4) = 88 W. Observe that
The sum of 0 W indicates that in this circuit the power absorbed by components is equal to the power delivered. This result is true for every circuit.
1.27 How long can a 12-V car battery supply 250 A to a starter motor if the battery has 4 × 106 J of chemical energy that can be converted to electric energy?
The best approach is to use t = W/P. Here,
And so
1.28 Find the current drawn from a 115-V line by a dc electric motor that delivers 1 hp. Assume 100 percent efficiency of operation.
From rearranging P = VI and from the fact that 1 W/V = 1 A,
1.29 Find the efficiency of operation of an electric motor that delivers 1 hp while absorbing an input of 900 W.
1.30 What is the operating efficiency of a fully loaded 2-hp dc electric motor that draws 19 A at 100 V? (The power rating of a motor specifies the output power and not the input power.)
Since the input power is
the efficiency is
1.31 Find the input power to a fully loaded 5-hp motor that operates at 80 percent efficiency.
For almost all calculations, the efficiency is better expressed as a decimal fraction that is the percentage divided by 100, which is 0.8 here. Then from η = Pout/Pin,
1.32 Find the current drawn by a dc electric motor that delivers 2 hp while operating at 85 percent efficiency from a 110-V line.
From 
1.33 Maximum received solar power is about 1 kW/m2. If solar panels, which convert solar energy to electric energy, are 13 percent efficient, how many square meters of solar cell panels are needed to supply the power to a 1600-W toaster?
The power from each square meter of solar panels is
So, the total solar panel area needed is
1.34 What horsepower must an electric motor develop to pump water up 40 ft at the rate of 2000 gallons per hour (gal/h) if the pumping system operates at 80 percent efficiency?
One way to solve for the power is to use the work done by the pump in 1 h, which is the weight of the water lifted in 1 h times the height through which it is lifted. This work divided by the time taken is the power output of the pumping system. And this power divided by the efficiency is the input power to the pumping system, which is the required output power of the electric motor. Some needed data are that 1 gal of water weighs 8.33 lb, and that 1 hp = 550 (ft º lb)/s. Thus,
1.35 Two systems are in cascade. One operates with an efficiency of 75 percent and the other with an efficiency of 85 percent. If the input power is 5 kW, what is the output power?
1.36 Find the conversion relation between kilowatthours and joules.
The approach here is to convert from kilowatthours to watt-seconds, and then use the fact that 1 J = 1 Ws:
1.37 For an electric rate of 7¢/kilowatthour, what does it cost to leave a 60-W light bulb on for 8 h?
The cost equals the total energy used times the cost per energy unit:
1.38 An electric motor delivers 5 hp while operating with an efficiency of 85 percent. Find the cost for operating it continuously for one day (d) if the electric rate is 6¢/kilowatthour.
The total energy used is the output power times the time of operation, all divided by the efficiency. The product of this energy and the electric rate is the total cost:
1.39 Find the charge in coulombs of (a) 6.28 × 1021 electrons and (b) 8.76 × 1020 protons.
Ans. (a) -1006 C, (b) 140 C
1.40 How many electrons have a total charge of –4 nC?
Ans. 2.5 × 1010 electrons
1.41 Find the current flow through a switch from a steady movement of (a) 90 C in 6 s, (b) 900 C in 20 min, and (c) 4 × 1023 electrons in 5 h.
Ans. (a) 15 A, (b) 0.75 A, (c) 3.56 A
1.42 A capacitor is an electric circuit component that stores electric charge. If a capacitor charges at a steady rate to 10 mC in 0.02 ms, and if it discharges in 1 μs at a steady rate, what are the magnitudes of the charging and discharging currents?
Ans. 500 A, 10 000 A
1.43 In a gas, if doubly ionized negative ions move to the right at a steady rate of 3.62 × 1020 ions per minute and if singly ionized positive ions move to the left at a steady rate of 5.83 × 1020 ions per minute, find the current to the right.
Ans. -3.49 A
1.44 Find the shortest time that 120 C can flow through a 20-A circuit breaker without tripping it.
Ans. 6 s
1.45 If a steady current flows to a capacitor, find the time required for the capacitor to (a) charge to 2.5 mC if the current is 35 mA, (b) charge to 36 pC if the current is 18υ A, and (c) store 9.36 × 1017 electrons if the current is 85.6 nA.
Ans. (a) 71.4 ms, (b) 2 μs, (c) 20.3 d
1.46 How long can a 4.5-Ah, 1.5-V flashlight battery deliver 100 mA?
Ans. 45 h
1.47 Find the potential energy in joules lost by a 1.2-lb book in falling off a desk that is 31 in high.
Ans. 4.2 J
1.48 How much chemical energy must a 1.25-V flashlight battery expend in producing a current flow of 130 mA for 5 min?
Ans. 48.8 J
1.49 Find the work done by a 9-V battery in moving 5 × 1020 electrons from its positive terminal to its negative terminal.
Ans. 721 J
1.50 Find the total energy available from a rechargeable 1.25-V flashlight battery with a 1.2-Ah rating.
Ans. 5.4 kJ
1.51 If all the energy in a 9-V transistor radio battery rated at 0.392 Ah is used to lift a 150-lb man, how high in feet will he be lifted?
Ans. 62.5 ft
1.52 If a charge of —4 C in moving from point a to point b gives up 20 J of energy, what is Vab?
Ans. – 5 V
1.53 Moving 6.93 × 1019 electrons from point b to point a requires 98 J of work. Find Vab.
Ans. –8.83 V
1.54 How much power does an electric clock require if it draws 27.3 mA from a 110-V line?
Ans. 3 W
1.55 Find the current drawn by a 1000-W steam iron from a 120-V line.
Ans. 8.33 A
1.56 For the circuit of Fig. 1-11, find the power absorbed by the current source for (a) V = 4 V, I = 2 mA; (b) V = –50 V, I = –150 μA; (c) V = 10 mV, I = –15 mA; (d) V = –120 mV, I = 80 mA.
Fig. 1-11
Ans. (a) –8 mW, (b) –7.5 mW, (c) 150 μW, (d) 9.6 mW
1.57 For the circuit of Fig. 1-12, determine P1, P2, P3, which are powers absorbed, for (a) I = 2 A, (b) I = 20 mA, and (c) I = –3 A.
Fig. 1-12
Ans. (a) P1 = 16 W, P2 = –24 W, P3 = –20 W; (b) P1 = 0.16 W, P2 = –2.4 mW, P3 = –0.2 W; (c) P1 = –24 W, P2 = –54 W, P3 = 30 W
1.58 Calculate the power absorbed by each component in the circuit of Fig. 1-13.
Fig. 1-13
Ans. P1 = 16 W, P2 = –48 W, P3 = –48 W, P4 = 80 W
1.59 Find the average input power to a radio that consumes 4500 J in 3 min.
Ans. 25 W
1.60 Find the voltage drop across a toaster that gives off 7500 J of heat when a 13.64-A current flows through it for 5 s.
Ans. 110 V
1.61 How many joules does a 40-W light bulb consume in 1 d?
Ans. 3.46 MJ
1.62 How long can a 12-V car battery supply 200 A to a starter motor if the battery has 28 MJ of chemical energy that can be converted to electric energy?
Ans. 3.24 h
1.63 How long does it take a 420-W color TV set to consume (a) 2 kWh and (b) 15 kJ?
Ans. (a) 4.76 h, (b) 35.7 s
1.64 Find the current drawn by a 110-V dc electric motor that delivers 2 hp. Assume 100 percent efficiency of operation.
Ans. 13.6 A
1.65 Find the efficiency of operation of an electric motor that delivers 5 hp while absorbing an input of 4190 W.
Ans. 89 percent
1.66 What is the operating efficiency of a dc electric motor that delivers 1 hp while drawing 7.45 A from a 115-V line?
Ans. 87 percent
1.67 Find the current drawn by a 100-V dc electric motor that operates at 85 percent efficiency while delivering 0.5 hp.
Ans. 4.39 A
1.68 What is the horsepower produced by an automobile starter motor that draws 250 A from a 12-V battery while operating at an efficiency of 90 percent?
Ans. 3.62 hp
1.69 What horsepower must an electric motor develop to operate a pump that pumps water at a rate of 24 000 liters per hour (L/h) up a vertical distance of 50 m if the efficiency of the pump is 90 percent? The gravitational force on 1 L of water is 9.78 N.
Ans. 4.86 hp
1.70 An ac electric motor drives a dc electric voltage generator. If the motor operates at an efficiency of 90 percent and the generator at an efficiency of 80 percent, and if the input power to the motor is 5 kW, find the output power from the generator.
Ans. 3.6 kW
1.71 Find the cost for one year (365 d) to operate a 20-W transistor FM-AM radio 5 h a day if electrical energy costs 8¢/kilowatthour.
Ans. $2.92
1.72 For a cost of $5, how long can a fully loaded 5-hp electric motor be run if the motor operates at an efficiency of 85 percent and if the electric rate is 6¢/kilowatthour?
Ans. 19 h
1.73 If electric energy costs 6¢kilowatthour, calculate the utility bill for one month for operating eight 100-W light bulbs for 50 h each, ten 60-W light bulbs for 70 h each, one 2-kW air conditioner for 80 h, one 3-kW range for 45 h, one 420-W color TV set for 180 h, and one 300-W refrigerator for 75 h.
Ans. $28.51.
