Engineering Circuit Analysis 9th Edition

1. Convert the following to engineering notation: (a) 0.045 W  45103 W = 45 mW (b) 2000 pJ  2000 1012  2 109 J = 2 nJ (c) 0.1 ns  0.1109  100 1012 s = 100 ps (d) 39,212 as = 3.9212×104×10-18 = 39.212×10-15 s = (f) 18,000 m  18 103m= 18 km (g) 2,500,000,000,000 bits  2.51012 bits = 2.5 terabits 1015 atoms  102 cm  3 (h)    1021 atoms/m3 (it’s unclear what a “zeta atom” is)  cm3  1 m  39.212 fs 2. Convert the following to engineering notation: (a) 1230 fs  1.23103 1015  1.231012 s = 1.23 ps (b) 0.0001 decimeter 1104 101 10106 m (c) 1400 mK  1.4 103 103  1.4 K = 10 m (d) 32 nm  32 109 m = (e) 13,560 kHz  1.356 104 103  13.56 106 Hz = (f) 2021 micromoles  2.021103 106  2.021103moles = (g) 13 deciliters  13101  1.3 liters (h) 1 hectometer 100 meters 2.021 millimoles 13.56 MHz 32 nm 3. Express the following in engineering units: (a) 1212 mV = 1.121 V (b) 1011 pA = 1011×10-12= 100 mA (c) 1000 yoctoseconds 1103 1024 11021 seconds = 1 zs (e) 13,100 attoseconds 1.311015 s = 1.31 fs (f) 10−14 zettasecond=10-14×1021=107=10×106 s = 10 Ms (g) 10−5 second 10106 seconds = (h) 10−9 Gs109 109 1 second (d) 33.9997 zeptoseconds 10 s 4. Expand the following distances in simple meters: (a) 1 Zm  11021 m (b) 1 Em  11018 m (c) 1 Pm  11015 m (d) 1 Tm  11012 m (e) 1 Gm  1109 m (f) 1 Mm  1106 m (c) 0 K 5. Convert the following to SI units, taking care to employ proper engineering notation: (a) 212F  212  32 5  273.15  373.15 K 9 (b) 0F  0 32 5  273.15  255.37 K 9 (d) 200 hp  200 745.7 W 1 hp  1.4914 105 W  149.14 103 W = 149.14 kW (e) 1 yard  0.9144 m 914.4 mm (f) 1 mile  1 1760 yards 0.9144 m  1,609.3 m  1.6093 km 1 mile 1 yard (a) 373.15 K (c) 4.2 K 6. Convert the following to SI units, taking care to employ proper engineering notation: (b) 273.15 K (already in SI) (d) 150 hp  150 745.7 W 1 hp  1.11855 105 W  111.855 103 W = 111.855 kW (e) 500 Btu  5001055 J  5.275 105 J  527.5 103 J  527.5 kJ 1 Btu (f) 100 J/s  100 W 7. It takes you approximately 2 hours to finish your homework on thermodynamics. Since it feels like it took forever, how many galactic years does this correspond to? (1 galactic year = 250 million years) 2 hours  1 day 24 hours  1 year 365 days  galactic years 250 106 years  913.24 1015 galactic years (or, 913.24 femto-galactic-years!)   8. A certain krypton fluoride laser generates 15 ns long pulses, each of which contains 550 mJ of energy. (a) Calculate the power using the pulse energy over the 15 ns duration. p  w  t 550 mJ  15 ns 550 103 J 7 15109 s 3.6667 10 W  36.667 MW (b) If up to 100 pulses can be generated per second, calculate the maximum average power output of the laser. In this case, look at the total energy of 100 pulses over the one second duration. p  w  100 pulses 550 mJ  55 W t 1 s 9. Your recommended daily food intake is 2,500 food calories (kcal). If all of this energy is efficiently processed, what would your average power output be? 2500 103 cal  4.187 J  10.4675 106 J cal p  w  t 10.4675 106 J 24 h / day60 min / h60 s / min  121.15 W 10. An electric vehicle is driven by a single motor rated at 40 hp. If the motor is run continuously for 3 h at maximum output, calculate the electrical energy consumed. Express your answer in SI units using engineering notation. w  pt   40 hp 745.7 W  3 h 3600 s  3.2214 108 J  322.14 106 J   1 hp     1 h   = 322.14 MJ 11. Under insolation conditions of 500 W/m2 (direct sunlight), and 10% solar cell efficiency (defined as the ratio of electrical output power to incident solar power), calculate the area required for a photovoltaic (solar cell) array capable of running the vehicle in Exercise 10 at half power. p  20 hp  20 hp  745.7 W 1 hp  14.914 103 W p  14.914 103 W   500 W  area10% efficiency   m 2  area  14.914 103 W 5000.1  298.28 m 2 m 2 12. A certain metal oxide nanowire piezoelectricity generator is capable of producing 100 pW of usable electricity from the type of motion obtained from a person jogging at a moderate pace. (a) 1 W 100 1012 W / nanowire (b) 1010 nanowires 5 nanowires/ m 2  2 109  2 1 m 2 106  m 2 This area would fit in a square that is approximately 4.5 cm x 4.5 cm, so very reasonable!  1010 nanowires  2 103m 2 13. Assuming a global population of 9 billion people, each using approximately 100 W of power continuously throughout the day, calculate the total land area that would have to be set aside for photovoltaic power generation, assuming 800 W/m2 of incident solar power and a conversion efficiency (sunlight to electricity) of 10%. Power needed  Power density from Sun AreaConversion Efficiency 100 W 9109 people Area  800 W / m 2 0.10 About the size of the state of Connecticut.  1.1251010 m 2  11,250 km2 14. The total charge flowing out of one end of a small copper wire and into an unknown device is determined to follow the relationship q(t) = 5e−t/2 C, where t is expressed in seconds. Calculate the current flowing into the device, taking note of the sign. dq d   t   t i    5e 2   2.5e 2 A dt dt   Note that the charge on the device starts positive, and then decreases. This means that current is flowing out of the device. The current flowing into the devices is therefore negative. 15. The current flowing into the collector lead of a certain bipolar junction transistor (BJT) is measured to be 1 nA. If no charge was transferred in or out of the collector lead prior to t = 0, and the current flows for 1 min, calculate the total charge which crosses into the collector. q   idt  109 A60 s  60 109 C  60 nC