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Circuit Variables
1
Assessment Problems
AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientiﬁc notation: $100 billion = $100 × 109 Now we determine the number of milliseconds in one year, again using a product of ratios: 1 year 1 hour 1 min 1 sec 1 year 1 day · · · = · 31.5576 × 109 ms 365.25 days 24 hours 60 mins 60 secs 1000 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: 1 year 100 $100 × 109 · = = $3.17/ms 1 year 31.5576 × 109 ms 31.5576 AP 1.2 First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal will travel in 10−9 s if it is traveling at 80% of the*…show more content…*

Thus, using the passive sign convention, p = −vi. Substituting the values of voltage and current given in the ﬁgure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line. Chapter Problems P 1.1 To begin, we calculate the number of pixels that make up the display: npixels = (1280)(1024) = 1,310,720 pixels Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel requires 3 bytes of information. We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by 3 bytes per pixel: nbytes = 1,310,720 pixels 3 bytes · = 3,932,160 bytes/display 1 display 1 pixel Finally, we use the fact that there are 106 bytes per MB: 3,932,160 bytes 1 MB = 3.93 MB/display · 6 1 display 10 bytes Problems P 1.2 c = 3 × 108 m/s so 1 c = 1.5 × 108 m/s 2 so x= 5 × 106 = 33.3 ms 1.5 × 108 1–5 1.5 × 108 m 5 × 106 m = 1s xs P 1.3 We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First, recall that

Thus, using the passive sign convention, p = −vi. Substituting the values of voltage and current given in the ﬁgure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line. Chapter Problems P 1.1 To begin, we calculate the number of pixels that make up the display: npixels = (1280)(1024) = 1,310,720 pixels Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel requires 3 bytes of information. We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by 3 bytes per pixel: nbytes = 1,310,720 pixels 3 bytes · = 3,932,160 bytes/display 1 display 1 pixel Finally, we use the fact that there are 106 bytes per MB: 3,932,160 bytes 1 MB = 3.93 MB/display · 6 1 display 10 bytes Problems P 1.2 c = 3 × 108 m/s so 1 c = 1.5 × 108 m/s 2 so x= 5 × 106 = 33.3 ms 1.5 × 108 1–5 1.5 × 108 m 5 × 106 m = 1s xs P 1.3 We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First, recall that

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