ch01_ssm (1) - Circuit Variables 1 Assessment Problems AP...

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1 Circuit Variables 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 scientiFc notation: $100 billion = $100 × 10 9 Now we determine the number of milliseconds in one year, again using a product of ratios: 1 year 365 . 25 days · 1 day 24 hours · 1 hour 60 mins · 1 min 60 secs · 1 sec 1000 ms = 1 year 31 . 5576 × 10 9 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: $100 × 10 9 1 year · 1 year 31 . 5576 × 10 9 ms = 100 31 . 5576 = $3 . 17 / ms AP 1.2 ±irst, we recognize that 1 ns = 10 9 s. The question then asks how far a signal will travelin10 9 s if it is traveling at 80% of the speed of light. Remember that the speed of light c = 3 × 10 8 m/s. Therefore, 80% of c is ( 0 . 8 )( 3 × 10 8 ) = 2 . 4 × 10 8 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 2 . 4 × 10 8 m 1s · 1s 10 9 ns · 100 cm 1m · 1in 2 . 54 cm = ( 2 . 4 × 10 8 )( 100 ) ( 10 9 )( 2 . 54 ) = 9 . 45 in 1ns Thus, a signal traveling at 80% of the speed of light will travel 9 . 45 00 in a nanosecond. 1–1
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1–2 CHAPTER 1. Circuit Variables AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq dt In this problem, we are given the current and asked to Fnd the total charge. To do this, we must integrate Eq. (1.2) to Fnd an expression for charge in terms of current: q(t) = Z t 0 i(x) dx We are given the expression for current, i , which can be substituted into the above expression. To Fnd the total charge, we let t →∞ in the integral. Thus we have q total = Z 0 20 e 5000 x dx = 20 5000 e 5000 x ± ± ± ± 0 = 20 5000 (e e 0 ) = 20 5000 ( 0 1 ) = 20 5000 = 0 . 004 C = 4000 µ C AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or i = .In this problem we are given an expression for the charge, and asked to Fnd the maximum current. ±irst we will Fnd an expression for the current using Eq. (1.2): i = = d ² 1 α 2 ³ t α + 1 α 2 ´ e αt µ = d ³ 1 α 2 ´ d ³ t α e ´ d ³ 1 α 2 e ´ = 0 ³ 1 α e α t α e ´ ³ α 1 α 2 e ´ = ³ 1 α + t + 1 α ´ e = te Now that we have an expression for the current, we can Fnd the maximum value of the current by setting the Frst derivative of the current to zero and solving for t : di = d (te ) = e + t( α)e = ( 1 αt)e = 0 Since e never equals 0 for a Fnite value of t , the expression equals 0 only when ( 1 αt) = 0. Thus, t = 1 will cause the current to be maximum. ±or this value of t , the current is i = 1 α e α/α = 1 α e 1 Remember in the problem statement, α = 0 . 03679. Using this value for α , i = 1 0 . 03679 e 1 = 10 A
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Problems 1–3 AP 1.5 Start by drawing a picture of the circuit described in the problem statement: Also sketch the four Fgures from ±ig. 1.6: [a] Now we have to match the voltage and current shown in the Frst Fgure with the polarities shown in ±ig. 1.6. Remember that 4A of current entering Terminal 2
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ch01_ssm (1) - Circuit Variables 1 Assessment Problems AP...

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