Electric Circuits Nilsson 7th Solution Manual
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 speed of light. Remember that the speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 100 cm 1 in (2.4 × 108 )(100) 9.45 in 2.4 × 108 m 1s · · = = · 9 1s 10 ns 1m 2.54 cm (109 )(2.54) 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond.
1–1
1–2 AP 1.3
CHAPTER 1. Circuit Variables Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq In dt this problem, we are given the current and asked to find the total charge. To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of current: q(t) = t 0
i(x) dx
We are given the expression for current, i, which can be substituted into the above expression. To find the total charge, we let t → ∞ in the integral. Thus we have qtotal = = AP 1.4
∞ 0
20e−5000x dx =
20 −5000x e −5000
∞ 0
=
20 (e∞ − e0 ) −5000
20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000
Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq . In dt this problem we are given an expression for the charge, and asked to find the maximum current. First we will find an expression for the current
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 speed of light. Remember that the speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 100 cm 1 in (2.4 × 108 )(100) 9.45 in 2.4 × 108 m 1s · · = = · 9 1s 10 ns 1m 2.54 cm (109 )(2.54) 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond.
1–1
1–2 AP 1.3
CHAPTER 1. Circuit Variables Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq In dt this problem, we are given the current and asked to find the total charge. To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of current: q(t) = t 0
i(x) dx
We are given the expression for current, i, which can be substituted into the above expression. To find the total charge, we let t → ∞ in the integral. Thus we have qtotal = = AP 1.4
∞ 0
20e−5000x dx =
20 −5000x e −5000
∞ 0
=
20 (e∞ − e0 ) −5000
20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000
Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq . In dt this problem we are given an expression for the charge, and asked to find the maximum current. First we will find an expression for the current