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Circuit Variables
Assessment Problems
AP 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 1 1 1 1 1
year · day · hour · min · sec = year
days 24 hours 60 mins 60 secs 1000 ms × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a product
of ratios:
$100 × 109 1 year 100
· = = $
1 year × 109 ms
AP 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 ()(3 × 108)= × 108
m/s. Now, we use a product of ratios to convert from meters/second to
inches/nanosecond:
× 108 1 100 1 ( × 108)(100)
m · s · cm · in = = in
1s 109 ns 1 m cm (109)() 1 ns
Thus, a signal traveling at 80% of the speed of light will travel in a
nanosecond.
1–1
1–2 CHAPTER 1. Circuit Variables
i = dq
AP Remember from Eq. (), current is the time rate of change of charge, or dt In
this problem, we are given the current and asked to find the total charge. To do this,
we must integrate Eq. () to find an expression for charge in terms of current:
t
q(t)= i(x) dx
0
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
∞ 20 ∞ 20
q = 20e−5000x dx = e−5000x = (e∞− e0)
total
0 −5000 0 −5000
20 20
= (0 − 1) = = = 4000 µ
−5000 5000 C C
i = dq
AP Recall from Eq. () that current is the time rate of change of charge, or dt .In
this problem we are given an expression for the charge, and asked to fi