Page 347 - Advanced engineering mathematics
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10.5 Applications and Illustrations of Techniques 327
Now use the initial condition to solve for the constants. Solve
⎛ ⎞ ⎛ ⎞
4 5c 2 + 10c 3 + 9
= c 1 + c 3 − 18/5 .
i(0+) = 4 ⎠ ⎝ ⎠
⎝
0 c 1 − 4c 2 − 9c 3 − 18/5
This can be written as
0 5 10 −5
⎛ ⎞ ⎛ ⎞
⎝ 1 0 1 ⎠ C = PC = 38/5 ⎠ .
⎝
1 −4 −9 18/5
Then
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
−5 1 4 5 5 −5 18/5
C = P −1 ⎝ 38/5 ⎠ = ⎝ 10 −10 10 ⎠⎝ 38/5 ⎠ = ⎝ −9 ⎠ .
10
18/5 −4 5 −5 18/5 4
The current is
⎛ −2t −20t/9 ⎞
9 − 4te + 40e
i = ⎝ 4e −20t/9 ⎠ .
36e −2t − 36e −20t/9
SECTION 10.5 PROBLEMS
1. Referring to the circuit of Figure 10.4, determine how pound of salt per gallon, tank 1 initially has 200
much time elapses between the time the switch is pounds of salt, and tank 2 initially has 150 pounds
closed and the time the charge on the capacitor is of salt. Determine the amount of salt in each tank at
a maximum. What is the maximum voltage on the time t > 0.
capacitor?
3. Two tanks are connected as shown in Figure 10.6.
2. Referring to Figure 10.5, tank 1 initially contains 200 Tank 1 initially contains 100 gallons of water in which
gallons of saltwater (brine), while tank 2 initially con- 40 pounds of salt are dissolved. Tank 2 initially con-
tains 300 gallons of brine. Beginning at time 0, brine tains 150 gallons of pure water. Beginning at t = 0, a
is pumped into tank 1 at the rate of 4 gallons per brine solution containing 1/5 pound of salt per gal-
minute, pure water is pumped into tank 2 at 6 gallons lon is pumped into tank 1 at the rate of 5 gallons
per minute, and the brine solutions are interchanged per minute. At this time, a solution which also con-
between the two tanks and also flow out of both tanks tains 1/5 pound of salt per gallon is pumped into
at the rates shown. The input to tank 1 contains 1/4 tank 2 at the rate of 10 gallons per minute. The
Brine 4 gal/min 12 gal/min Water 6 gal/min
Tank 1 Tank 2
6 gal/min
4 gal/min 12 gal/min
FIGURE 10.5 Connected tank system for Problem 2, Section 10.5.
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