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See Fig. 2-13. Wmax 1:875 104 2 10 3 2 75:0 mJ 1 1 Wmax CV2 60:0 10 6 50:0 2 75:0 mJ max 2 2

t>2)

Section 12. Supplier Control (for outsourcing)

Fig. 2-13

1, mr2 33 a) (b) [1

3.75. (a) 11 > 144 (b) 11 > 144 (c) !11>12 (d) !11>12 (e) 1 > 144 (f) 1 > 11 3.76. (a) 1 (b) 1 (c) 1 (d) 1 (e) 0 (f) 0 3.77. (a) 73 > 960 (b) 73 > 960 (c) !73>960 (d) !73>960 (e) 1 > 64 (f) 15 > 73 3.78. (a) 233 > 324 (b) 233 > 324 (c) !233>18 (d) !233>18 (e) 91 > 324 (f) 91 > 233 3.79. (a) 4 (b) 4> !35 3.81. (a) (3x 2) > (6x 3.80. 3) for 0 !15>4 x 0 (b) 1 (b) (3y 2) > (6y 3) for 0 y 1

A 20.0-mF capacitance is linearly charged from 0 to 400 mC in 5.0 ms. Find the voltage function and Wmax .

3.82. (a) 1 > 2 for x 3.83. (a)

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! 400 10 6 q t 8:0 10 2 t C 5:0 10 3 v q=C 4:0 103 t V Vmax 4:0 103 5:0 10 3 20:0 V Wmax 1 CV2 4:0 mJ max 2

X E(Y u X)

0 4>3

A series circuit with R 2 , L 2 mH, and C 500 mF has a current which increases linearly from zero to 10 A in the interval 0 t 1 ms, remains at 10 A for 1 ms t 2 ms, and decreases linearly from 10 A at t 2 ms to zero at t 3 ms. Sketch vR , vL , and vC .

N/A 8. Error 1 occurred after 4 days, and error 2 occurred 5 days later. Plot the error rate versus error number graph and the error rate versus time graph, and estimate the number of errors in the system and the time to remove all the errors. As shown in Fig. 7-3, plotting error rate versus error number shows an intercept about 6. Thus, there are 4 errors left in the system.

2 5>7

vR must be a time function identical to i, with Vmax 2 10 20 V. For 0 < t < 1 ms, di 10 103 A=s dt When di=dt 0, for 1 ms < t < 2 ms, vL 0. and vL L di 20 V dt

Y E(X u Y)

0 4>3

CHAP. 2]

1 7>6

104 t dt 107 t2 V For 1 ms < t < 2 ms,

2 1>2

As shown in Fig. 7-4, plotting error rates versus time elapsed shows an intercept around 29 to 30. Thus, it should take about 20 more units to completely remove the errors.

3.84. (a)

vC 20 10 t 10 3 10 V See Fig. 2-14.

6x2 6x 1 18(2x 1)2

Fig. 2-14

1 (b)

6y2 6y 1 18(2y 1)2

A single circuit element has the current and voltage functions graphed in Fig. 2-15. Determine the element.

3.85. (a) 1 > 9 (b) 1 3.86. (a) (b)

Fig. 2-15

Fig. 7-4.

X Var(Y u X)

0 5>9

The element cannot be a resistor since v and i are not proportional. v is an integral of i. For 2 ms < t < 4 ms, i 6 0 but v is constant (zero); hence the element cannot be a capacitor. For 0 < t < 2 ms, di 5 103 A=s dt Consequently, 0 L v di 3 mH dt and v 15 V

1 4>5

(Examine the interval 4 ms < t < 6 ms; L must be the same.)

2 24 > 49

Y Var(X u Y)

(a) i2 1 A, (b) i2 2 A,

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