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principios de analise instrumental 6ed skoog resolucoes

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A 
 i = 2.4 × 10–6 e–t/2.00 A or 2.4 e–t/2.0 μA 
 
t, s i, μA t, s i, μA 
0.00 2.40 1.0 1.46 
0.010 2.39 10 0.0162 
0.10 2.28 
 
2-14. vC = VC e–t/RC (Equation 2-40) 
 vC/VC = 1.00/100 for discharge to 1% 
 0.0100 = e–t/RC = 
6/ 0.015 10t Re
−− × × 
 ln 0.0100 = –4.61 = –t/1.5 × 10–8R 
 t = 4.61 × 1.5 × 10–8R = 6.90 × 10–8R 
 (a) When R = 10 MΩ or 10 × 106 Ω, t = 0.69 s 
 (b) Similalry, when R = 1 MΩ, t = 0.069 s 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 6
 (c) When R = 1 kΩ, t = 6.9 × 10–5 s 
2-15. (a) When R = 10 MΩ, RC = 10 × 106 Ω × 0.015 × 10–6 F = 0.15 s 
 (b) RC = 1 × 106 × 0.015 × 10–6 = 0.015 s 
 (c) RC = 1 × 103 × 0.015 × 10–6 = 1.5 × 10–5 s 
2-16. Parts (a) and (b) are given in the spreadsheet below. For part (c), we calculate the 
quantities from 
 i = Iinit e-t/RC, vR = iR, and vC = 25 – vR 
 For part (d) we calculate the quantities from 
 / = t RCCvi e
R
−− , vR = iR, and vC = –vR 
 The results are given in the spreadsheet. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 7
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 
 
7
 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 8
2-17. Proceeding as in Problem 2-16, the results are in the spreadsheet 
 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 9
2-18. In the spreadsheet we calculate XC, Z, and φ from 
 XC = 2/2πfC, 2 = + C2Z R X , and φ = arc tan(XC/R) 
 
 
2-19. Let us rewrite Equation 2-54 in the form 
 o
2
( ) 1 = = 
( ) (2π ) 1
p
p i
V
y
V fRC + 
 y2(2πfRC)2 + y2 = 1 
 
2
2 2
1 1 1 1 = 1 = 
2πRC 2πRC
yf
y y
−− 
 The spreadsheet follows 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 10
 
2-20. By dividing the numerator and denominator of the right side of Equation 2-53 by R, we 
obtain 
 o
2
( ) 1 = = 
( ) 1 + (1/ 2π )
p
p i
V
y
V fRC
 
Squaring this equation yields 
 y2 + y2/(2πfRC)2 = 1 
 
2
22π = 1
yfRC
y− 
 
2
2
1 = 
2π 1
yf
RC y− 
The results are shown in the spreadsheet that follows. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 2
 
 11
 
 
 
Skoog/Holler/Crouch Chapter 3
Principles of Instrumental Analysis, 6th ed. Instructor’s Manual
 
 
 1 
CHAPTER 3 
3-1. vo = Avs = A(v+ – v–) For + limit, +13 V =A(v+ – v–) For – limit, –14 V =A(v+ – v–) 
(a). If A = 200,000, for + limit, (v+ – v–) = 13 V/200, 000 = 65 μV. So v+ must exceed 
v– by 65 μV for + limit to be reached. 
 For – limit, (v+ – v–) = – 14 V/200,000 = – 75 μV, so v– must exceed v+ by 75 μV. 
(b). For + limit (v+ – v–) = 13 V/500,000 = 26 μV. For – limit (v+ – v–) = 
–14 V/500,000 = – 28 μV. 
(c). + limit, (v+ – v–) = 13 V/1.5 × 106 = 8.7 μV; – limit, (v+ – v–) = –14 V/1.5 × 106 = 
–9.3 μV 
3-2. d
cm
CMRR = A
A
 The difference gain, Ad = 10 V/500 μV = 2.0 × 104 
The common mode gain, Acm = 1 V/500 mV = 2.0 
CMRR = 2.0 × 104/ 2.0 = 1.0 × 104 or in dB, CMRR = 20 log(1.0 × 104) = 80 dB 
3-3. For vs = 5.0 μV, A = vo/vs = 13/5 × 10–6 = 2.6 × 106. 
3-4. (a). From Equation 3-2, o i = 1
Av v
A
⎛⎜ +⎝ ⎠
⎞⎟ = 2.0 × 1.0 × 105/( 1 + 1.0 × 105) = 1.99998 V 
 Error = 2.0 – 1.99998 V = 0.00002 V or 20 μV 
 Rel Error = (20 × 10–6/ 2.0) × 100%= 0.001% 
 (b). Current drawn from the source, i, is 124 12
2.0 V= = 1.99999998 10 A
(10 10 ) 
i −×+ Ω 
 The IR drop across the source resistance is 1.99999998 × 10–12 A × 104 Ω = 20 nV 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 2
 Rel Error = (20 nV/2.0 V) × 100% = 1.0 × 10–6% 
3-5. Resistors R1 and R2 form a voltage divider. A fraction of the output voltage 1 o
1 2
R v
R R
⎛ ⎞⎜ ⎟+⎝ ⎠
 
is feedback to the inverting input. The amplifier maintains the voltage at the 
noninverting input, vi equal to the voltage at the inverting input 1 o
1 2
R v
R R
⎛ ⎞⎜ +⎝ ⎠⎟
. Hence, if 
1
i
1 2
 = Rv
R R
⎛ ⎞⎜ +⎝ ⎠ o
v⎟ , it follows that 1 2o
1
 = R Rv
R
⎛ ⎞+⎜⎝ ⎠ i
v⎟ . This is a voltage follower 
configuration, but since (R1 + R2) > R1, there will be gain. 
3-6. R1 + R2 = 3.5 × R1 = 10.0 × 103 Ω. So, R1 = 10.0 × 103 Ω/3.5 = 2.86 × 103 Ω (2.9 kΩ). 
 R2 = 10.0 × 103 Ω – R1 = 10.0 × 103 – 2.9 × 103 = 7.1 k Ω. 
3-7. vo = –ix and v+ = v– = 0 
vi = i(R – x) 
( )i = = 
ov ix x
v i R x R x
− −− − 
o i = 
xv v
R x
⎛ ⎞− ⎜ ⎟−⎝ ⎠ 
3-8. (a). The output voltage vo = –iRf. So Rf = –vo/I = 1.0 V/10.0 × 10–6 A = 1.0 × 105 Ω. 
 Use the circuit of Figure 3-6 with Rf = 100 kΩ. 
(b). From Equation 3-5, Ri = Rf/A = 100 kΩ/2 × 105 = 0.5 Ω. 
(c). Equation 3-6 states o f i b = ( ) 1
Av R i i
A
⎛ ⎞− − ⎜ ⎟+⎝ ⎠ 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 3
 ( ) 5o 52 10 = 100 k 25 μA 2.5 nA = 2.4997 V1 + 2 10v
⎛ ⎞×− Ω − ⎜ ⎟×⎝ ⎠
 
 Rel error = 2.5000 2.4997 100% = 0.01%
2.5000
−⎛ ⎞ ×⎜ ⎟⎝ ⎠ 
3-9. (a). Gain = Rf/Ri so if Ri = 10 kΩ and the gain = 25, Rf= 25 × 10 kΩ = 250 kΩ. 
 Use the circuit of Figure 3-7, with Rf = 250 kΩ and Ri = 10 kΩ. 
(b). fo
i
 = Rv
R
⎛ ⎞− ⎜ ⎟⎝ ⎠ i
v so if vo = ± 10 V, and gain = 25, i 10 V= = 0.40 V25v
± ± 
(c). Ri is determined by the input resistor, so Ri = 10 kΩ. 
(d). Insert a voltage follower between the input voltage source and Ri as in Figure 3-
11. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 4
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 
 
4
3-10. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 5
3-11. The rise time tr is given by Equation 3-9 
 9r 6
1 1 = = = 6.7 10 s = 6.7 ns
3 3 50 10
t
f
−×Δ × × 
 The slew rate is given by 910 V = = 1.5 10 V/s = 1500 V/μs
6.7 ns
v
f
Δ ×Δ 
3-12. Several resistor combinations will work including the following 
 
31 2
o f
1 2 3
 = VV VV R
R R R
⎛ ⎞′− + +⎜ ⎟⎝ ⎠
 3 3
1000 = = 
1000
V V′ 3V− − 
3 31 2 1 2
o = 3000 = 30001000 600 500 1000 600 500
V VV V V VV
′⎛ ⎞ ⎛− + + − + −⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠ 
Vo = –3V1 –5V2 + 6V3 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 6
3-13. Several combinations will work, including the following 
 
 31 2o = 1 3 3 3
VV VV ⎛ ⎞′ − × + +⎜ ⎟⎝ ⎠ 
 1 2 3o o
1000 = 1000
1 3
V V VV V + +⎛ ⎞′− × = ⎜ ⎟⎝ ⎠ 
3-14. Several resistor combinations will work including 
 
1 2 1 2
o f
1 2
 = = 3000
6000 10000
V V V VV R
R R
⎛ ⎞ ⎛− + − +⎜ ⎟ ⎜⎝ ⎠⎝ ⎠
⎞⎟ = –(0.500 V1 + 0.300 V2) 
( )o 11 = 5 310V V− + 2V 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 7
3-15. The circuit is shown below. Resistances values that work include Rf = 1.00 kΩ and R1 = 
250 Ω 
 
 I1 + I2 = If Note I2 = Ii 
 o1 i
1 f
 = VV I
R R
+ − 
 1 fo i
1
 = + V RV I
R
− fR or 1 fo i
1
= V R fRR
− −V I 
 Substituting the vales of Rf and Ri gives 
 1o 1
1000 = 1000 = 4 1000
250 i
VV I×− − − − iV I 
3-16. (a). Let vx be the output from the first operational amplifier, 
 1 f1 2 f1
1 2
 = x
v R v Rv
R R
− − 
and 
 f2 3 f21 f1 f2 2 f1 f2o
4 1 4 2 4
 = = + xv R v Rv R R v R Rv
3R R R R R R
− − 
(b). o 1 2 3 1 2
200 400 200 400 400 = + + 4 40
200 400 50 400 10
v v v v v v× × − = −× × 3v 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 3
 
 8
3-17. Let vx be the output of the first operational amplifer, 
1 2 1
15 15 = = 5 3
3 5x
v v v v− − − − 2v 
o 3 4 3 4 1 2
12 12 12 = = 2 3 2 = 10 + 6 3 2
6 4 6x x
v v v v v v v v v v− − − − − − − −3