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Skoog/Holler/Crouch
Principles of Instrumental Analysis, 7th ed.
Chapter 2
CHAPTER 2
2-1.
(a) Applying the voltage divider equation (2-10)
R1
1.0
=
10
R1 + R2 + R3
R2
4.0
=
10
R1 + R2 + R3
V3 = 10.0V โ 1.0 V โ 4.0 V = 5.0 V
R3
5.0
=
10
R1 + R2 + R3
Dividing the first equation by the second, gives
R1/R2 = 1.0/4.0
Similarly,
R2/R3 = 4.0/5.0
Letting R1 = 250 ฮฉ , R2 = 250 ร 4.0 = 1.0 kฮฉ,
and R3 = 1.0 kฮฉ ร 5.0/4.0 = 1.25 kฮฉ . Use a 1.0 kฮฉ resistor and a 250 kฮฉ resistor in
series. The 500 kฮฉ resistor is not used.
(b) V3 = IR3 = 10.0 V โ 1.0 V โ 4.0 V = 5.0 V
(c) I = V/( R1 + R2 + R3) = 10.0 V/(250 ฮฉ + 1000 ฮฉ + 1250 ฮฉ) = 0.004 A (4.0 mA)
(d) P = IV = 0.004 A ร 10.0 V = 0.04 W
2-2.
(Equation 2-2)
(a) From Equation 2-10, V2 = 15 ร 400/(200 + 400 + 2000) = 2.31 V
(b) P = V22 /R2 = (2.31)2/400 = 0.013 W
(c) Total P = V2/Rs = (15)2/2600 = 0.087 W
1
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
Percentage loss in R2 = (0.013/0.087) ร 100 = 15%
2-3.
V2,4 = 24.0 ร [(2.0 + 4.0) ร 103]/[(1.0 + 2.0 + 4.0) ร 103] = 20.6 V
With the meter in parallel across contacts 2 and 4,
R + 6.0 kฮฉ
1
1
1
=
+
= M
R2,4
(2.0 + 4.0) kฮฉ
RM
RM ร 6.0 kฮฉ
R2,4 = (RM ร 6.0 kฮฉ)/(RM + 6.0 kฮฉ)
(a) R2,4 = (4.0 kฮฉ ร 6.0 kฮฉ)/(4.0 kฮฉ + 6.0 kฮฉ) = 2.40 kฮฉ
VM = (24.0 V ร 2.40 kฮฉ)/(1.00 kฮฉ + 2.40 kฮฉ) = 16.9 V
rel error =
16.9 V โ 20.6 V
ร 100% = โ 18%
20.6 V
Proceeding in the same way, we obtain (b) โ1.2% and (c) โ0.2%
2-4.
Applying Equation 2-19, we can write
(a)
โ1.0% = โ
1000 ฮฉ
ร100%
(RM โ 1000 ฮฉ)
RM = (1000 ร 100 โ 1000) ฮฉ = 99000 ฮฉ or 99 k ฮฉ
(b)
โ0.1% = โ
1000 ฮฉ
ร100%
(RM โ 1000 ฮฉ)
RM = 999 k ฮฉ
2-5.
Resistors R2 and R3 are in parallel, the parallel combination Rp is given by Equation 2-17
Rp = (500 ร 250)/(500 + 250) = 166.67 ฮฉ
(a) This 166.67 ฮฉ Rp is in series with R1 and R4. Thus, the voltage across R1 is
V1 = (15.0 ร 250)/(250 + 166.67 + 1000) = 2.65 V
2
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
V2 = V3 = 15.0 V ร 166.67/1416.67 = 1.76 V
V4 = 15.0 V ร 1000/1416.67 = 10.59 V
(b)
I1R1 = V1 = 2.647 V
I1 = 2.647/250 = 0.01059A (1.06 ร 10โ2 A)
I2 = 1.76 V/500 ฮฉ= 3.5 ร 10โ3 A
I3 = 1.76 V /250 ฮฉ = 7.0 ร 10โ3 A
I4 = 10.59 V/ 1000 ฮฉ = 0.01059 ฮ (1.06 ร 10โ2 A)
(c)
P = IV = 1.76 V ร 7.0 ร 10โ3 A = 1.2 ร 10โ2 W
(d)
Since point 3 is at the same potential as point 2, the voltage between points 3 and
4 (Vโฒ) is the sum of the drops across the 166.67 ฮฉ and the 1000 ฮฉ resistors. Or,
Vโฒ = 1.76 V + 10.59 V = 12.35 V. It is also the source voltage minus the V1
Vโฒ = 15.0 โ 2.65 = 12.35 V
2-6.
The resistance between points 1 and 2 is the parallel combination or RB and RC
R1,2 = 3.0 kฮฉ ร 4.0 kฮฉ/(3.0 kฮฉ + 4.0 kฮฉ) = 1.71 kฮฉ
Similarly the resistance between points 2 and 3 is
R2,3 = 2.0 kฮฉ ร 1.0 kฮฉ/(2.0 kฮฉ + 1.0 kฮฉ) = 0.667 kฮฉ
These two resistors are in series with RA for a total series resistance RT of
RT = 1.71 kฮฉ + 0.667 kฮฉ + 2.0 kฮฉ = 4.38 kฮฉ
I = 24/(4380 ฮฉ) = 5.5 ร 10โ3 A
(a) P1,2 = I2R1,2 = (5.5 ร 10โ3)2 ร 1.71 ร 103 = 0.052 W
(b) As above I = 5.5 ร 10โ3 A (5.5 mA)
(c) VA = IRA = 5.5ร 10โ3 A ร 2.0 ร 103 ฮฉ = 11.0 V
3
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
(d) VD = 24 ร R2,3/RT = 24 ร 0.667/4.38 = 3.65 V
(e) V5,4 = 24 โ VA = 24 โ 11.0 = 13 V
2-7.
With the standard cell in the circuit,
Vstd = Vb ร AC/AB where Vb is the battery voltage
1.018 = Vb ร 84.3/AB
With the unknown voltage Vx in the circuit,
Vx = Vb ร 44.2/AB
Dividing the third equation by the second gives,
1.018 V
84.3 cm
=
Vx
44.3 cm
Vx = 1.018 ร 44.3 cm/84.3 cm = 0.535 V
2-8.
Er = โ
RS
ร 100%
RM + RS
For RS = 20 ฮฉ and RM = 10 ฮฉ, Er = โ
Similarly, for RM = 50 ฮฉ, Er = โ
20
ร 100% = โ 67%
10 + 20
20
ร 100% = โ 29%
50 + 20
The other values are shown in a similar manner.
2-9.
Equation 2-20 is Er = โ
Rstd
ร 100%
RL + Rstd
For Rstd = 1 ฮฉ and RL = 1 ฮฉ, Er = โ
1ฮฉ
ร 100% = โ 50%
1 ฮฉ +1 ฮฉ
4
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
Similarly for RL = 10 ฮฉ, Er = โ
1ฮฉ
ร 100% = โ 9.1%
10 ฮฉ + 1 ฮฉ
The other values are shown in a similar manner.
2-10. (a) Rs = V/I = 1.00 V/20 ร 10โ6 A = 50000 ฮฉ or 50 kฮฉ
(b) Using Equation 2-19
โ1% = โ
50 kฮฉ
ร 100%
RM + 50 kฮฉ
RM = 50 kฮฉ ร 100 โ 50 kฮฉ = 4950 kฮฉ or โ 5 Mฮฉ
2โ11.
ฮ1 = 90/(25 + 5000) = 1.791 ร 10โ2 A
I2 = 90/(45 + 5000) = 1.784 ร 10โ2 A
% change = [(1.784 ร 10โ2 A โ 1.791 ร 10โ2 A)/ 1.791 ร 10โ2 A] ร 100% = โ 0.4%
2-12.
I1 = 12.5/420 = 2.976 ร 10โ2 A
I2 = 12.5/440 = 2.841ร 10โ2 A
% change = [(2.841ร 10โ2 โ 2.976 ร 10โ2)/ 2.976 ร 10โ2] ร 100% = โ4.5%
2-13.
i = Iinit eโt/RC
(Equation 2-35)
RC = 25 ร 106 ฮฉ ร 0.2 ร 10โ6 F = 5.00 s
Iinit = 24V/(25 ร 106 ฮฉ) = 9.6 ร 10โ7 A
i = 9.6 ร 10โ7 eโt/5.00 A or 0.96 ร eโt/5.0 ยตA
t, s
0.00
0.010
0.10
i, ยตA
0.96
0.958
0.941
t, s
1.0
10
i, ยตA
0.786
0.130
5
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
2-14. vC = VC eโt/RC
Chapter 2
(Equation 2-40)
vC/VC = 1.00/100 for discharge to 1%
โ6
0.0100 = eโt/RC = eโt /( R ร 0.025 ร 10 )
ln 0.0100 = โ4.61 = โt/(2.5 ร 10โ8R)
t = 4.61 ร 2.5 ร 10โ8R = 1.15 ร 10โ7R
(a) When R = 10 Mฮฉ or 10 ร 106 ฮฉ, t = 1.15 s
(b) Similarly, when R = 1 Mฮฉ, t = 0.115 s
(c) When R = 1 kฮฉ, t = 1.15 ร 10โ4 s
2-15. (a) When R = 10 Mฮฉ, RC = 10 ร 106 ฮฉ ร 0.025 ร 10โ6 F = 0.25 s
(b) RC = 1 ร 106 ร 0.025 ร 10โ6 = 0.025 s
(c) RC = 1 ร 103 ร 0.025 ร 10โ6 = 2.5 ร 10โ5 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 โ vR
For part (d) we calculate the quantities from
i=
โvC โt / RC
, vR = iR, and vC = โvR
e
R
The results are given in the spreadsheet.
6
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
2-17. Proceeding as in Problem 2-16, the results are in the spreadsheet
7
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
8
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
2-18. In the spreadsheet we calculate XC, Z, and ฯ from
XC = 2/2ฯfC, Z =
R 2 + X C2 , and ฯ = arc tan(XC/R)
2-19. Let us rewrite Equation 2-54 in the form
y=
(V p )o
(V p )i
=
1
(2ฯfRC ) 2 + 1
y2(2ฯfRC)2 + y2 = 1
f =
1
2ฯRC
1
1
1 โ y2
โ
1
=
y2
2ฯRC
y2
The spreadsheet follows
9
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
2-20. By dividing the numerator and denominator of the right side of Equation 2-53 by R, we
obtain
y=
(V p )o
(V p )i
=
1
1 + (1/ 2ฯfRC ) 2
Squaring this equation yields
y2 + y2/(2ฯfRC)2 = 1
2ฯfRC =
f =
y2
1โ y2
1
y2
2ฯRC 1 โ y 2
The results are shown in the spreadsheet that follows.
10
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Principles of Instrumental Analysis, 7th ed.
Chapter 2
11
2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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