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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
C =12µF
Time (µs)
𝑖= 𝐶
𝑑𝑣
𝑑𝑡
𝑑𝑣
𝑑𝑡
(V/µs)
i(t) (A)
0 ≤t≤6 2
24
6 ≤t ≤10 −5
−60
10 ≤t ≤16 1.33
16
t >16
0
0
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
𝑖= 𝐶
C =12µF
𝑑𝑣
Time(ms)
0 ≤t ≤2
2 ≤t ≤4
4 ≤t ≤7
𝑑𝑡
𝑑𝑣
𝑑𝑡
(V/µs)
2
−1
0
7 ≤t ≤9
9 ≤t ≤10
t >0
0
i(t) (A)
6
−3
0
3
− −4.5
2
1
3
0
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
From the graph we know that
𝑣(𝑡)
𝑖(𝑡)
𝑝(𝑡)
𝑝(𝑡)
= −2𝑡 + 24 0 ≤ 𝑡 ≤ 6
= 24
= 𝑣(𝑡) × 𝑖(𝑡)
= −48𝑡 + 576
𝑡
𝐸(𝑡) = ∫ 𝑝(𝑥)𝑑𝑥
0
𝐸(𝑡) = −24𝑡 2 + 576𝑡 𝑚𝐽
= 3.455 𝑚𝐽 𝑎𝑡 𝑡 = 6𝑚𝑠
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
1
∫ 𝑖𝑑𝑡 + 𝑉(𝑡0)
𝐶
1
∫ 4𝑡 × 10−3 𝑑𝑡 + 10
=
50 × 10−3
2𝑡 2
=
+ 10
50
𝑣=
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
𝑡
𝑤 = 𝐿 ∫ 𝐿𝑖 2 𝑑𝑡 =
0
1 2
1
𝐿𝑡 (𝑡) − 𝐿𝑖 2 (−∞)
2
2
1
= × 80 × 10−3 × (60 × 10−3 )2 − 0
2
= 144µ𝐽
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
IR =
𝑉
=
6
𝑅
12
𝑑𝑉
IC= 𝐶
𝑑𝑡
𝑒 −2000𝑡 = 0.5𝑒 −2000𝑡
= 100 × 10−3 × 6(−2000)𝑒 −2000𝑡
= −120010−2000𝑡
I=IR+IC=-1199.5𝑒 −2000𝑡
𝑝 = 𝑣𝑖 = −7197𝑒 −4000𝑡 𝑊
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
1 𝑡
∫ 6(1 − 𝑒 −𝑡 )𝑑𝑡 + 0
1 0
2
= 12(𝑡 + 𝑒 −𝑡 ) − 12
𝑣(2) = 12(2 + 𝑒 −𝑡 ) − 12 = 13.624𝑉
𝑝 = 𝑖𝑣 = [12(𝑡 + 𝑒 −𝑡 ) − 12]6(1 − 𝑒 −𝑡 )
𝑝(2) = [12(2 + 𝑒 −2 ) − 12]6(1 − 𝑒 −2 ) = 70.66 𝑊
𝑣(𝑡) =
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(a) 3μF is in series with 6μF 3x6/(9) = 2μF
V4µF= 1/2 x 120 = 60V
V2µF = 60V
V6µF=
3
6+3
(60) = 20𝑉
V3µF = 60 − 20 = 40 𝑉
1
(b) Hence 𝑤 = 𝐶𝑣 2
1
2
W4µF = × 4 × 10−6 × 3600 = 7.2𝑚𝐽
2
1
W2µF= × 2 × 10−6 × 3600 = 3.6𝑚𝐽
2
1
W6µF= × 2 × 10−6 × 400 = 1.2𝑚𝐽
2
1
W3µ= × 2 × 10−6 × 1600 = 2.4𝑚𝐽
2
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The equivalent capacitor can be evaluated as following
8 × 6µ𝐹
8+6
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
2µ𝐹 +
24µ𝐹
7
So the Cequivalent= 2.85µF
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
C in series with C = C/(2)
C/2 in parallel with C = 3C/2
3
in series with C =
2
3𝐶
2
3𝐶
2
𝐶
5
2
𝐶×
In parallel with C = C+
3𝐶
5
= 1.6 𝐶
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We combine 10, 20, and 30μF capacitors in parallel to have 60μF. Then 60μF capacitor in series with
another 60μF capacitor gives 30μF. 30 + 50 = 80μF, 80 + 40 = 120μF the circuit is reduced to that shown
below.
120-μF capacitor in series with 80μF gives (80x120)/200 = 48μF
48 + 12 = 60μF
60μF capacitor in series with 12μF gives
(60x12)/72 = 10μF
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The has been modified as following
b
From the figure in P6.53,
4C1=2C2
From figure above
3(C1+C2) = 3C2
Solving for C1 and C2.yields
C1= 2µF
C2 = 4µF
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We can simplify the above figure by as following
L1=L3=L6= 1 mH
L2= 12 mH, L4 = 6mH
L5=4 mH, L7= L8= L9=2 mH
Leq1 = L6 + L7 = 3 Mh
Leq2 = L4Lwq1/(L4 + Leq1) = 2Mh
Leq3 = L8 + L9 + 4 Mh
Leq4 = Leq3L5/(Leq3+Leq5)= 2Mh
Leq5 =
𝐿2(𝐿𝑒𝑞2+𝐿𝑒𝑞4)
𝐿2+𝐿𝑒𝑞2+𝐿𝑒𝑞4
= 5𝑚𝐻
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
LAB = 5 mH
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The figure in the question can be drawn as follows
4𝐿
+ 2) 𝐿
4+𝐿
=2
4𝐿
+2+𝐿
4+𝐿
6𝐿2 + 8𝐿 = 2𝐿2 + 20𝐿 + 16
(𝑙 + 4)(𝐿 + 1) = 0
𝐿 = 4𝑚𝐻
(
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(a) VL= 𝐿
= −2𝑒
𝑑𝑖
𝑑𝑡
−400𝑡
= 100 × 10−3 (−400) × 50 × 10−3 𝑒 −400𝑡
𝑉)
(b) R and L are parallel to each other so VR = VL
1
(C) 𝑤 = 𝐿𝑖 2 = 0.5 × 100 × 10−3 (0.05)2
2
= 125µ𝐽
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Rs = 10KΩ
𝑣 = −10 ∫ 𝑣𝑠 𝑑𝑡
Req = Rs + 70K = 80k
Considering i1 = i2 (ideal op-amp)
𝑣𝑠 − 0
𝑑𝑣
= −𝐶
𝑅𝑒𝑞
𝑑𝑡
1
∫ 𝑣𝑠 𝑑𝑡
𝑣𝑜 = −
𝑅𝑒𝑞 𝐶
1
𝑅𝑒𝑞 𝐶 =
10
𝐶 = 1.25µ𝐹
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
1
1
2
2
(a) W5 = 𝐿𝑖 2 = × 5 × (4)2 = 40 𝐽
1
W20 = (20)(−2)2 = 40𝐽
2
(b) w= w5+w20 = 80J
(c) i1 =
I2 ==
1 𝑡
∫ −50𝑒 −200𝑡
𝐿2 0
𝑑𝑡 + 𝑖1(0)
1 𝑡
∫ −50𝑒 −200𝑡 𝑑𝑡
𝐿2 0
= 5 × 10−5 (𝑒 −200𝑡 − 1) + 4 𝐴
+ 𝑖2(0)
= 1.25 × 10−5 (𝑒 −200𝑡 − 1) − 2 𝐴
(d) I = i1+i2 = 6.25 × 10−5 (𝑒 −200𝑡 − 1) + 2 𝐴
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
If we differentiate the wave form of I it will give the wave form similar to the voltage wave form So
𝑑𝑖
𝑑𝑡
−4𝑡,
0 < 𝑡 < 1 𝑚𝑠
𝑖={
8 − 4𝑡,
1 < 𝑡 < 2 𝑚𝑠
4000𝐿, 0 < 𝑡 < 1 𝑚𝑠
𝑣={
−4000𝐿, 1 < 𝑡 < 2 𝑚𝑠
𝑣=𝐿
From the graph
4000L = 5
L = 1.25 mH
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SOLUTION:
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
i = iR+ iC
𝑣𝑖 − 0 0 − 𝑣0
𝑑(0 − 𝑣0 )
=
+𝐶
𝑅
𝑅𝐹
𝑑𝑡
𝑅𝐹 = 106 × 10−6 = 1
𝑣𝑖 = −(𝑣0 +
The graph of 𝑣0 is shown below
Chapter 6: Capacitance and Inductance
𝑑𝑣0
)
𝑑𝑡
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 6: Capacitance and Inductance
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The given equation can be modeled with a op amp summer, integrator and an inverter.
Summer equation
𝑣0 = −(
𝑅𝐹
𝑅𝐹2
𝑣1 +
𝑣 )
𝑅1
𝑅2 2
Integrator equation
𝑣0 = −
1
∫ 𝑣𝑠 𝑑𝑡
𝑅𝐶
Inverter equation
𝑣0 = −
𝑅𝐹
𝑣
𝑅 𝑠
Comparing the above equation with the given equation we get
𝑅𝐹 = 10𝑅
1
𝐶=
2𝑅
and for inverter 𝑅𝐹 = 𝑅
So the circuit of the given equation is following
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Chapter 6: Capacitance and Inductance