,RadioFrequencyIntegrated
t t
Circuits and Systems
t t t
SolutionManual
t
Hooman Darabi
t
,Solutions to Problem Sets t t t
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem sets
t t t t t t t t t t t t t t t t
depict examples of practical applications of the concepts described in the book, more detailed
t t t t t t t t t t t t t t
analysis of some of the ideas, or in some cases present a new concept.
t t t t t t t t t t t t t t
Note that selected problems have been given answers already in the book.
t t t t t t t t t t t
, 1 Chapter One t
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
t t t t t t t t t t
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
t t t t t t t t t t t t t t t t t
diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
t t t t t t t t t t t t t t t t t
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
t t t t t t t t t t t t t t
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge the
t t t t t t t t t t t t t t t
same.
t
-
+
+S - + a + -
b
+
-
From Gauss’s law:
t t
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2 t t t t t t
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
t t t t t t t t
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 t t
𝑟 t
t
Assuming a potential of 𝑉0 between the𝑎inner and
𝑎 2 1 we
2 outer surfaces, 1 have:
𝑑𝑟 = 𝜌𝑆 𝑎
t
1
t t t t t t t t t t t t t
𝑉 =−
t
𝜌 t
t
t t t t t t
0 𝑆 ( − ) t t t t t
𝑏 𝑟2 𝜖 𝑎 𝑏 t
Thus: 𝜖 t
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 t t
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 t t t
t
𝑎 −𝑏
t
0
𝜖 𝑎 (𝑎 𝑏− ) t
t
t
t
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 =
t t t t t t t t t t t t t t t t
t t t t ×
36𝜋
5t
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 =
t t t t t t t t t
𝑝𝐹 = 0.55𝑝𝐹. t t
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
t t t t t t t t t t t
capacitance as a function of the parameters shown in the figure.
t t t t t t t t t t t
t t
Circuits and Systems
t t t
SolutionManual
t
Hooman Darabi
t
,Solutions to Problem Sets t t t
The selected solutions to all 12 chapters problem sets are presented in this manual. The problem sets
t t t t t t t t t t t t t t t t
depict examples of practical applications of the concepts described in the book, more detailed
t t t t t t t t t t t t t t
analysis of some of the ideas, or in some cases present a new concept.
t t t t t t t t t t t t t t
Note that selected problems have been given answers already in the book.
t t t t t t t t t t t
, 1 Chapter One t
1. Using spherical coordinates, find the capacitance formed by two concentric spherical
t t t t t t t t t t
conducting shells of radius a, and b. What is the capacitance of a metallic marble with a
t t t t t t t t t t t t t t t t t
diameter of 1cm in free space? Hint: let 𝑏 → ∞, thus, 𝐶 = 4𝜋𝜀𝜀0𝑎 = 0.55𝑝𝐹.
t t t t t t t t t t t t t t t t t
Solution: Suppose the inner sphere has a surface charge density of +𝜌𝑆. The outer surface
t t t t t t t t t t t t t t
charge density is negative, and proportionally smaller (by (𝑎/𝑏)2) to keep the total charge the
t t t t t t t t t t t t t t t
same.
t
-
+
+S - + a + -
b
+
-
From Gauss’s law:
t t
ф𝐷 ⋅ 𝑑𝑆 = 𝑄𝑄 = +𝜌𝑆4𝜋𝑎2 t t t t t t
𝑆
Thus, inside the sphere (𝑎 ≤ 𝑟 ≤ 𝑏):
t t t t t t t t
𝑎2
𝐷 = 𝜌𝑆 2 𝑎𝑟 t t
𝑟 t
t
Assuming a potential of 𝑉0 between the𝑎inner and
𝑎 2 1 we
2 outer surfaces, 1 have:
𝑑𝑟 = 𝜌𝑆 𝑎
t
1
t t t t t t t t t t t t t
𝑉 =−
t
𝜌 t
t
t t t t t t
0 𝑆 ( − ) t t t t t
𝑏 𝑟2 𝜖 𝑎 𝑏 t
Thus: 𝜖 t
𝜌𝑆4𝜋𝑎2 = 4𝜋𝜖
𝑄𝑄 t t
𝐶 =𝑉 =
𝜌 𝑆 21 1 1 1 t t t
t
𝑎 −𝑏
t
0
𝜖 𝑎 (𝑎 𝑏− ) t
t
t
t
1
In the case of a metallic marble, 𝑏 → ∞, and hence: 𝐶 = 4𝜋𝜀𝜀0 𝑎. Letting 𝜀𝜀0 =
t t t t t t t t t t t t t t t t
t t t t ×
36𝜋
5t
10−9, and 𝑎 = 0.5𝑐𝑚, it yields 𝐶 =
t t t t t t t t t
𝑝𝐹 = 0.55𝑝𝐹. t t
9
2. Consider the parallel plate capacitor containing two different dielectrics. Find the total
t t t t t t t t t t t
capacitance as a function of the parameters shown in the figure.
t t t t t t t t t t t
