ECE 2006 HW#01 Solution
Problem source: the textbook (4th Edition)
Chapter 01: Problems 1.1, 1.3 (a-c), 1.7, 1.13, 1.17, 1.19
Chapter 02: Problems 2.9, 2.21
Chapter 1, Problem 1.
How many coulombs are represented by these amounts of electrons:
(a) 6.482 × 1017
(b) 1.24 × 1018
(c) 2.46 × 1019
(d) 1.628 × 10 20
Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
Chapter 1, Problem 3. (a-c)
Find the charge q(t) flowing through a device if the current is:
(a) i (t ) = 3A, q(0) = 1C
(b) i (t ) = ( 2t + 5) mA, q(0) = 0
(c) i (t ) = 20 cos(10t + π / 6) µA, q(0) = 2 µ C
Chapter 1, Solution 3. (a-c)
(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C
(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC
(c) q(t)
=
=
∫ 20 cos (10t + π / 6 ) + q(0)
(2sin(10t + π / 6) + 1) µ C
Chapter 1, Problem 7.
The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding
current.
Chapter 1, Solution 7
 25A,
dq 
i=
= - 25A,
dt 
 25A,
0<t<2
2<t<6
6<t<8
which is sketched below:
Chapter 1, Problem 13.
The charge entering the positive terminal of an element is
q = 10 sin 4π t mC
while the voltage across the element (plus to minus) is
v = 2 cos 4π t V
(a) Find the power delivered to the element at t = 0.3 s
(b) Calculate the energy delivered to the element between 0 and 0.6s.
Chapter 1, Solution 13
dq
(a) =
i = 40π cos 4π t mA
dt
p= vi= 80π cos 2 4π t mW
At t=0.3s,
=
p 80
=
π cos 2 (4π x0.3) 164.5 mW
(b)=
W
pdt
∫=
0.6
0.6
0
0
80π ∫ cos 2 4π
=
tdt 40π ∫ [1 + cos8π t ]dt mJ

0.6 
1
40π 0.6 +
sin 8π t
78.34 mJ
W=
=
0 
8π

Chapter 1, Problem 17.
Figure 1.28 shows a circuit with five elements. If
p1 = −205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W,
calculate the power p3 received or delivered by element 3.
Figure 1.28
Chapter 1, Solution 17
Σ p=0
→ -205 + 60 + 45 + 30 + p3 = 0
p3 = 205 – 135 = 70 W
Thus element 3 receives 70 W.
Chapter 1, Problem 19.
Find I in the network of Fig. 1.30.
I
1A
+
+
+
3V
4A
9V
9V
–
+
–
–
–
Figure 1.30
For Prob. 1.19.
6V
Chapter 1, Solution 19
I = 4 –1 = 3 A
Or using power conservation,
9x4 = 1x9 + 3I + 6I = 9 + 9I
4 = 1 + I or I = 3 A
Chapter 2, Problem 9.
Find i1 , i 2 , and i3 in Fig. 2.73.
8A
2A
10 A
i2
12 A
B
A
14 A
Figure i2.73
For Prob. 2.9.
1
At A,
=
2 + 12 i1
At B,
12= i2 + 14
At C,
14
= 4 + i3
C
4A
Chapter 2, Solution 9
i3

→=
i1 14 A

→

→
i2= −2 A
=
i3 10 A
Chapter 2, Problem 21.
Find Vx in the circuit of Fig. 2.85.
2 Vx
1Ω
+
15 V
–
+
+
_
5Ω
Vx
_
2Ω
Figure 2.85 For Prob. 2.21.
Chapter 2, Solution 21
Applying KVL,
-15 + (1+5+2)I + 2 Vx = 0
But Vx = 5I,
-15 +8I + 10I =0,
I = 5/6
Vx = 5I = 25/6 = 4.167 V