UNIVERSITY OF CALIFORNIA
College of Engineering
Department of Electrical Engineering
And Computer Sciences
B.E. BOSER
Problem Set 1 Solutions
EE42/100
Spring 2008
1.
2.
!
!
!
a. Use Kirchoff’s Voltage Law (KVL).
v x + "v1 = 0
v x = v1
b. v x = v1 = "3.4V
a. Use Kirchoff’s Current Law (KCL).
ix + i1 = 0
ix = "i1
b. ix = "i1 = "1.2µA
3.
!
a. 3 nodes
b. In our nomenclature, all circuit elements are "branches" with terminal
connected to wires. In this scenario there are 4 branches in this circuit, one for
each circuit element.
We will also accept 3 branches if you used the methodology discussed in discussion
sections.
c. Use Kirchoff’s Current Law.
ix + i1 + i3 = 0
ix = "(i1 + i3 )
d. ix = "(i1 + i3 ) = 2.8µA
!
4.
!
!
! 5.
!
!
a. Use Kirchoff’s Voltage Law. (KVL)
v1 " v x + v 2 = 0
v x = v1 + v 2
b. Use Kirchoff’s Voltage Law (KVL)
vy + v2 = 0
v y = "v 2
c. v x = "4mV
v y = "4mV
a. Use Kirchoff’s Current Law
ix + i1 = 0 " ix = #i1
iy " i1 + i2 = 0 # iy = i1 " i2
iz + i2 = 0 " iz = #i2
!
!
!
b.
ix = "5mA
iy = 2mA
iz = "3mA
c. Use the equations found in problem 4a) and problem 5a)
!
ix = "5mA
iy = 7mA
v x = "6V
v y = "2V
iz = 2mA
Use the proper sign conventions for calculating power. See Nilsson Figure 1.6.
P(X n ) = v n in (Power Equation)
P(X1 ) = v1 * i!
1 = "8V * 5mA = "40mW
P(X 2 ) = v 2 * iy = 2V * 7mA = 14mW
!
P(X 3 ) = v y * iz = "2V * 2mA = "4mW
!
P(X 4 ) = v x * ix = "6V * "5mA = 30mW
Remember, if the power is positive (p>0), power is being dissipated by the circuit
element. If the power is negative (p<0), power is being delivered to the circuit by the
! circuit element.
d. " P(X n ) = " v n in = 0
In a closed circuit, the total power dissipated by the circuit should sum to zero.
6.
!
!
!
a. 6 nodes
b.
P(X 6 ) = v 6i6
v 6 + v 3 + v 5 = 0 " v 6 = #(v 3 + v 5 )
i6 = i5 + i3
P(X 6 ) = #(v 3 + v 5 )(i5 + i3 )
c.
P(X 3 ) = v 7i7
v 7 + v 2 = 0 " v 7 = #v 2
i7 + i2 + i4 = 0 " i7 = #(i2 + i4 )
P(X 3 ) = (v 2 )(i2 + i4 )
d. The trick to this problem is to label all branch currents and node voltages
needed to solve the power equation. Using KCL and KVL, you should be able to
determine all new currents and voltages from the given ones.
Corrected Problem Set with i3=4mA:
P(X 3 ) = 8mW
P(X 6 ) = 18mW
!
Uncorrected Problem Set with i3=-4mA:
P(X 3 ) = 8mW
P(X 6 ) = "2mW
" P(X
!
!
n
) = " v n in = 0
7. Recall that a Joule is the work done to produce power of one watt continuously for
one second; or one watt second, symbol W·s.
1
t = 22kJ "
50mW
1
1min ute 1hour
t = 22 "10 3 Ws "
#3
50 "10 W 60s 60min utes
t = 122hours
8.
!
!
capacity total = 50A " hours "12V = 600W " hours
capacity available = .8 " capacity total = 480W " hours
capacity available
1
t=
480W " hours "
powerconsumed
40W
t = 12hours
9. Recall that one ampere represents the rate of 1 coulomb of charge per second.
n electrons
1electron
= 1µA *
s
1.602 176 487 "10 -19 C
n electrons 1"10#6 C
1electron
=
*
s
s
1.602 176 487 "10 -19 C
n electrons 6.241"1012 electrons
=
s
s
!
10. Determine the carrier velocity, we should first determine the number of free electrons
per unit length as well the number of electrons passing per second.
n electrons
1electron
= 1A *
s
1.602 176 487 "10 -19 C
n electrons 1" C
1electron
=
*
s
s 1.602 176 487 "10 -19 C
n electrons 6.241"1018 electrons
=
s
s
3"10 24
.1mm 2
" # *(
)
3
cm
2
3"10 24 electrons
.1"10$3 m 2
nx =
"
#
*
(
)
(10$2 m) 3
2
nx =
!
n x = 2.356 "10 24 electrons /m
n electrons
6.241"1018 electrons
m
nx =
"
s
s
2.356 "10 24 electrons
v = 2.648 "10#6 m /s = 2.648 "10#9 km /s
v = 9.536km /h
v=
!
!