The Analysis and Design of Linear Circuits
Seventh Edition
2 Basic Circuit Analysis
2.1 Exercise Solutions
Exercise 2–1.
A 6V lantern battery powers a light bulb that draws 3 mA of current. What is the resistance
of the lamp? How much power does the lantern use?
Using Ohm’s law, we have
v
=
iR
or
R
=
v
i
, so we can compute the resistance as
R
=
6 V
3 mA
= 2 kΩ. The
power is
p
=
vi
= (6 V)(3 mA) = 18 mW.
Exercise 2–2.
What is the maximum current that can flow through a
1
8
W, 6.8kΩ resistor? What is the
maximum voltage that can be across it?
The resistor can dissipate up to 0.125 W of power. We have
p
MAX
=
i
2
MAX
R
, which we can solved for
i
MAX
and then substitute in values for the power and resistance
i
MAX
=
radicalbigg
p
MAX
R
=
radicalbigg
0
.
125
6800
= 4
.
2875 mA
Similarly, we can use
p
MAX
=
v
2
MAX
R
to solve for the maximum voltage as follows:
v
MAX
=
radicalbig
R p
MAX
=
radicalbig
(6800)(0
.
125) = 29
.
155 V
Exercise 2–3.
A digital clock is a voltage that switches between two values at a constant rate that is used
to time digital circuits. A particular clock switches between 0 V and 5 V every 10
μ
s. Sketch the clock’s
i

v
characteristics for the times when the clock is at 0 V and at 5 V.
When the clock has a value of 0 V, its voltage is constant and zero for a wide range of currents. In this
case, the
i

v
characteristic is a vertical line at 0 V. Likewise, when the clock has a value of 5 V, the voltage
is constant at 5 V for a wide range of currents. In this case, the
i

v
characteristic is a vertical line at 5 V.
Exercise 2–4.
Refer to Figure 2–12.
Figure 2–12
(a). Write KCL equations at nodes A, B, C, and D.
KCL states that the sum of the currents entering a node is zero at every instant.
As we sum the
currents at a node, if the current enters that node, it is positive and if the current leaves the node, it is
negative. At node A, both currents
i
1
and
i
2
are leaving the node, so the equation is
−
i
1
−
i
2
= 0. At
node B, current
i
2
enters the node and currents
i
3
and
i
4
leave the node, so we have
i
2
−
i
3
−
i
4
= 0.
At node C, current
i
4
enters the node and currents
i
5
and
i
6
leave the node, so we have
i
4
−
i
5
−
i
6
= 0.
At node D, currents
i
1
,
i
3
,
i
5
, and
i
6
enter the node, so we have
i
1
+
i
3
+
i
5
+
i
6
= 0.
(b). Given
i
1
=
−
1 mA,
i
3
= 0
.
5 mA,
i
6
= 0
.
2 mA, find
i
2
,
i
4
, and
i
5
.
Applying the KCL equation for node A, we can find
i
2
=
−
i
1
= 1 mA. Applying the KCL equation
for node B, we have
i
4
=
i
2
−
i
3
= 1
−
0
.
5 = 0
.
5 mA. Finally, applying the KCL equation for node C,
we have
i
5
=
i
4
−
i
6
= 0
.
5
−
0
.
2 = 0
.
3 mA.
Solution Manual Chapter 2
Page 21