(b) If the diameter of the wire is doubled, the cross sectional area
A
is
P1.56 The resistance is proportional to the resistivity and inversely
P1.57*
100
100
1002
2
1
P
V
R
P1.58
P1.59
P1.60 The power delivered to the resistor is
P1.61 The power delivered to the resistor is
and the energy delivered is
P1.62* (a) Not contradictory.
(b) A 2-A current source in series with a 3-A current source is
P1.63*
Applying Ohm’s law, we have
V 5A 1 5
2
v
. However,
2
v
Thus, parallel. in are that resistors three all across voltage the is
P1.64*
As shown above, the 2 A current circulates clockwise through all three
elements in the circuit. Applying KVL, we have
P1.65 (a) The voltage across the voltage source is 10 V independent of the
P1.66 (a) The 1- resistance, the 2- resistance, and the voltage source
vx
are
in series.
(b) The 6– resistance and the 3- resistance are in parallel.
(c) Refer to the sketch of the circuit. Applying Ohm’s law to the 6-
P1.67 The power for each element is 120 W in magnitude. The voltage source
P1.68 This is a parallel circuit, and the voltage across each element is 12 V
positive at the bottom end. Thus, the current flowing through the
resistor is
P1.69
Ohm’s law for the 4– resistor yields:
V. 4
1
v
Then, we have
P1.70* (a) Applying KVL, we have
xx vv
510
, which yields
V 667.16/10
x
v
P1.71* We have a voltage-controlled current source in this circuit.
P1.72 (a) No elements are in series.
(c)
V 422
2
v
P1.74 This is a voltage-controlled current source. First, we have
P1.75 This circuit contains a voltage-controlled voltage source.
Applying KVL around the periphery of the circuit, we have
P1.76 Consider the series combination shown below on the left. Because the
P1.77 Consider the parallel combination shown below on the left. Because the
voltage for parallel elements must be the same, the voltage
vab
must be
P1.78 (a)
21
4
ii
P1.79 (a)
21
vvv
P1.80 The source labeled
Is
is an independent current source. The source
labeled
aix
is a current-controlled current source. Applying ohm’s law to
P1.81 (a)
213
iii
P1.82 The source labeled 24 V is an independent voltage source. The source
Practice Test
T1.2 (a) The current
Is
3 A circulates clockwise through the elements
(b) Because
Is
enters the negative reference for
Vs
, we have
PV
VsIs
(d) First, we must find the voltage
vI
across the current source. We
choose the reference shown:
T1.3 (a) The currents flowing downward through the resistances are
vab
/
R
1 and
vab
/
R
2. Then, the KCL equation for node
a
(or node
b
) is
8 V.
(b) The power for current source
I
1 is
W 2438
11
IvP ab
I
.
(c) The power absorbed by
R
1 is
W. 33.512/)8(/ 2
1
2
1
RvP ab
R
The
T1.4 (a) Applying KVL, we have
.0
21
vvVs
Substituting values given in
T1.5 Applying KVL, we have
.0
xs vV
Thus,
V. 15
sx Vv
Next Ohm’s law