Chapter 2, Solution 15.
(a)
P
abs
=
(+3.2 V)(2 mA) = –6.4 mW
(or +6.4 mW
supplied
)
(b)
P
abs
=
(+6 V)(20 A) = –120 W
(or +120 W
supplied
)
(a)
P
abs
=
(+6 V)(2
i
x
) = (+6 V)[(2)(5 A)] = +60 W
(e)
P
abs
=
(4 sin 1000
t
V)(8 cos 1000
t
mA)

t
= 2 ms
=
+12.11 W
Chapter 2, Solution 22.
We are told that V
x
= 1 V, and from Fig. 2.33 we see that the current flowing
through the dependent source (and hence through each element of the circuit) is
5V
x
= 5 A. We will compute
absorbed
power by using the current flowing
into
the positive reference terminal of the appropriate voltage (passive sign
convention), and we will compute
supplied
power by using the current flowing
out of
the positive reference terminal of the appropriate voltage.
(a) The power absorbed by element “A” = (9 V)(5 A) = 45 W
(b) The power supplied by the 1V source = (1 V)(5 A) = 5 W, and
the power supplied by the dependent source = (8 V)(5 A) = 40 W
(c)
The sum of the supplied power = 5 + 40 = 45 W
The sum of the absorbed power is 45 W, so
yes, the sum of the power supplied = the sum of the power absorbed, as
we expect from the principle of conservation of energy.
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 Fall '05
 n/a
 Dependent source, Resistor, Electrical resistance, Pabs, positive reference terminal

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