Unformatted text preview: ng Workbook Contents 48 PART 1
i 4Ω io a +
vo
− 6Ω 12 V +
− 3Ω 12
=2A
4+2
and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage
division, since the 12 V in Fig. 2.42(b) is divided between the 4 and
2 resistors. Hence,
2
vo =
(12 V) = 4 V
2+4
Similarly, io can be obtained in two ways. One approach is to apply
Ohm’s law to the 3 resistor in Fig. 2.42(a) now that we know vo ; thus, (a)
4Ω a
+
vo
− 12 V +
− Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is
not affected by the combination of the resistors because the resistors are
in parallel and therefore have the same voltage vo . From Fig. 2.42(b), we
can obtain vo in two ways. One way is to apply Ohm’s law to get
i= b i DC Circuits 2Ω b 4
A
3
Another approach is to apply current division to the circuit in Fig. 2.42(a)
now that we know i , by writing
vo = 3io = 4 (b) Figure 2.42 For Example 2.12: (a) original
circuit, (b) its equivalent circuit. ⇒ io = 6
2
4
i = (2 A) = A
6+3
3
3
The power dissipated in the 3 resistor is
io = po = vo io = 4 4
3 = 5.333 W PRACTICE PROBLEM 2.12
i1 Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and i2
and the power dissipated in the 12 and 40 resistors. 12 Ω
+ v1 − Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V,
i2 = 250 mA, p2 = 2.5 W. 6Ω
i2
+
15 V 10 Ω +
− Figure 2.43 v2
− 40 Ω For Practice Prob. 2.12. EXAMPLE 2.13  v v For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo , (b)
the power supplied by the current source, (c) the power absorbed by each
resistor.
Solution:
(a) The 6k and 12k resistors are in series so that their combined
value is 6 + 12 = 18 k . Thus the circuit in Fig. 2.44(a) reduces to that  eText Main Menu  Textbook Table of Contents  Problem Solving Workbook Contents CHAPTER 2 Basic Laws 49 shown in Fig. 2.44(b). We now apply the current division technique to
ﬁnd i1 and i2 .
18,000
(30 mA) = 20 mA
i1 =
9000 + 18,000
i2 = 6 kΩ 30 mA...
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 Spring '12
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 Resistor, Electrical impedance, Series and parallel circuits, DC Circuits, eText Main Menu

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