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Unformatted text preview: 6×3
=2
6+3
(The symbol is used to indicate a parallel combination.) Also, the 1and 5 resistors are in series; hence their equivalent resistance is
6 3 2Ω
Req +2 Figure 2.34 =6 Req = 4 For Example 2.9. 2Ω Req 6Ω =4
resistor in Fig. 2.35(a); (a)
4Ω = + 2.4 +8 Req 2.4 Ω
8Ω = 14.4 (b) Figure 2.35
PRACTICE PROBLEM 2.9
2Ω By combining the resistors in Fig. 2.36, ﬁnd Req .
Answer: 6 .
Req v v Figure 2.36  eText Main Menu  Textbook Table of Contents  Equivalent circuits for
Example 2.9. 3Ω 6Ω
1Ω  2Ω 8Ω 4×6
= 2.4
4+6
The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig.
2.35(b), the three resistors are in series. Hence, the equivalent resistance
for the circuit is
6 3Ω 4Ω This 4 resistor is now in parallel with the 6their equivalent resistance is
4 6Ω 8Ω Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig.
2.35(a), we notice that the two 2 resistors are in series, so the equivalent
resistance is
2 5Ω = +5 1 1Ω 4Ω 4Ω
3Ω For Practice Prob. 2.9. Problem Solving Workbook Contents 5Ω 46 PART 1 DC Circuits EXAMPLE 2.10
Calculate the equivalent resistance Rab in the circuit in Fig. 2.37.
10 Ω c a 1Ω 1Ω d 6Ω Rab 4Ω 3Ω 5Ω 12 Ω
b
b Figure 2.37 10 Ω
a c 1Ω d 2Ω
b b 3Ω 6Ω b b (a)
10 Ω c a 3Ω 2Ω
b
b b (b) Figure 2.38 Equivalent circuits for
Example 2.10. b For Example 2.10. Solution:
The 3 and 6 resistors are in parallel because they are connected to
the same two nodes c and b. Their combined resistance is
3×6
3
6=
(2.10.1)
=2
3+6
Similarly, the 12 and 4 resistors are in parallel since they are connected to the same two nodes d and b. Hence
12 × 4
12
4=
(2.10.2)
=3
12 + 4
Also the 1 and 5 resistors are in series; hence, their equivalent
resistance is
1 +5 =6
(2.10.3)
With these three combinations, we can replace the circuit in Fig. 2.37 with
that in Fig. 2.38(a). In Fig. 2.38(a), 3 in parallel with 6 gives 2 , as
calculated in Eq. (2.10.1). This 2 equivalent resistance is now in series
with the 1...
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