This preview shows page 1. Sign up to view the full content.
Unformatted text preview: . We express v1 and v2
in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes
6i3 = 3i2 ⇒ i3 = i2
2 (2.8.5) Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives
i2
30 − 3i2
− i2 − = 0
8
2
or i2 = 2 A. From the value of i2 , we now use Eqs. (2.8.1) to (2.8.5) to
obtain
i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V @
Network Analysis PRACTICE PROBLEM 2.8
Find the currents and voltages in the circuit shown in Fig. 2.28. 2Ω Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A,
i3 =1.25 A. i1 i3 + v1 − i2 + v3 −
+
v2
− +
− 5V Figure 2.28 4Ω −
+ 8Ω For Practice Prob. 2.8. 2.5 SERIES RESISTORS AND VOLTAGE DIVISION
The need to combine resistors in series or in parallel occurs so frequently
that it warrants special attention. The process of combining the resistors
is facilitated by combining two of them at a time. With this in mind,
consider the singleloop circuit of Fig. 2.29. The two resistors are in
series, since the same current i ﬂows in both of them. Applying Ohm’s
law to each of the resistors, we obtain
v1 = iR1 , v2 = iR2 v v   eText Main Menu  Textbook Table of Contents  v R1 R2 + v1 − a + v2 − +
− b (2.24) Figure 2.29 If we apply KVL to the loop (moving in the clockwise direction), we have
−v + v1 + v2 = 0 i A singleloop circuit
with two resistors in series. (2.25) Problem Solving Workbook Contents 3V 42 PART 1 DC Circuits Combining Eqs. (2.24) and (2.25), we get
v = v1 + v2 = i(R1 + R2 )
i a Req
+v − v +
− (2.26) or
v
R1 + R 2
Notice that Eq. (2.26) can be written as
i= (2.27) v = iReq
b Figure 2.30 Equivalent circuit
of the Fig. 2.29 circuit. (2.28) implying that the two resistors can be replaced by an equivalent resistor
Req ; that is,
Req = R1 + R2 (2.29) Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The
two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the
same voltagecurrent relationships at the terminals a b. An equivalent
circuit such as the one in Fig. 2.30 is useful in simplifying the analysis
of a circuit. In general, The equivalent resistance of any numbe...
View
Full
Document
This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.
 Spring '12
 bkav

Click to edit the document details