# From the value of i2 we now use eqs 281 to 285 to

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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 single-loop 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 | | e-Text 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 single-loop 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 voltage-current 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...
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## 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.

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