# 1 2 3vo 4 i2 i1 24 v v v figure 334 2 8 4 i3 vo for

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Unformatted text preview: e currents i1 , i2 , and i3 . Solution: The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35 includes the output results, implying that it is the schematic displayed on the screen after the simulation.) Notice that the voltage-controlled voltage source E1 in Fig. 3.35 is connected so that its input is the voltage across the 4resistor; its gain is set equal to 3. In order to display the required currents, we insert pseudocomponent IPROBES in the appropriate branches. The schematic is saved as exam311.sch and simulated by selecting Analysis/Simulate. The results are displayed on IPROBES as shown in Fig. 3.35 and saved in output ﬁle exam311.out. From the output ﬁle or the IPROBES, we obtain i1 = i2 = 1.333 A and i3 = 2.667 A. 1Ω 2Ω 3vo +− 4Ω i2 i1 24 V + − | v v Figure 3.34 | 2Ω 8Ω 4Ω i3 + vo − For Example 3.11. e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 101 102 PART 1 DC Circuits E E1 −+ −+ 2 R5 1 R1 R6 4 + 24 V − R2...
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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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