However kcl must be satised at a supernode like any

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Unformatted text preview: ted differently. Why? Because an essential component of nodal analysis is applying KCL, which requires knowing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the supernode in Fig. 3.7, i1 + i4 = i2 + i3 (3.11a) or v 1 − v2 v 1 − v3 v2 − 0 v3 − 0 (3.11b) + = + 2 4 8 6 To apply Kirchhoff ’s voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8. Going around the loop in the clockwise direction gives −v2 + 5 + v3 = 0 ⇒ v2 − v3 = 5 | v v From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. | e-Text Main Menu | Textbook Table of Contents | 5V + +− + v2 − (3.12) Figure 3.8 v3 − Applying KVL to a supernode. Problem Solving Workbook Contents 84 PART 1 DC Circuits Note the following properties of a supernode: 1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages. 2. A supernode has no voltage of its...
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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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