Applying kcl to the supernode as shown in fig 310a

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Unformatted text preview: own. 3. A supernode requires the application of both KCL and KVL. EXAMPLE 3.3 10 Ω 2V v1 For the circuit shown in Fig. 3.9, ﬁnd the node voltages. Solution: The supernode contains the 2-V source, nodes 1 and 2, and the 10- resistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives +− 2Ω 2A v2 4Ω 2 = i1 + i 2 + 7 7A Expressing i1 and i2 in terms of the node voltages 2= Figure 3.9 For Example 3.3. v1 − 0 v2 − 0 + +7 2 4 8 = 2v1 + v2 + 28 ⇒ or v2 = −20 − 2v1 (3.3.1) To get the relationship between v1 and v2 , we apply KVL to the circuit in Fig. 3.10(b). Going around the loop, we obtain −v1 − 2 + v2 = 0 ⇒ v2 = v1 + 2 (3.3.2) From Eqs. (3.3.1) and (3.3.2), we write v2 = v1 + 2 = −20 − 2v1 or 3v1 = −22 ⇒ v1 = −7.333 V and v2 = v1 + 2 = −5.333 V. Note that the 10- resistor does not make any difference because it is connected across the supernode. 2 v2 2A 2A 2Ω 4Ω 2V 1 i2 7 A i1 + 7A +− 1 v1 2 + v1 v2 − − (b) (a) | v v Figure 3.10 | Applyi...
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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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