kcl_kvl_problem3 - Sample problems Topic: KCL and KVL...

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Sample problems Topic: KCL and KVL Statement of problem The variable resistor R in the circuit below is adjusted until a v equals 60 V. 240V R 45 Ω 10 Ω 180 Ω 18 Ω 12 Ω a v + Find the value of R. Solution The current flowing through the 12 Ω resistor is easily obtained from a v 1 60 5 12 12 a v V iA === ΩΩ In this circuit, there are six branches in total, which are shown in the figure below. 1 5 = 2 i 3 i 4 i 5 i 6 i 1 n 2 n 3 n Therefore, five currents are unknown. Together with the value of R, there are six unknowns in total, so six equations are needed. However, only three nodes (see figure above) are ready for INDEPENDENT KCL equations. Obviously, the other three equations are KVL equations in three loops: the upper one, the lower left one and the
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This note was uploaded on 10/30/2011 for the course EE 215 taught by Professor Unknown during the Spring '05 term at University of Washington.

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kcl_kvl_problem3 - Sample problems Topic: KCL and KVL...

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