5263a plugi1backinto4yields i203947a

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Unformatted text preview: o (2): ‐ 6*I2 – 18*(I1 + I2) = 0 ‐18*I1 – 24*I2 = 0 Simultaneously solve (4) and (1): I1 = = 0.5263 A Plug I1 back into (4) yields: I2 = ‐ 0.3947 A Using the values of I1 and I2 in (3): IX = 0.1315 A b) To find Vab, you can apply KVL to the following loop: ‐10 ‐ V1 + V2 + Vab = 0 Vab = 10 + V1 – V2 = 10 + (I1 * 5) ‐ (I2 * 3) = 13.82 V c) Applying KVL, we have: ‐ V3 ‐ Vda + V4 = 0 Vda = V4 ‐ V3 = (‐ IX * 6) – (I2 * 3) = 0.39 V d) Vdb = Vda + Vab = 14.21 V e) Power associated...
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This note was uploaded on 06/25/2013 for the course ECE 100 taught by Professor Pister during the Spring '13 term at University of California, Berkeley.

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