5.5 The Phasor Circuit cont.¾Example 5-1 cont.

KCL at node n:

(10/-90° V. - Vn) / 3W - Vn / (-j3 W) + 3/30° A. = 0 A.

Which has solution Vn » 6.745/-80.21° V.

Then I = Vn / (-j3 W) » 2.25/+9.79° A.

\ i(t) » 2.25cos(3t + 9.79°) A.

Voila!

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