Find the current i2 for the network of the following Fig. i1 = 80 !! 10-3 sin ?t iT = 120 !! 10-3 sin (?t + 60°) i2 = ? i2 (t) = 105.8 !! 10-3 sin (? t + 100.89°) i2 (t) = 0.4!! 10-3 sin (? t + 75°) i2 (t) = 22.3 sin (? t + 17°) i2 (t) = 10-3 sin (? t + 89°)
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i₁ = 80 × 10³ sin ωt iₓ = 120 × 10³ sin (ωt + 60°) Show more…
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