Solution: u (phasor) = 100 ∠ 45 V, ω = 5 rad/s.xl1= ω l1= 5× 2 =10 ω, XL = ω L2 = 5×/kloc.

The phasor diagram of that circuit thus obtain is shown in the above figure.

Circuit impedance is: z = j10+(5+j5) ∨ (-j10) = j10+10 =10 ∠ 2 ∠ 45.

So: I (phasor) =U (phasor)/z =100 ∠ 45/10 ∠ 2 ∠ 45 = 5 ∠ 2 ∠ 0 (a).

That is, I (t) = 5 √ 2×√ 2sin (5t+0) =10sin (5t).

φ u = 45, φ I = 0, so: φ = φ u-φ I = 45.

Active power: p = uicosφ =100× 5 √ 2× cos45 = 500 (w).