Let I 1 (phasor) = 10 ∠ 0 (a), then: U2 (phasor) =I 1 (phasor) × (-jxc) =10 ∠ 0× xc.

I2 (phasor) =U2 (phasor)/Z2 =10xc ∞-90/∠ 2R2 ∠ 45 = 5 ∠ 2 (xc/R2) ∞-135 =/kloc-0 ∞

Where: 5√2(Xc/R2)= 10√2, so: Xc=2R2.

According to KCl: I 1 (phasor) +I2 (phasor) =I (phasor), I (phasor) =10-j10 =-j10 (.

I= 10(A)。

U 1 (phasor) =I 1 (phasor) × r1=-j10× 5 =-j50 (v).

U (phasor) =U 1 (phasor) +U2 (phasor) =-j50+10xc ∞-90 =-j50-j10xc =-j (50+10xc

And: U=|U (phasor) |=50+ 10Xc=200, so: xc = 15 (ω).

Therefore: R2 = XL = XC/2 = 15/2 = 7.5 (ω).