Angular frequency.
ω = 2 π f = 2 π ⋅ 50 ≈ 314 rad/s \omega=2\pi f=2\pi\cdot 50\approx 314\ \text{rad/s} ω = 2 π f = 2 π ⋅ 50 ≈ 314 rad/s
Reactances.
X L = ω L = 314 ⋅ 0,5 ≈ 157 Ω X_L=\omega L=314\cdot 0{,}5\approx 157\ \Omega X L = ω L = 314 ⋅ 0 , 5 ≈ 157 Ω X C = 1 ω C = 1 314 ⋅ 10 ⋅ 10 − 6 = 1 3,14 ⋅ 10 − 3 ≈ 318 Ω X_C=\frac{1}{\omega C}=\frac{1}{314\cdot 10\cdot 10^{-6}}=\frac{1}{3{,}14\cdot 10^{-3}}\approx 318\ \Omega X C = ω C 1 = 314 ⋅ 10 ⋅ 1 0 − 6 1 = 3 , 14 ⋅ 1 0 − 3 1 ≈ 318 Ω X L − X C ≈ 157 − 318 = − 161 Ω X_L-X_C\approx 157-318=-161\ \Omega X L − X C ≈ 157 − 318 = − 161 Ω
Impedance.
Z = R 2 + ( X L − X C ) 2 = 50 2 + 161 2 = 2500 + 25921 = 28421 ≈ 169 Ω ≈ 1,7 ⋅ 10 2 Ω Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{50^2+161^2}=\sqrt{2500+25921}=\sqrt{28421}\approx 169\ \Omega\approx 1{,}7\cdot 10^{2}\ \Omega Z = R 2 + ( X L − X C ) 2 = 5 0 2 + 16 1 2 = 2500 + 25921 = 28421 ≈ 169 Ω ≈ 1 , 7 ⋅ 1 0 2 Ω
Phase shift.
tan φ = X L − X C R = − 161 50 ≈ − 3,22 ⟹ φ ≈ − 72 ∘ \tan\varphi=\frac{X_L-X_C}{R}=\frac{-161}{50}\approx -3{,}22\quad\Longrightarrow\quad \varphi\approx -72^\circ tan φ = R X L − X C = 50 − 161 ≈ − 3 , 22 ⟹ φ ≈ − 7 2 ∘
Since φ < 0 \varphi<0 φ < 0 X C > X L X_C>X_L X C > X L leads the voltage.
Z ≈ 1,7 ⋅ 10 2 Ω , φ ≈ − 72 ∘ (current leading) \ev{Z\approx 1{,}7\cdot 10^{2}\ \Omega,\quad \varphi\approx -72^\circ\ \text{(current leading)}} Z ≈ 1 , 7 ⋅ 1 0 2 Ω , φ ≈ − 7 2 ∘ (current leading)