A series LCR circuit is connected across a source of emf ε=10sin(100πt-π/6) . . .

Question : A series LCR circuit is connected across a source of emf ε=10sin(100πt-π/6). The current from the supply is I=2sin(100πt+π/2). Draw the impedance triangle for the circuit.

Doubt by Mohini

Solution :

ε=10sin(100πt-π/6)

I=2sin(100πt+π/2)

Compare the above two equations with 

ε=ε₀sin(ωt+Φ₁)

I=I₀sin(ωt+Φ₂)

ε₀ =10 V

I₀ = 2 A

Φ₁ = -π/6

Φ₂ = π/2

Φ = Φ₂-Φ₁
Φ = π/2-(-π/6)
Φ = 3π/6+π/6
Φ = 4π/6
Φ = 2π/3 = 120°

From Ohm's law

ε₀ = I₀z

z = ε₀/I₀
z = 10/2

z = 5 Ω

From Power Factor 

cosΦ = R/z

R = z cosΦ

R = z cos(2π/3)
R = 5×cos(π-π/3)
R = 5×[-cosπ/3] [∵cos(π-θ) = -cosθ]

R = 5×[-1/2]
R = -5/2
R = -2.5 Ω
|R| = 2.5 Ω

Also

z=√[R²+(X_L-X_C)²]
z²=R²+(X_L-X_C)²
z²-R² = (X_L-X_C)²
(5)²-(2.5)²=(X_L-X_C)²

25-6.25 = (X_L-X_C)²
18.75=(X_L-X_C)²
X_L-X_C=√18.75

X_L-X_C=4.33 Ω

So, 

Now You can draw the Impedance Triangle by Yourself with the given below details.

Φ = π/3 = 60°

X_L-X_C=4.33 Ω
R=2.5 Ω

z = 5 Ω