Answer:
(a) [tex]X_{C}[/tex] = 139.8 Ω
(b) [tex]X_{C}[/tex] = 245 Ω
(c) [tex]X_{C}[/tex] = -24113.8 Ω
Explanation:
Given that: R = 245 Ω, [tex]X_{L}[/tex] = 385 Ω.
But,
cos(ϕ) = [tex]\frac{R}{Z}[/tex]
Where Z is the impedance in the circuit.
(a) when cos(ϕ) = 0.707,
0.707 = [tex]\frac{245}{Z}[/tex]
⇒ Z = [tex]\frac{245}{0.707}[/tex]
= 346.5346
The impedance of the circuit, Z, is 346.6 Ω.
But,
Z = [tex]\sqrt{R^{2} + (X_{L} - X_{C})^{2} }[/tex]
[tex]Z^{2}[/tex] = [tex]R^{2}[/tex] + [tex](X_{L} - X_{C})^{2}[/tex]
[tex]Z^{2}[/tex] - [tex]R^{2}[/tex] = [tex](X_{L} - X_{C})^{2}[/tex]
[tex]\sqrt{Z^{2} - R^{2} }[/tex] = [tex]X_{L} - X_{C}[/tex]
[tex]X_{C}[/tex] = [tex]X_{L}[/tex] - [tex]\sqrt{Z^{2} - R^{2} }[/tex]
= 385 - [tex]\sqrt{346.6^{2} - 245^{2} }[/tex]
= 385 - 245.2
= 139.8
Therefore, [tex]X_{C}[/tex] is 139.8 Ω.
(b) When cos(ϕ) = 1.00, then;
cos(ϕ) = [tex]\frac{R}{Z}[/tex]
1.00 = [tex]\frac{245}{Z}[/tex]
Z = 245 Ω
The impedance of the circuit is 245 Ω.
So that;
[tex]X_{C}[/tex] = [tex]X_{L}[/tex] - [tex]\sqrt{Z^{2} - R^{2} }[/tex]
= 245 - 0
= 245 Ω
The capacitive reactance is 245 Ω. In this circuit, resonance occurs since [tex]X_{L}[/tex] = [tex]X_{C}[/tex].
(c) When cos(ϕ) = 1.00 x [tex]10^{-2}[/tex],
cos(ϕ) = [tex]\frac{R}{Z}[/tex]
1.00 x [tex]10^{-2}[/tex] = [tex]\frac{245}{Z}[/tex]
Z = 24500 Ω
So that:
[tex]X_{C}[/tex] = [tex]X_{L}[/tex] - [tex]\sqrt{Z^{2} - R^{2} }[/tex]
= 385 - [tex]\sqrt{24500^{2} - 245^{2} }[/tex]
= 385 - 24498.8
= -24113.8 Ω
The capacitive reactance is -24113.8 Ω. This implies that the voltage lags behind the current.
