An RLC circuit has resistance R = 245 Ω and inductive reactance XL = 385 Ω. HINT (a) Calculate the circuit's capacitive reactance XC (in Ω) if its power factor is cos(ϕ) = 0.707. Ω (b) Calculate the circuit's capacitive reactance XC (in Ω) if its power factor is cos(ϕ) = 1.00. Ω (c) Calculate the circuit's capacitive reactance XC (in Ω) if its power factor is cos(ϕ) = 1.00 ✕ 10−2.

(a) = 139.8 Ω

(b) = 245 Ω

(c) = -24113.8 Ω

Explanation:

Given that: R = 245 Ω, = 385 Ω.

But,

cos(ϕ) =

Where Z is the impedance in the circuit.

(a) when cos(ϕ) = 0.707,

0.707 =

⇒ Z =

       = 346.5346

The impedance of the circuit, Z, is 346.6 Ω.

But,

Z =

= +

- =

=

= -

     = 385 -

     = 385 - 245.2

     = 139.8

Therefore, is 139.8 Ω.

(b) When cos(ϕ) = 1.00, then;

cos(ϕ) =

1.00 =

Z = 245 Ω

The impedance of the circuit is 245 Ω.

So that;

= -

     = 245 - 0

     = 245 Ω

The capacitive reactance is 245 Ω. In this circuit, resonance occurs since = .

(c) When cos(ϕ) = 1.00 x ,

cos(ϕ) =

1.00 x =

Z = 24500 Ω

So that:

= -

     = 385 -

     = 385 -  24498.8

     = -24113.8 Ω

The capacitive reactance is -24113.8 Ω. This implies that the voltage lags behind the current.

b) phosphorus pentoxide