Question

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.

Answer 1

(a)
`X_(C)` = 139.8 Ω

(b)
`X_(C)` = 245 Ω

(c)
`X_(C)` = -24113.8 Ω

Step-by-step explanation:

Given that: R = 245 Ω,
`X_(L)` = 385 Ω.

But,

cos(ϕ) =
`(R)/(Z)`

Where Z is the impedance in the circuit.

(a) when cos(ϕ) = 0.707,

0.707 =
`(245)/(Z)`

⇒ Z =
`(245)/(0.707)`

= 346.5346

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

But,

Z =
`\sqrt{R^(2) + (X_(L) - X_(C))^(2) }`

`Z^(2)` =
`R^(2)` +
`(X_(L) - X_(C))^(2)`

`Z^(2)` -
`R^(2)` =
`(X_(L) - X_(C))^(2)`

`\sqrt{Z^(2) - R^(2) }` =
`X_(L) - X_(C)`

`X_(C)` =
`X_(L)` -
`\sqrt{Z^(2) - R^(2) }`

= 385 -
`\sqrt{346.6^(2) - 245^(2) }`

= 385 - 245.2

= 139.8

Therefore,
`X_(C)` is 139.8 Ω.

(b) When cos(ϕ) = 1.00, then;

cos(ϕ) =
`(R)/(Z)`

1.00 =
`(245)/(Z)`

Z = 245 Ω

The impedance of the circuit is 245 Ω.

So that;

`X_(C)` =
`X_(L)` -
`\sqrt{Z^(2) - R^(2) }`

= 245 - 0

= 245 Ω

The capacitive reactance is 245 Ω. In this circuit, resonance occurs since
`X_(L)` =
`X_(C)`.

(c) When cos(ϕ) = 1.00 x
`10^(-2)`,

cos(ϕ) =
`(R)/(Z)`

1.00 x
`10^(-2)` =
`(245)/(Z)`

Z = 24500 Ω

So that:

`X_(C)` =
`X_(L)` -
`\sqrt{Z^(2) - R^(2) }`

= 385 -
`\sqrt{24500^(2) - 245^(2) }`

= 385 - 24498.8

= -24113.8 Ω

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