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An RLC circuit has resistance R = 245 Ω and inductive reactance XL = 385 Ω. HINT

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.
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carlosiscr7
carlosiscr7
5,0(99 marks)

(a) X_{C} = 139.8 Ω

(b) X_{C} = 245 Ω

(c) X_{C} = -24113.8 Ω

Explanation:

Given that: R = 245 Ω, X_{L} = 385 Ω.

But,

cos(ϕ) = \frac{R}{Z}

Where Z is the impedance in the circuit.

(a) when cos(ϕ) = 0.707,

0.707 = \frac{245}{Z}

⇒ Z = \frac{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) = \frac{R}{Z}

1.00 = \frac{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) = \frac{R}{Z}

1.00 x 10^{-2} = \frac{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.

angelaOR
angelaOR
4,5(16 marks)

good question

Step-by-step explanation:

123

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