Question

A series circuit has a 69 ohms resistor, an inductor with reactance 116 ohms and a capacitor with reactance 102 Ohms. They are connected to a 120 V power source. What is the power factor of this circuit?

Answer 1

The power factor of an RLC series circuit consisting of a 69 ohms resistor, an inductor with reactance 116 ohms, and a capacitor with reactance 102 Ohms connected to a 120 V power source is approximately 0.98.

Step-by-step explanation:

To find the power factor of the given RLC series circuit, we need to consider both the resistive (real) and reactive (imaginary) components of impedance. In a series circuit, the total impedance (Z) is the vector sum of resistance (R), inductive reactance (XL), and capacitive reactance (XC). Inductive reactance adds to the impedance along the imaginary axis, while capacitive reactance subtracts from it.

• Resistance (R): 69 ohms
• Inductive reactance (XL): 116 ohms
• Capacitive reactance (XC): 102 ohms

The net reactance (X) is XL - XC, which is 116 ohms - 102 ohms = 14 ohms. The power factor (pf) is then given by R/Z, where Z is the square root of (R2 + X2). Substituting the given values:

1. Z = √(692 + 142) ohms
2. Z = √(4761 + 196) ohms
3. Z = √(4957) ohms
4. Z = 70.4 ohms approximately
5. pf = R/Z = 69/70.4
6. pf ≈ 0.98

The power factor of the circuit is approximately 0.98, indicating that the circuit is mostly resistive with a very small phase difference between voltage and current.