An inductor of reactance 100Ω, a capacitor of reactance 50Ω and a resistor of 50Ω are connected in series. What is the power factor of the circuit?
An inductor of reactance 100Ω, a capacitor of reactance 50Ω and a resistor of 50Ω are connected in series. What is the power factor of the circuit?
Right Answer is:
1/√2
SOLUTION
Let’s analyze the given circuit:
- Inductive reactance (XL) = 100 Ω
- Capacitive reactance (XC) = 50 Ω
- Resistance (R) = 50 Ω
Calculating Impedance (Z):
In a series RLC circuit, the impedance Z is given by:
Z = √(R² + (XL – XC)²)
Substituting the values: Z = √(50² + (100 – 50)²)
Z = √(2500 + 2500)
Z = √5000 Z = 50√2 Ω
Calculating Power Factor (cos φ):
Power factor (cos φ) = R / Z
cos φ = 50 / (50√2)
cos φ = 1/√2
Therefore, the power factor of the circuit is 1/√2.