Q13Basic Electrical Engineering

Question

Q.3. Determine the power factor of a RLC series circuit with R = 5 Ω, XL = 8 Ω and XC = 12 Ω.

Answer 1

In an RLC series alternating current circuit, the total opposition to the current flow is called Impedance ( $Z$ ). It is calculated using the Resistance ( $R$ ), Inductive Reactance ( $X_{L}$ ), and Capacitive Reactance ( $X_{C}$ ).

Given values: Resistance ( $R$ ) = $5 Ω$ Inductive Reactance ( $X_{L}$ ) = $8 Ω$ Capacitive Reactance ( $X_{C}$ ) = $12 Ω$

Step 1: Calculate Total Impedance ( $Z$ ) The formula for impedance in a series circuit is:

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

Z = 5^{2} + (8 - 12)^{2} = 25 + (- 4)^{2} = 25 + 16 = 41 \approx 6.403 Ω

Step 2: Calculate Power Factor (pf) The power factor is the cosine of the phase angle ( $ϕ$ ) and is defined as the ratio of true resistance to total impedance.

p f = cos (ϕ) = \frac{R}{Z} = \frac{5}{6.403} \approx 0.7808

Nature of the Power Factor: Because the capacitive reactance ( $X_{C} = 12 Ω$ ) is strictly greater than the inductive reactance ( $X_{L} = 8 Ω$ ), the overall circuit behaves as a capacitive circuit. Therefore, the current leads the voltage, meaning the power factor is $0.78$ leading. (Correction: If XC > XL, circuit is capacitive, so it's leading, not lagging as previously summarized).

Q13Basic Electrical Engineering

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