Using the formulas found in question 1, compute the amount of capacitive current needed in the circuit to correct the power factor to 90%. ### Educational Content: Power Factor Correction in Electrical Circuits#### Instructions for Experimentation1. **Measure the Amount of True Power in the Circuit.** - Record the true power measured in watts (W). - Measurement: __________ W2. **Measure the Amount of Voltage Across the Motor and the Amount of Circuit Current.** - Record the voltage (E) in volts (V) and current (I) in amperes (A). - E = __________ V - I = __________ A 3. **Compute the Circuit Power Factor.** - Calculate the power factor as a percentage (%). - PF = __________ %4. **Conclude the Experiment Safely.** - Turn off the power supply and disconnect the circuit. - Return all components to their proper place.---#### Review Questions1. **Problem Scenario:** Assume that a single-phase motor is connected to a 240-VAC line and has a current draw of 6 A. A wattmeter connected in the circuit indicates a true power value of 864 W. Compute the following values: - **Apparent Power (VA):** - VA = __________ - **Power Factor (PF):** - PF = __________ % - **Reactive Power (VARS):** - VARs = __________2. **Capacitive Current Calculation:** Using the values found in question 1, compute the amount of capacitive current needed in the circuit to correct the power factor to 90%. - \( I_C = \) __________ A3. **Capacitive Reactance Calculation:** Compute the capacitive reactance needed to correct the power factor. - \( X_C = \) __________ Ω4. **Capacitance Calculation:** Compute the amount of capacitance needed to correct the power factor to 90%. - \( C = \) __________ μF### ExplanationThis exercise focuses on understanding the process of power factor correction in electrical circuits. Power factor is a crucial aspect in AC power systems, influencing the efficiency and stability of electrical networks. The experiment and review questions guide students through the calculations needed to determine actual power, apparent power, reactive power, and the correction needed to optimize the circuit’s power factor.

Power Factor Correction is a technique that uses capacitors to reduce the reactive power component of an AC circuit in order to improve its efficiency and reduce current. Given Data: Voltage = 240 A Current = 6 A Power = 864 W Power factor = PowerVA = 8646 x 240 = 0.6Cos∅ = 0.6 ==> ∅ = 53.13Required power factor = 90% = 0.9cos∅1 = 0.9 ==> ∅1 = 24.84Required VARS = P (tan∅ - tan∅1) = 0.864 (tan53.13 - tan24.84) = 0.752 kVARVAR = V * Ic Ic = 0.752 * 1000240 = 3.14 A