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A single-phase motor takes 10 A at a power factor of 0.776 lagging when connected to a 230V, 50 Hz supply. A capacitor bank is connected in parallel with the motor to raise the power factor to 0.95 a) Calculate the motor equivalent impedance, the currents in the circuit, and draw the phasor diagram before and after power correction.

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Final answer:

The student's question involves the calculation of equivalent impedance, currents before and after power correction, and the creation of phasor diagrams for a single-phase motor connected to a 230V, 50 Hz supply.

Step-by-step explanation:

The problem requires us to find the equivalent impedance of a motor and calculate the currents in the circuit before and after power factor correction. To begin with, the motor's power factor is 0.776, and we know the voltage (230V) and the current (10A). The apparent power (S) is given by S = V * I, and the real power (P) can be found as P = V * I * power factor. With these values, the equivalent impedance (Z) can be calculated as Z = V / I.

After finding the impedance, to correct the power factor to 0.95, we would need to calculate the required capacitive reactance (Xc) which can be connected in parallel to the motor to achieve the desired power factor. The change in the reactive power (Q) required to alter the power factor from 0.776 to 0.95 can be found using the initial and final power factors. Then, using the formula for capacitive reactance (Xc = 1 / (2 * π * f * C)), we can find the value of the required capacitance.

Creating phasor diagrams can visualize the difference before and after power factor correction, showing the original current, the added capacitive current, and the resultant corrected current.

User Ievgen Martynov
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