Final answer:
The equivalent resistance of the load is 12.97 Ω and the inductance is 72.83 Ω. A capacitance of 4.3 mF is required to raise the power factor.
Step-by-step explanation:
To calculate the equivalent resistance and inductance of the load, we need to use the concept of power factor and impedance in AC circuits.
The apparent power is given as 2.43 KVA, which can be written as 2430 VA. The load has a lagging power factor, so the current lags behind the voltage. The current magnitude is given as 11.320 A, and its angle is -11.320°. We can use this information to calculate the equivalent resistance and inductance of the load.
First, we need to calculate the real power, which is the product of the apparent power and the power factor.
Real power (P) = Apparent Power (S) x Power Factor (pf).
So, P = 2430 VA x 0.683 = 1661.59 W.
The real power is also given by the formula P = I² x R, where I is the current magnitude and R is the resistance.
Rearranging the formula, we can solve for R:
R = P / I²
Substituting the given values, we get R = 1661.59 W / (11.320 A)²
= 12.97 Ω.
Next, we need to find the reactance (X) of the load, which is the imaginary component of the impedance.
The impedance (Z) can be calculated using the formula Z = Vrms / Irms, where Vrms is the root mean square (RMS) voltage and Irms is the RMS current.
The magnitude of the impedance can also be calculated as Z = Vrms / |I|, where |I| is the absolute value of the current magnitude.
Rearranging the formula, we can solve for X:
Substituting the given values, we get
=
=
= 72.83 Ω.
Therefore, the equivalent resistance of the load is 12.97 Ω and the inductance is 72.83 Ω.
To determine the capacitance required to raise the power factor, we need to calculate the reactance of the required capacitor, which is given by Xc = 1 / (2πfC), where f is the frequency and C is the capacitance.
Rearranging the formula, we can solve for C:
C = 1 / (2πfXc)
Substituting the given values, we get C = 1 / (2πfXc)
= 1 / (2π x 60 Hz x 72.83 Ω)
= 0.0043 F
= 4.3 mF.