Question

THE LOAD APPARENT POWER AT .6 PF LAG IS GIVEN BY SL=57∠-53.13 mva,

What is the load impedance in ohms if Zl=(10.45)²
divided by 57∠-53.13
Handwritten steps thanks

Answer 1

The load impedance is approximately 1.872 + j0.258 Ω.

Step-by-step explanation:

The load impedance in ohms can be found by dividing the square of the load impedance magnitude by the load apparent power. In this case, the load impedance is given as Zl=(10.45)^2 divided by 57∠-53.13.

Substituting the given values, the load impedance is:

Zl = (10.45)^2 / 57∠-53.13 = 1.872 + j0.258 Ω

Therefore, the load impedance is approximately 1.872 + j0.258 Ω.

Answer 2

The student needs to calculate the load impedance given the apparent power and a formula. By applying the given formula and power in polar form, the load impedance in ohms can be found.

Step-by-step explanation:

The student asked about determining the load impedance of an AC circuit when the apparent power (SL) is given as 57∠-53.13 MVA with a power factor (PF) of 0.6 lagging. To calculate the load impedance ZL, the apparent power is divided by the square of the current, or in this context, ZL is calculated using the given apparent power and a formula where ZL = (10.45)2 / SL. Considering the apparent power is represented in polar form, and we're given the magnitude and phase angle, the calculation for ZL will involve both real and imaginary parts derived from SL. After computing ZL, we will obtain the load impedance in ohms (Ω).