Final answer:
The impedance of the inductive coil is 20Ω in Cartesian form and 20Ω ∠ 60 degrees in polar form. The power factor is 0.5, indicating a phase shift due to reactive elements in the circuit. The reactive power is approximately 1732.05 VAR, and the apparent power is 2000 VA.
Step-by-step explanation:
The problem presented involves an inductive coil connected to a 200 V, 50 Hz AC supply which draws a current of 10 A and consumes 1 kW of power. To solve for impedance in both Cartesian and polar forms, the power factor, and the reactive and apparent power, we first use the formula for power (P = VIcosφ) to find the power factor, cosφ. We know that P = 1 kW and I = 10 A, so:
1000 W = 200 V × 10 A × cosφ
cosφ = 1000 W / (200 V × 10 A) = 0.5
This means the power factor is 0.5. To find the impedance (Z), we use the formula Z = V / I, where V = 200 V and I = 10 A:
Z = 200 V / 10 A = 20 Ω
The impedance in Cartesian form is simply 20 Ω, because there is only a real component (resistance) and an imaginary component (reactance) which is zero in this case. To express it in polar form, we take into account the power factor angle (φ), which is the arccosine of the power factor. Hence:
φ = arccos(0.5) = 60 degrees
The polar form of the impedance (Z) is 20 Ω ∠ 60 degrees. Since the power factor is less than 1, this indicates that there is a phase difference between the voltage and current, which suggests the presence of inductance or capacitance in the circuit that causes reactive power (Q). Reactive power can be found using the formula Q = VI sinφ:
Q = 200 V × 10 A × sin(60 degrees) = 1732.05 VAR
Apparent power (S) can be calculated using the formula S = VI:
S = 200 V × 10 A = 2000 VA
Which concludes that the apparent power is 2000 VA and the reactive power is approximately 1732.05 VAR.