Question 2.(a) A coil of resistance 10Ω and inductance 140mH is connected in parallel with a 260Ωresistor across a 230V, 50Hz supply. Calculate the following(i) Current in the coil and phase angle of this current (4 Marks)(ii) Supply current (4 Marks)(iii) Circuit impedance (3 Marks)(iv) Power factor (3 Marks)(v) Power consumed (3 Marks)(b) Explain what is meant by the term “ Power Factor Correction”. Using a phasor diagramshow the impact of a 47μF capacitor placed in parallel with the coil described above.

Sure, let's solve this step by step:

(i) Current in the coil and phase angle of this current:

First, we need to calculate the inductive reactance (XL) of the coil. The formula for inductive reactance is XL = 2πfL, where f is the frequency and L is the inductance.

XL = 2π * 50Hz * 140mH = 44Ω

The coil is a combination of resistance (R) and inductive reactance (XL), so we can use Pythagoras' theorem to find the total impedance (Z) of the coil: Z = √(R² + XL²)

Z = √(10Ω² + 44Ω²) = 45.2Ω

The current in the coil (I) can be found using Ohm's law: I = V/Z

I = 230V / 45.2Ω = 5.08A

The phase angle (φ) can be found using the formula: φ = arctan(XL/R)

φ = arctan(44Ω / 10Ω) = 77.47°

(ii) Supply current:

The supply current is the same as the current in the coil, so I = 5.08A.

(iii) Circuit impedance:

The total circuit impedance (ZT) can be found using the formula for parallel resistances: 1/ZT = 1/Z1 + 1/Z2

1/ZT = 1/45.2Ω + 1/260Ω

ZT = 40.8Ω

(iv) Power factor:

The power factor (PF) can be found using the formula: PF = cos(φ)

PF = cos(77.47°) = 0.22

(v) Power consumed:

The power consumed (P) can be found using the formula: P = IVcos(φ)

P = 5.08A * 230V * 0.22 = 258.4W

(b) Power Factor Correction:

Power factor correction is the process of improving a power factor closer to 1 by using capacitors or inductors to counteract the reactive power in the system. A power factor of 1 means that the voltage and current are in phase and there's no reactive power. This makes the system more efficient because less current is needed to deliver a certain amount of power.

When a 47μF capacitor is placed in parallel with the coil, it creates a capacitive reactance that counteracts the inductive reactance of the coil. This brings the phase angle closer to 0°, which improves the power factor. The phasor diagram would show the voltage and current vectors closer together.