Answer:
a)
i) To find the resistance of the circuit, we can use the formula:
Power = (Voltage)^2 / Resistance
Rearranging the formula, we get:
Resistance = (Voltage)^2 / Power
Substituting the given values, we get:
Resistance = (240)^2 / 450 = 127.2 ohms
Therefore, the resistance of the circuit is 127.2 ohms.
ii) To find the reactive power of the circuit, we can use the formula:
Reactive power = (Voltage)^2 x sin(θ)
where θ is the angle between the voltage and current phasors.
Since the load current is leading, the angle θ is negative. We can find the value of sin(θ) using the power factor:
Power factor = cos(θ)
cos(θ) = resistance / impedance
impedance = resistance / cos(θ) = 127.2 / cos(-cos⁻¹(0.8)) = 223.4 ohms
sin(θ) = √(1 - cos²(θ)) = √(1 - 0.64) = 0.8
Substituting the given values, we get:
Reactive power = (240)^2 x 0.8 = 46,080 VAR (volt-ampere reactive)
Therefore, the reactive power of the circuit is 46,080 VAR.
iii) To find the capacitance of the circuit, we can use the formula:
Capacitance = Reactive power / (ω x Voltage^2)
where ω is the angular frequency of the AC supply and is given by 2πf, where f is the frequency of the supply.
Substituting the given values, we get:
ω = 2π x 50 = 314.16 rad/s
Capacitance = 46,080 / (314.16 x 240^2) = 1.53 x 10^-6 F (farads)
Therefore, the capacitance of the circuit is 1.53 x 10^-6 F.