Answer to Question #158280 in Electric Circuits for Theo

Question

A single phase load draws 60 amps at a power factor of 0.9 lagging from a 230 v 50 hertz supply.

calculate

-apparent power

-true power

phase angle

the reactive power

Answer 1

Explanations & Calculations

• When a circuit is not purely resistive, the power distribution occurs both as reactive & real power.
• Real power is what accessible by the user whereas the reactive power stores inside the system during the operation.
• The relationship between these is given by

True power = Apparent power $\small \times$ power factor

Power factor = cousin of phase angle

• Then,

• Apparent power = $\small Vi=230V\times60A=13800=13.8 kVA$

Units are expressed in volt-ampere instead of Watts

• True power = $\small 13.8\times 0.9=12.42 kW$

Units are expressed in watts

• If the phase angle is $\small \theta$,

$\qquad\qquad \cos\theta=0.9\to\theta=0.451 rad$

• Reactive power can be calculated in 2 ways.
• First, R.Power = Apparent power $\small \times \sin(phase angle)$

= $\small 13.8\times \sin(0.451)=6.015 \,kVAR$

• Second, R.Power = $\small \sqrt{(Apparent\,power)^2-(True\,power)^2}=\sqrt{13.8^2-12.42^2}= 6.015\,kVAR$

Units are expressed as VA reactive.