Question

A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at a PF of 0.8 lagging, and the third requires 150 W at a unity PF. The parallel load is supplied by a source Vs in series with a 10- resistor. The loads must all operate at 115 V rms. Determine

(a) the rms current through the source;
(b) the PF of the composite load.

Answer 1

Answer:

a) $I_{RMS} = 4.79 A$

b) $PF = 0.908$

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, $cos \theta_{1} = 0.92$

$\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074$

$Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W$

For load 2

Active power, P₂ = 250 W

Power factor, $cos \theta_{2} = 0.8$

$\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87$

$Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W$

For load 3

Active power, P₃ = 250 W

Power factor, $cos \theta_{3} = 1$

$\theta_{3} = cos^{-1} 1\\\theta_{3} =0$

$Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W$

Calculate the total reactive power, $Q_{net} = 42.6 + 187.5 + 0$

$Q_{net} = 230.1 W$

Calculate the total active power, $P_{net} = 100 + 250 + 150 = 500 W$

$S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1$

$P_{net} = IVcos \theta_{net}$

$\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712$

V = 115 $V_{rms}$

$500 = I_{RMS} * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A$

b) Power factor of the composite load is $cos\theta_{net}$

$\theta_{net} = 24.712\\PF = cos 24.712\\PF = 0.908$