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

a)
`I_(RMS) = 4.79 A`

b)
`PF = 0.908`

Step-by-step 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) (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`