Three loads are connected in parallel across a 216 volts (rms) line, as shown in

Question

Three loads are connected in parallel across a 216 volts (rms) line, as shown in circuit above.
Load 1 absorbs 5.6 kW and 7.8 kVAR.
Load 2 absorbs 3.1 kVA at 0.6 pf leading.
Load 3 absorbs 3.8 kW at unity power factor.

1. Find the real part of the impedance of Load 1.
ohms.

2. Find the imaginary part of the impedance of Load 1.
ohms.

3. Find the real part of the impedance of Load 2.
ohms.

4. Find the imaginary part of the impedance of Load 2.
ohms.

5. Find the real part of the impedance of Load 3.
ohms.

6. Find the imaginary part of the impedance of Load 3.
ohms.

7. Find the total average power due to all three parallel loads above.
watts.

8. Find the total reactive power due to all three parallel loads above.
vars.

9. Find the power factor of the equivalent load as seen from the line's input terminals.

Explanation / Answer

1+2)
Total power = 9.6 KVA at 54.3 degree leading

So impedance = 216^2/9600 =4.86

real part = 4.86 cos54.3
= 2.83
imaginary part = 3.94
(unit is ohm)

3+4)
total impedance =216^2/3100
=15.05
real part= 15.05 * power factor
=9.03
imaginary part = 15.05*(1-.62)

=12.04

5+6)

real part = 216^2 /3800 = 12.3

imaginary part =0 (unity power factor means purely resistive circuit)

total active(real part/average power) power = 5.6 + 3.1*.6 + 3.8

= 11.26 KW

total reactive power =7.8 + 3.1 *.8 +0

=10.28 KVAR

[check ur definition of average power , if it means apparent power then take modulus of the power vector

i.e (real ^2 + im^2)]

p.f = cos(tan-1 im/re) = .739 leading(as im part of power is +ve)

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