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• Notes

$\begin{array}{l}{\text { The two loads in the circuit shown can be described as follows: }} \\ {\text { Load } 1 \text { absorbs } 8 \mathrm{kW} \text { at a leading p.f. of } 0.8 \text { . }} \\ {\text { Load2 absorbs } 20 \mathrm{kVA} \text { at a lagging p.f. of } \mathrm{O} \text { . } 6 \text { . }} \\ {\text { a) Determine the power factor of the two loads in parallel. }} \\ {\text { b) Determine the apparent power required to supply the loads, }} \\ {\text { the magnitude of the current, } I_{s} \text { , and power loss in T.L. }(0.05+\mathrm{j} 0.5)}\end{array}$

Load (1) $L_{1} \quad P_{1}=8 K w$

$PF=0.8$    leading     $Q=-v e$

$Q=S \sin \phi=\frac{P}{cos \phi}*\sqrt{1-(PF)^2}$

$Q=\frac{8000}{0.8} * \sqrt{1-(0.8)^2}=6000 \ VAR$

$S_{1}=P+J Q=8000-J6000$

Load (2) $L_{2} \quad 20 \mathrm{KVA}\ , \mathrm{PF}=0.6$

$P=S \cos \phi=20000 \times 0.6$

$P=12000 \mathrm{w}$

$Q=S \sin \phi=20000 \sqrt{1-(0.6)^{2}}$

$Q=16000 \mathrm{VAR}$

$S_{2}=P+J Q=12000 +J 16000$

$\overline{S}_{t o t a l}=\overline{S}_{1}+\overline{S}_{2}=8000-J{6000}+12000+J16000$

$\overline{S}_{t o t a l}=(20000+J10000) \ VA$

$I_s^*=\frac{\overline{S}_{total}}{\overline{V}_{r ms}}=\frac{20000+J10000}{250 \ \angle 0}=(80+J40) \ A$

$\left.I_{s}\right|_{r ms}=(80-J40) A \ , \ |I_{rms}|=\sqrt{(80)^2+(40)^2}=89.44 A$

$P_{loss}=I^{2} R=(89.44)^2*0.05=400 w$