# Power Calculation of Delta Connected Balanced Load

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Please refer to Fig. 1 for the following discussion
Fig. 1: $\Delta$-connected balanced load.
Average Power
$$\begin{array}{c}P_{\phi}=V_{\phi} I_{\phi} \cos \theta_{I_{\phi}}^{V_{\phi}}=I_{\phi}^{2} R_{\phi}=\frac{V_{R}^{2}}{R_{\phi}} \\P_{T}=3 P_{\phi} \quad \text { (W) }\end{array}$$
Reactive Power
$$\begin{array}{c}Q_{\phi}=V_{\phi} I_{\phi} \sin \theta_{I_{\phi}}^{V_{\phi}}=I_{\phi}^{2} X_{\phi}=\frac{V_{X}^{2}}{X_{\phi}} \\Q_{T}=3 Q_{\phi} \quad \text { (VAR) }\end{array}$$
Apparent Power
$$\begin{array}{c}S_{\phi}=V_{\phi} I_{\phi} \\S_{T}=3 S_{\phi}=\sqrt{3} E_{L} I_{L}\end{array}$$
Power Factor
$$F_{p}=\frac{P_{T}}{S_{T}}$$
Example 1: For the Y-connected load of Fig. 2, find the total average, reactive, and apparent power. In addition, find the power factor of the load.
Fig. 2: Example 1.
Solution: Consider the $\Delta$ and $Y$ separately. For the $\Delta$ :
\begin{aligned}\mathbf{Z}_{\Delta} &=6 \Omega-j 8 \Omega=10 \Omega \angle-53.13^{\circ} \\I_{\phi} &=\frac{E_{L}}{Z_{\Delta}}=\frac{200 \mathrm{~V}}{10 \Omega}=20 \mathrm{~A} \\P_{T_{\Delta}} &=3 I_{\phi}^{2} R_{\phi}=(3)(20 \mathrm{~A})^{2}(6 \Omega)=\mathbf{7 2 0 0} \mathrm{W} \\Q_{T_{\Delta}} &=3 I_{\phi}^{2} X_{\phi}=(3)(20 \mathrm{~A})^{2}(8 \Omega)=\mathbf{9 6 0 0} \mathrm{VAR}(C) \\S_{T_{\Delta}} &=3 V_{\phi} I_{\phi}=(3)(200 \mathrm{~V})(20 \mathrm{~A})=\mathbf{1 2 , 0 0 0 ~ V A}\end{aligned}
For the Y:
$$\begin{array}{l}\mathbf{Z}_{\mathrm{Y}}=4 \Omega+j 3 \Omega=5 \Omega \angle 36.87^{\circ} \\I_{\phi}=\frac{E_{L} / \sqrt{3}}{Z_{\mathrm{Y}}}=\frac{200 \mathrm{~V} / \sqrt{3}}{5 \Omega}=\frac{116 \mathrm{~V}}{5 \Omega}=23.12 \mathrm{~A} \\P_{T_{\mathrm{Y}}}=3 I_{\phi}^{2} R_{\phi}=(3)(23.12 \mathrm{~A})^{2}(4 \Omega)=\mathbf{6 4 1 4 . 4 1 \mathrm { W }} \\Q_{T_{\mathrm{Y}}}=3 I_{\phi}^{2} X_{\phi}=(3)(23.12 \mathrm{~A})^{2}(3 \Omega)=\mathbf{4 8 1 0 . 8 1} \mathrm{VAR}(L) \\S_{T_{\mathrm{Y}}}=3 V_{\phi} I_{\phi}=(3)(116 \mathrm{~V})(23.12 \mathrm{~A})=\mathbf{8 0 4 5 . 7 6 ~ V A}\end{array}$$
For the total load:
\begin{aligned}P_{T} &=P_{T_{\Delta}}+P_{T_{\mathrm{Y}}}=7200 \mathrm{~W}+6414.41 \mathrm{~W}=13,614.41 \mathrm{~W} \\Q_{T} &=Q_{T_{\Delta}}-Q_{T_{\mathrm{Y}}}=9600 \mathrm{VAR}(C)-4810.81 \mathrm{VAR}(I) \\&=\mathbf{4 7 8 9 . 1 9} \mathrm{VAR(C)}\\S_{T} &=\sqrt{P_{T}^{2}+Q_{T}^{2}}=\sqrt{(13,614.41 \mathrm{~W})^{2}+(4789.19 \mathrm{VAR})^{2}} \\&=\mathbf{1 4 , 4 3 2 . 2 \mathrm { VA }} \\F_{p} &=\frac{P_{T}}{S_{T}}=\frac{13,614.41 \mathrm{~W}}{14,432.20 \mathrm{VA}}=\mathbf{0 . 9 4 3} \text { leading }\end{aligned}

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