Let's solve **Problem 1** **step-by-step** with explanations and formulas: --- ### **Given:** - **Load 1:** 60 kVA at PF 0.80 lagging - **Load 2:** 30 kVA at PF 0.75 lagging - **Line Voltage at Loads:** \( V_{LL} = 208~\text{V}_{rms} \) (line-to-line voltage) - **Frequency:** 60 Hz ### **Required:** a) Magnitude of the line currents (\( I_L \)) b) Power factor at which the source is supplying power (\( PF_S \)) --- ## **Step 1: Find Total Apparent Power (\( S_{total} \))** \[ S_{total} = S_1 + S_2 = 60~\text{kVA} + 30~\text{kVA} = 90~\text{kVA} \] --- ## **Step 2: Find Real Power (\( P \)) and Reactive Power (\( Q \)) for Each Load** ### **Load 1:** \[ P_1 = S_1 \times PF_1 = 60 \times 0.80 = 48~\text{kW} \] \[ Q_1 = S_1 \times \sin(\cos^{-1}(0.80)) = 60 \times \sin(36.87^\circ) = 60 \times 0.6 = 36~\text{kVAR} \] ### **Load 2:** \[ P_2 = S_2 \times PF_2 = 30 \times 0.75 = 22.5~\text{kW} \] \[ Q_2 = S_2 \times \sin(\cos^{-1}(0.75)) = 30 \times \sin(41.41^\circ) = 30 \times 0.6614 = 19.84~\text{kVAR} \] --- ## **Step 3: Find Total Real and Reactive Power** \[ P_{total} = P_1 + P_2 = 48 + 22.5 = 70.5~\text{kW} \] \[ Q_{total} = Q_1 + Q_2 = 36 + 19.84 = 55.84~\text{kVAR} \] --- ## **Step 4: Find Total Apparent Power (\( S_{total} \))** \[ S_{total} = \sqrt{P_{total}^2 + Q_{total}^2} \] \[ S_{total} = \sqrt{(70.5)^2 + (55.84)^2} = \sqrt{4970.25 + 3120.71} = \sqrt{8090.96} = 89.96~\text{kVA} \] --- ## **Step 5: Find Power Factor at Source** \[ PF_S = \frac{P_{total}}{S_{total}} = \frac{70.5}{89.96} = 0.784 \text{ (lagging, since both loads are lagging)} \] --- ## **Step 6: Find Line Current (\( I_L \))** For a balanced three-phase system: \[ S = \sqrt{3} V_{LL} I_L \] \[ I_L = \frac{S_{total}}{\sqrt{3} V_{LL}} \] Convert \( S_{total} \) to VA: \[ S_{total} = 89,960~\text{VA} \] \[ I_L = \frac{89,960}{\sqrt{3} \times 208} \] \[ I_L = \frac{89,960}{360.624} = 249.641~\text{A} \] --- ## **Final Answers** **a) Magnitude of the line currents:** \[ \boxed{I_L = 249.641~\text{A}_{rms}} \] **b) Source power factor:** \[ \boxed{PF_S = 0.784~\text{lagging}} \] --- ### **Summary Table of Formulas Used** | Quantity | Formula | Description | |------------------|-----------------------------------------------------|-----------------------------------------------| | Real Power (\(P\)) | \(P = S \times PF\) | Real power (kW) | | Reactive Power (\(Q\)) | \(Q = S \times \sin(\cos^{-1}(PF))\) | Reactive power (kVAR) | | Apparent Power (\(S\)) | \(S = \sqrt{P^2 + Q^2}\) | Total apparent power (kVA) | | Power Factor (\(PF\)) | \(PF = \frac{P_{total}}{S_{total}}\) | Power factor at source | | Line Current (\(I_L\)) | \(I_L = \frac{S_{total}}{\sqrt{3} \times V_{LL}}\) | Magnitude of the line current (A) | --- Let me know if you want step-by-step solutions for Problems 2 or 3!