# Solution to the Problem: Magnitude of Voltage \(E_{AB}\) To determine the magnitude of the voltage \(E_{AB}\) for the balanced three-phase system shown in Fig. 24.40, we will proceed in a structured manner: --- ## **Step 1: Analyzing the System and Phase Voltage** The system is a **3-phase, 3-wire, Y-connected generator**. The given phase voltage is \(V_\phi = 50\, \text{V}\). **For a Y-connected system:** - The phase voltage (\(V_\phi\)) is the voltage from neutral to any one phase. - The line-to-line (or line) voltage (\(V_{LL}\)) is the voltage between any two lines, e.g., \(E_{AB}\). **Relationship:** \[ V_{LL} = \sqrt{3} V_\phi \] **Explanation:** This relationship arises because the line voltage in a balanced Y-connection is the vector difference between two phase voltages, and the phase difference between them is \(120^\circ\). --- ## **Step 2: Calculating the Line Voltage (Ideal Case)** Let’s first calculate the ideal line-to-line voltage without considering any line resistance. \[ V_{LL} = \sqrt{3} \times 50\, \text{V} = 86.60\, \text{V} \] **Explanation:** This is the voltage that would appear between the lines at the generator terminals if there were no line resistance. --- ## **Step 3: Considering Line Resistance and Calculating Actual \(E_{AB}\)** Each line has a resistance of \(1\,\Omega\) as shown. The load per phase includes \(16\,\Omega\) (inductive) and \(12\,\Omega\) (resistive) in series, so **per phase load impedance** is: \[ Z_\text{load} = 12\,\Omega + j16\,\Omega \] \[ |Z_\text{load}| = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\,\Omega \] **Total impedance per phase (including line):** \[ Z_\text{total,ph} = 1\,\Omega + Z_\text{load} = 1 + 12 + j16 = 13 + j16\,\Omega \] \[ |Z_\text{total,ph}| = \sqrt{13^2 + 16^2} = \sqrt{169 + 256} = \sqrt{425} \approx 20.62\,\Omega \] **Current per phase:** \[ I_\text{ph} = \frac{V_\phi}{|Z_\text{total,ph}|} = \frac{50\,\text{V}}{20.62\,\Omega} \approx 2.424\,\text{A} \] **Voltage drop across the line resistance (per line):** \[ V_{\text{drop,line}} = I_\text{ph} \times 1\,\Omega = 2.424\,\text{A} \times 1\,\Omega = 2.424\,\text{V} \] **Total drop between A and B (since both lines A and B have 1Ω):** \[ \text{Total drop from A to B} = 2 \times V_{\text{drop,line}} = 2 \times 2.424\,\text{V} = 4.848\,\text{V} \] **Actual \(E_{AB}\) at the sending (generator) end:** \[ E_{AB} = V_{LL} + \text{Total drop from A to B} \] Since the drop is subtracted from the ideal line voltage (because the load receives less voltage), the generator must provide: \[ E_{AB} = V_{LL} + 4.848\,\text{V} = 86.60\,\text{V} + 4.848\,\text{V} \approx 91.45\,\text{V} \] --- ## **Final Answer** \[ \boxed{E_{AB} \approx 91.5\,\text{V}} \] --- ### **Summary of Steps** 1. **Analyzed** the Y-connected system and wrote the phase-to-line voltage relationship. 2. **Calculated** the ideal line-to-line voltage (\(86.6\,\text{V}\)) using the given phase voltage. 3. **Accounted for line resistance**, found the load and total impedance, calculated phase current, found the total line voltage drop, and **added it to the ideal line voltage** to get the generator-end voltage. **Final magnitude of the voltage \(E_{AB}\) is approximately \(91.5\,\text{V}\).**