Let's solve the problem step by step as requested: --- ## **Given:** - **System:** 3-phase, 4-wire, 208 V system (line-to-line voltage). - **Load:** Balanced, star (Y)-connected. - **Load Impedance per phase:** \( Z = 20∠-30^\circ\ \Omega \) (for each phase). - **Phase voltage:** \( V_{AN} = 120.09∠-90^\circ \) V. We are to **find the line currents** and **draw the phasor diagram**. --- ## **Step 1: Understanding the System** For a **star (Y) connection**: - **Phase Voltage:** Voltage between line and neutral, \( V_{ph} = V_{AN}, V_{BN}, V_{CN} \). - **Line Current:** Current in each line is the same as the phase current, \( I_A = I_{AN}, I_B = I_{BN}, I_C = I_{CN} \). --- ## **Step 2: Find the Phase Voltages** Given \( V_{AN} = 120.09∠-90^\circ \) V. For a balanced 3-phase system: - \( V_{BN} \) lags \( V_{AN} \) by 120°. - \( V_{CN} \) lags \( V_{BN} \) by 120°, or \( V_{CN} \) lags \( V_{AN} \) by 240°. Thus, \[ \begin{align*} V_{AN} &= 120.09∠-90^\circ \text{ V} \\ V_{BN} &= 120.09∠(-90^\circ - 120^\circ) = 120.09∠-210^\circ \text{ V} \\ V_{CN} &= 120.09∠(-90^\circ - 240^\circ) = 120.09∠-330^\circ \text{ V} \end{align*} \] **Or,** \[ \begin{align*} V_{BN} &= 120.09∠+150^\circ \text{ V} \\ V_{CN} &= 120.09∠+30^\circ \text{ V} \end{align*} \] (Since \(-210^\circ = +150^\circ\) and \(-330^\circ = +30^\circ\)) --- ## **Step 3: Find the Phase Currents** The current for each phase is: \[ I_{ph} = \frac{V_{ph}}{Z} \] Given \( Z = 20∠-30^\circ \ \Omega \): - \( 1/Z = 1/20∠+30^\circ = 0.05∠+30^\circ \) So: ### **For Phase A:** \[ \begin{align*} I_A &= \frac{120.09∠-90^\circ}{20∠-30^\circ} \\ &= 6.0045∠(-90^\circ - (-30^\circ)) \\ &= 6.0045∠(-60^\circ) \\ &\approx 6∠-60^\circ\ \text{A} \end{align*} \] ### **For Phase B:** \[ \begin{align*} I_B &= \frac{120.09∠+150^\circ}{20∠-30^\circ} \\ &= 6.0045∠(150^\circ - (-30^\circ)) \\ &= 6.0045∠180^\circ \\ &\approx 6∠180^\circ\ \text{A} \end{align*} \] ### **For Phase C:** \[ \begin{align*} I_C &= \frac{120.09∠+30^\circ}{20∠-30^\circ} \\ &= 6.0045∠(30^\circ - (-30^\circ)) \\ &= 6.0045∠60^\circ \\ &\approx 6∠60^\circ\ \text{A} \end{align*} \] --- ## **Step 4: Final Answers** - \( I_A = 6∠-60^\circ \) A - \( I_B = 6∠180^\circ \) A - \( I_C = 6∠60^\circ \) A **Or, equivalently:** - \( I_A = 6∠300^\circ \) A (since \(-60^\circ = 300^\circ\)) - \( I_B = 6∠180^\circ \) A - \( I_C = 6∠60^\circ \) A --- ## **Step 5: Phasor Diagram** ### **How to Draw:** - Draw a horizontal axis (real axis). - Draw \( I_A \) at \(-60^\circ \) (or \( 300^\circ \)), which is below the axis, 60° clockwise from the positive real axis. - Draw \( I_B \) at \( 180^\circ \), to the left along the negative real axis. - Draw \( I_C \) at \( 60^\circ \), above the axis, 60° counterclockwise from the positive real axis. **All vectors have equal magnitudes (6 A) and are separated by 120°.** --- ## **Step 6: Boxed Final Answers** \[ \boxed{ \begin{align*} I_A &= 6∠-60^\circ\ \text{A} \\ I_B &= 6∠180^\circ\ \text{A} \\ I_C &= 6∠60^\circ\ \text{A} \end{align*} } \] **Or in alternate notation:** \[ \boxed{ \begin{align*} I_A &= 6∠300^\circ\ \text{A} \\ I_B &= 6∠180^\circ\ \text{A} \\ I_C &= 6∠60^\circ\ \text{A} \end{align*} } \] --- ## **Phasor Diagram:** (ASCII art for quick sketch) ``` I_C(60°) * / / / * I_A(300°)*---------* I_B(180°) ``` - Each arrow has length proportional to 6 A, spaced by 120°. --- ### **Summary Table** | Current | Magnitude (A) | Angle (deg) | |---------|---------------|-------------| | \(I_A\) | 6 | \(-60\) | | \(I_B\) | 6 | \(180\) | | \(I_C\) | 6 | \(60\) | --- **This matches your provided answer. If you need a hand-drawn phasor diagram, just ask!**