Question 16.15: In Figure 16–21, the load impedance and line impedances are ......
In Figure 16–21, the load impedance and line impedances are Z_{Y} = 10 + j5 Ω per phase and Z_{W} = 0.15 + j0.85 Ω per phase, respectively. The magnitude of the line voltage at the source is V_{L} = 208 V (rms). Using ∠ \pmb{I}_{A} = 0° as the phase reference, find the line current phasors and the line voltage phasors at the load for a positive phase sequence.
In this example the line voltage at the source is V_{L} = 208 V (rms). The phase voltage magnitude at the source is V_{P} = V_{L}/\sqrt{3} = 120 V (rms). This phase voltage appears across the combined impedance Z_{W} + Z_{Y}, so the line current magnitude is found to be
I_{L} = \frac{V_{P}}{|Z_{Y} + Z_{W}|}= \frac{120}{|10.15 + j5.85|}= 10.24 A (rms)
The specified phase reference means that \pmb{I}_{A} = 10.24∠0° A (rms). For a positive phase sequence the other two line currents are \pmb{I}_{B} = 10.24 ∠ – 120° A (rms) and \pmb{I}_{C} = 10.24 ∠ – 240° A (rms). The phase voltages at the load are balanced and denoted \pmb{V}_{an}, \pmb{V}_{bn}, and \pmb{V}_{cn}. The first of these is found as
\pmb{V}_{an} = \pmb{I}_{A}Z_{Y} = 10.24 ∠0° × (10 + j5) = 115∠26.6° V (rms)
Line voltages are √3 times as large as phase voltages and lead the phase voltages by 30°. Hence, we have
\pmb{V}_{ab} = \pmb{V}_{an} × \sqrt{3}∠30° = 199 ∠56.6° V (rms)
The other two line voltages are \pmb{V}_{bc} = 199∠ – 63.4° and \pmb{V}_{ca} = 199∠ – 183.4° V (rms). The line voltage magnitude at the load (199 V) is less than the line voltage at the source (208 V) due to the voltage drop in the line impedances.