Question 5 of 7: AC Power Flow on a Line-Load System, With and Without a Shunt Capacitor

Question 5: AC Power Flow on a Line-Load System, With and Without a Shunt Capacitor (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Figure 5 — redrawn: the source feeds a line impedance (X_Line then R_Line); a switched capacitive branch (X_C) shunts the line-load junction; the load (X_Load in series with R_Load) is always connected. See the official exam paper.]

Given.

Find. With the switch open, then closed: source-current magnitude and the real power absorbed by the line and by the load.

Approach. Vs is written as $\sqrt2\cdot100\cos(120\pi t)$, i.e. an rms value of 100 V at $\omega=120\pi$ rad/s (60 Hz) — take $\underline{V}_s=100\angle0^\circ$ V (rms) as the power-system reference phasor, so that $P=\text{Re}(\underline{S})=|\underline{I}|^2\,\text{Re}(\underline{Z})$ directly with no extra factor of ½. With the switch open, the circuit is a simple series R–L loop; closing the switch adds $\underline{Z}_C=-j100\ \Omega$ in parallel with the load impedance.

1. Switch open — total impedance and source current. $$\underline{Z}_{total} = (R_{Line}+jX_{Line})+(R_{Load}+jX_{Load}) = (2+j2)+(6+j4) = 8+j6\ \Omega,\quad |\underline{Z}_{total}|=10\ \Omega$$ $$\underline{I}_s = \dfrac{100\angle0^\circ}{8+j6} = 8-j6\ \text{A} \quad\Rightarrow\quad \boxed{|I_s|=10\ \text{A}}$$
2. Switch open — power supplied and absorbed. Since the whole current flows through both the line and the load (a series loop): $$P_{source}=|I_s|^2\,\text{Re}(\underline{Z}_{total})=(10)^2(8)=800\ \text{W}$$ $$P_{line}=|I_s|^2R_{Line}=(10)^2(2)=\boxed{200\ \text{W}},\qquad P_{load}=|I_s|^2R_{Load}=(10)^2(6)=\boxed{600\ \text{W}}$$ (200+600 = 800 W ✓, all the source's real power is dissipated resistively.)
3. Switch closed — parallel combination. The capacitor now shunts the load: $$\underline{Z}_{par} = \dfrac{\underline{Z}_C\,\underline{Z}_{Load}}{\underline{Z}_C+\underline{Z}_{Load}} = \dfrac{(-j100)(6+j4)}{-j100+6+j4}$$ $$\underline{Z}_{total,cl} = \underline{Z}_{Line}+\underline{Z}_{par}$$ $$\underline{I}_s = \dfrac{100\angle0^\circ}{\underline{Z}_{total,cl}} \quad\Rightarrow\quad \boxed{|I_s| = 9.75\ \text{A}}$$
4. Switch closed — power in the line and the load. The line still carries the full source current, but the load current is now only the fraction of $\underline{I}_s$ that flows through $\underline{Z}_{Load}$ (current divider between $\underline{Z}_C$ and $\underline{Z}_{Load}$): $$P_{line} = |I_s|^2R_{Line} = (9.75)^2(2) = \boxed{190.1\ \text{W}}$$ $$P_{load} = |I_{Load}|^2R_{Load},\quad |I_{Load}|=10.14\ \text{A}\ \Rightarrow\ P_{load}=\boxed{616.5\ \text{W}}$$