Given. The source feeds the line impedance $Z_{Line}=R_{Line}+jX_{Line}$, which reaches a load impedance $Z_{Load}=R_{Load}+jX_{Load}$. A switch in series with capacitor $X_C$ places that branch in parallel with the load when closed (a shunt power-factor-correction capacitor -- a standard configuration, Sadiku Ch. 11); $V_s(t)$ has a $100\text{ V}$ rms amplitude at $60\text{ Hz}$.
Figure 5: Line + load with switched PF-correction capacitor (RLine=2Ω, XLine=2Ω, RLoad=6Ω, XLoad=4Ω, Xc=100Ω)
Approach. Work in rms phasors throughout (the given $V_s(t)$ amplitude is $\sqrt2\times100$, i.e. $100\text{ V}$ rms). Switch open: simple series circuit. Switch closed: the capacitor branch is a shunt in parallel with the load; recombine the total impedance and re-split the current to the load.
a) Switch open -- source current and power. $$Z_{tot} = Z_{Line}+Z_{Load} = (2+j2)+(6+j4) = 8+j6\,\Omega = 10\angle{36.87^\circ}\,\Omega$$ $$I_s = \dfrac{V_s}{Z_{tot}} = \dfrac{100\angle0^\circ}{10\angle{36.87^\circ}} = 10\angle{-36.87^\circ}\text{ A} \;\Rightarrow\; \boxed{|I_s| = 10\text{ A}}$$ $$P_{source} = |V_s||I_s|\cos(36.87^\circ) = (100)(10)(0.8) = \boxed{800\text{ W}}$$
b) Switch open -- line and load power. The same current flows through both impedances in this series loop: $$P_{line} = |I_s|^2R_{Line} = (10)^2(2) = \boxed{200\text{ W}}, \qquad P_{load} = |I_s|^2R_{Load} = (10)^2(6) = \boxed{600\text{ W}}$$ (check: $200+600=800\text{ W} = P_{source}$ ✓).
c) Switch closed -- source current. The capacitor ($Z_C=-j100\,\Omega$) now parallels the load: $$Z_{par} = \dfrac{Z_CZ_{Load}}{Z_C+Z_{Load}} = \dfrac{(-j100)(6+j4)}{-j100+6+j4} = 6.485+j3.761\,\Omega$$ $$Z_{tot} = Z_{Line}+Z_{par} = 8.485+j5.761\,\Omega = 10.256\angle{34.18^\circ}\,\Omega$$ $$I_s = \dfrac{100\angle0^\circ}{10.256\angle{34.18^\circ}} = 9.750\angle{-34.18^\circ}\text{ A} \;\Rightarrow\; \boxed{|I_s| = 9.750\text{ A}}$$ (lower than the open-switch $10\text{ A}$, and closer to unity power factor -- the classic effect of shunt PF correction.)
d) Switch closed -- line and load power. Line power uses the full source current; load power needs the current split into the $Z_{Load}$ branch: $$P_{line} = |I_s|^2R_{Line} = (9.750)^2(2) = \boxed{190.1\text{ W}}$$ $$V_{par} = I_sZ_{par} = (9.750\angle{-34.18^\circ})(7.497\angle{30.11^\circ}) = 73.10\angle{-4.06^\circ}\text{ V}$$ $$I_{load} = \dfrac{V_{par}}{Z_{Load}} = \dfrac{73.10\angle{-4.06^\circ}}{7.211\angle{33.69^\circ}} = 10.14\angle{-37.75^\circ}\text{ A}$$ $$P_{load} = |I_{load}|^2R_{Load} = (10.14)^2(6) = \boxed{616.5\text{ W}}$$ (check: $P_{source}=|V_s||I_s|\cos(34.18^\circ)=806.6\text{ W}\approx190.1+616.5=806.6\text{ W}$ ✓.)
Final Results -- Question 5
Quantity
Switch open
Switch closed
$|I_s|$
$10.0\text{ A}$
$9.750\text{ A}$
$P_{source}$
$800\text{ W}$
$806.6\text{ W}$
$P_{line}$
$200\text{ W}$
$190.1\text{ W}$
$P_{load}$
$600\text{ W}$
$616.5\text{ W}$
Closing the switch reduces the source current from $10.0\text{ A}$ to $9.750\text{ A}$ while the load actually absorbs slightly more real power ($616.5\text{ W}$ vs. $600\text{ W}$), because less real power is now lost across the line impedance -- exactly the economic motivation for shunt power-factor correction on a distribution feeder.
