Given. $V_s(t)$ drives $R_1$ into a node shared by a $C_1\parallel L_1$ branch and a series $R_2$-$L_2$-$C_2$ branch (whose terminal voltage is $V_2(t)$, current $I_2(t)$); $I_1(t)$ is the current into the $C_1\parallel L_1$ branch.
Given data
Quantity
Value
$R_1,\ R_2$
$5\,\Omega,\ 10\,\Omega$
$L_1,\ L_2$
$10\text{ mH},\ 5\text{ H}$
$C_1,\ C_2$
$10\,\mu\text{F},\ 200\text{ pF}$
$V_s(t)$
$100\cos(\omega t)\text{ V}$
Find. $P,Q$ supplied by the source at $60\text{ Hz}$; the resonant frequency of the $R_2$-$L_2$-$C_2$ branch; $I_1(t),I_2(t)$ and $Q$ at that frequency.
Figure 4: R1=5Ω, L1=10mH, C1=10μF || branch, then R2=10Ω-L2=5H-C2=200pF series branch
Approach. Use phasor impedances for each branch, combine $C_1\parallel L_1$ and the series $R_2$-$L_2$-$C_2$ branch in parallel, add $R_1$ in series, then apply $S=\tfrac12 V_s I_s^{*}$ for power (using peak-valued phasors, consistent with the peak-amplitude $V_s(t)$ given).
a) Power at $60\text{ Hz}$. With $\omega=2\pi(60)=377.0\text{ rad/s}$: $$Z_{L_1}=j\omega L_1 = j3.77\,\Omega, \quad Z_{C_1}=\dfrac{1}{j\omega C_1}=-j265.3\,\Omega \;\Rightarrow\; Z_1 = Z_{L_1}\parallel Z_{C_1} = j3.82\,\Omega$$ $$Z_{L_2}=j1885\,\Omega,\quad Z_{C_2}=\dfrac{1}{j\omega C_2}=-j13.26\text{ M}\Omega \;\Rightarrow\; Z_2 = R_2+Z_{L_2}+Z_{C_2} \approx 10-j13.26\text{ M}\Omega$$ Because $|Z_2|\gg|Z_1|$, the parallel combination $Z_{par}=Z_1\parallel Z_2$ is close to $Z_1$; combining exactly and adding $R_1$ in series (carried out numerically) gives a total impedance whose current from the $100\text{ V}$ (peak) source yields $$\boxed{P \approx 630.9\text{ W}}, \qquad \boxed{Q \approx 482.6\text{ VAR (inductive)}}$$
b) Resonant frequency of the $R_2$-$L_2$-$C_2$ branch. $I_2$ is in phase with $V_2$ exactly when that branch's impedance is purely resistive, i.e. $\omega L_2 = 1/(\omega C_2)$: $$\omega_0 = \dfrac{1}{\sqrt{L_2C_2}} = \dfrac{1}{\sqrt{(5)(200\times10^{-12})}} = \boxed{31{,}623\text{ rad/s}}$$ $$f_0 = \dfrac{\omega_0}{2\pi} = \boxed{5032.9\text{ Hz}}$$ This is the (series) resonant frequency of the $R_2$-$L_2$-$C_2$ branch.
c) $I_1(t)$ and $I_2(t)$ at $\omega_0$. At resonance $Z_2=R_2=10\,\Omega$ (purely resistive); recomputing $Z_1=C_1\parallel L_1$ at $\omega_0$ gives $Z_1=3.19\angle{-90^\circ}\,\Omega$ (net capacitive, since $\omega_0$ is far above $C_1\parallel L_1$'s own resonance). Combining $Z_{par}=Z_1\parallel Z_2 = 0.926-j2.898\,\Omega$ and $Z_{tot}=R_1+Z_{par}=5.93-j2.90\,\Omega$, the peak source current is $I_s=100/Z_{tot}=15.16\angle{26.06^\circ}\text{ A}$, giving the parallel-node voltage $V_{load}=I_sZ_{par}=46.13\angle{-46.22^\circ}\text{ V}$. Dividing by each branch impedance: $$I_1 = \dfrac{V_{load}}{Z_1} = 14.44\angle{43.78^\circ}\text{ A} \;\Rightarrow\; \boxed{i_1(t) = 14.44\cos(\omega_0 t+43.78^\circ)\text{ A}}$$ $$I_2 = \dfrac{V_{load}}{Z_2} = 4.613\angle{-46.22^\circ}\text{ A} \;\Rightarrow\; \boxed{i_2(t) = 4.613\cos(\omega_0 t-46.22^\circ)\text{ A}}$$ Note $I_2$ and $V_{load}$ (=$V_2$) share the same $-46.22^\circ$ angle, confirming the resonance condition from part (b).
d) Reactive power at $\omega_0$. $$S = \tfrac12 V_sI_s^{*} = \tfrac12(100)(15.16\angle{-26.06^\circ}) = 758.0\angle{-26.06^\circ} = 680.9-j333.0 \;\Rightarrow\; \boxed{Q \approx -333.0\text{ VAR}}$$ The negative sign shows the source is now supplying a net capacitive reactive power at this higher frequency (the source current leads the source voltage), the opposite sign from part (a).
