Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 2016 — 04-BS-4 Electric Circuits and Power. Three hours duration, closed book (one aid sheet permitted). Seven questions are printed; any five constitute a complete paper, but all seven are solved below as a complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits, 7th ed. (circuit analysis, transients, AC power, rectifiers – Questions 1–6); Chapman, Electric Machinery Fundamentals, 5th ed. (magnetic circuits – Question 7).
Given. A source $v_s(t)$ in series with $R$ and a switch S. Position 1 places $L_1$ and $C_1$ in a single series loop with $v_s$ and $R$; position 2 instead places a parallel $C_2\Vert L_2$ tank in that same series loop.
Given data
R
10 Ω
L1
10 mH
C1
10 μF
L2
0.5 H
C2
200 pF
vs(t) amplitude
100 V
Find. (a) $P$, $Q$ at $f=60$ Hz, position 1; (b) the frequency of maximal source-current amplitude in position 1, and its name; (c) $P$ at that frequency; (d) the frequency in position 2 at which the source supplies zero reactive power.
Figure 5 — switched resonant network: position 1 is a series R–L1–C1 loop; position 2 is R in series with a parallel C2‖L2 tank.
Approach. Position 1 is a simple series RLC loop, so use $Z=R+j(\omega L_1-1/\omega C_1)$ and rms phasors ($V_{s,rms}=100/\sqrt2$ V). Current amplitude is maximal where $|Z|$ is minimal, i.e. at series resonance $\omega_0=1/\sqrt{L_1C_1}$. Position 2 places an ideal (lossless) parallel tank in series with $R$; the tank's impedance $Z_{tank}=j\omega L_2/(1-\omega^2L_2C_2)$ is purely imaginary at every frequency and diverges to infinity only at the tank's own anti-resonance $\omega_0'=1/\sqrt{L_2C_2}$ — the one frequency at which the source truly delivers zero reactive (and, in the ideal case, zero active) power, because the tank blocks all current.
(a) Active/reactive power at 60 Hz, position 1. $\omega=2\pi(60)=377.0\text{ rad/s}$: $\omega L_1=3.77\;\Omega$, $1/(\omega C_1)=265.26\;\Omega$, so
$$ Z=10+j(3.77-265.26)=10-j261.49\;\Omega $$
$$ I=\frac{V_{s,rms}}{Z}=\frac{70.71\angle0^\circ}{Z}\;\Rightarrow\;S=V_{s,rms}I^*=P+jQ $$
$$ P=\boxed{0.730\text{ W}},\qquad Q=\boxed{-19.09\text{ VAR}} $$
(the negative sign shows the loop is net capacitive at 60 Hz — $C_1$'s reactance dominates $L_1$'s far below resonance.)
(b) Frequency of maximal current, position 1. $|I|$ is maximal where $|Z|$ is minimal, which for a series RLC loop is where the reactance cancels ($\omega L_1=1/\omega C_1$):
$$ \omega_0=\frac{1}{\sqrt{L_1C_1}}=\frac{1}{\sqrt{(0.010)(10\times10^{-6})}}\;\Rightarrow\;f_0=\boxed{503.29\text{ Hz}} $$
This is called the (series) resonant frequency of the loop.
(c) Active power at the resonant frequency. At $\omega_0$ the reactance cancels exactly, leaving $Z=R$ (purely resistive), so all of the apparent power is active:
$$ P=\frac{V_{s,rms}^2}{R}=\frac{70.71^2}{10}=\boxed{500\text{ W}}\qquad(Q=0) $$
(d) Zero reactive power, position 2. The ideal tank admittance is $Y_{tank}=j(\omega C_2-1/(\omega L_2))$, purely imaginary for every $\omega\gt0$; its impedance $Z_{tank}=1/Y_{tank}$ is likewise purely imaginary everywhere except at the anti-resonance $\omega_0'$ where $Y_{tank}\to0$ and $Z_{tank}\to\infty$. At that single frequency the tank presents an open circuit, current from the source drops to zero, and consequently both the active and reactive power supplied vanish — the only way an ideal series $R$ + lossless-tank loop can show zero net reactive power at a nonzero frequency:
$$ \omega_0'=\frac{1}{\sqrt{L_2C_2}}=\frac{1}{\sqrt{(0.5)(200\times10^{-12})}}\;\Rightarrow\;f_0'=\boxed{15{,}915.49\text{ Hz}} $$
