Question 3 of 7: AC Steady State — Series and Parallel Resonance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 — Electric Circuits and Power — National Exam, May 2014.
Closed book; one aid sheet permitted; Casio/Sharp approved calculator only.
Any five of the seven questions constitute a complete paper — every question is answered.
Reference texts: Sadiku & Alexander, Fundamentals of
Electric Circuits (DC/AC network analysis, Thevenin, resonance);
Chapman, Electric Machinery Fundamentals (magnetic circuits);
Boylestad, Electronic Devices and Circuit Theory (diode rectifiers);
Mano, Digital Design (combinational logic).
Question 3: AC Steady State — Series and Parallel Resonance (20 marks)
Given. Switch position 1 places $R$, $L_1$ and $C_1$ in
series across the source (current $i_1$). Switch position 2 places $C_2$ and
$L_2$ in parallel across the source, with $i_2$ the total current drawn and
$i_{2L}$ the share through $L_2$.
Given data
Quantity
Value
$R$
10 Ω
$L_1$
10 mH
$L_2$
0.5 H
$C_1$
10 μF
$C_2$
200 pF
$v_s(t)$
$100\cos(\omega t)$ V (peak 100 V)
Find. $P$, $Q$ at 60 Hz (position 1); the series-resonant frequency
and the resulting $i_1(t)$, $P$, $Q$; and the parallel(tank)-resonant frequency at
which $i_2(t)=0$ (position 2).
[Figure not reproduced: Figure 3 — redrawn from the printed figure. Position 1: series $R$–$L_1$–$C_1$. Position 2: $C_2 \parallel L_2$ (tank), matching the two separately-named currents $i_2(t)$ and $i_{2L}(t)$ in the original figure. See the official exam paper.]
Approach. In position 1 the branch is a series RLC driven by a
single source — compute $Z_1(\omega)$, then find where $|Z_1|$ is minimized
(series resonance). In position 2 the branch is a lossless parallel LC tank —
its impedance is maximal (ideally infinite) at the parallel-resonant frequency, which
is exactly where the source current nulls.
(a) Power at 60 Hz, position 1. $\omega=2\pi(60)=376.99$ rad/s.
$$Z_1=R+j\omega L_1+\dfrac{1}{j\omega C_1}=10+j(3.770-265.26)=10-j261.49\,\Omega$$
$$I_1=\dfrac{V_{s,pk}}{Z_1}=\dfrac{100\angle0^\circ}{261.68\angle{-87.81^\circ}}
=0.3822\angle87.81^\circ\text{ A (peak)}$$
Using $P=\tfrac12|I_1|^2R$ and $Q=\tfrac12|I_1|^2X_{net}$ (peak-phasor convention,
$X_{net}=\omega L_1-1/(\omega C_1)$):
$$\boxed{P=0.730\text{ W}}\qquad\boxed{Q=-19.09\text{ VAR (net capacitive)}}$$
(b) Series-resonant frequency. $|Z_1|$ is minimized (purely
resistive, $=R$) when the reactances cancel:
$$\omega_0=\dfrac{1}{\sqrt{L_1C_1}}=\dfrac{1}{\sqrt{(0.01)(10\times10^{-6})}}
=3162.3\text{ rad/s}$$
$$\boxed{f_0=\dfrac{\omega_0}{2\pi}=503.29\text{ Hz}}$$
This is called the (series) resonant frequency.
(c) At resonance. $Z_1=R=10\,\Omega$ exactly (purely resistive),
so the current is in phase with the source and at its maximum:
$$\boxed{i_1(t)=10\cos(3162.3\,t)\text{ A}}$$
$$\boxed{P=\tfrac12(10)^2(10)=500\text{ W}}\qquad\boxed{Q=0\text{ VAR}}$$
(d) Position 2 — frequency for $i_2(t)=0$. $C_2$ and $L_2$
are in parallel, so the combined impedance is
$Z_2(\omega)=\dfrac{j\omega L_2}{1-\omega^2L_2C_2}$, which is a lossless tank:
its magnitude diverges (ideally infinite) exactly at the parallel-resonant frequency,
at which point the source can supply zero net current (all the energy circulates
internally between $L_2$ and $C_2$):
$$\omega_0'=\dfrac{1}{\sqrt{L_2C_2}}=\dfrac{1}{\sqrt{(0.5)(200\times10^{-12})}}
=1.000\times10^5\text{ rad/s}$$
$$\boxed{f_0'=\dfrac{\omega_0'}{2\pi}=15\,915.5\text{ Hz}}$$
