Question 3 of 7: Switched RLC Circuit – Resonance and AC Power
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — December 2014 national exam. 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; the first five as they appear in the answer book are marked. All seven questions are solved below as a complete study resource.
Reference texts: Sadiku & Alexander, Fundamentals of Electric Circuits (6th ed.) — DC circuit analysis, Thevenin/Norton equivalents, first-order transients, AC steady-state phasors and power, magnetic circuits, diode rectifiers (Ch. 2–4, 9–11, 13); Mano & Ciletti, Digital Design — combinational logic design (Ch. 2–4), cited inline for Question 7.
Question 3: Switched RLC Circuit – Resonance and AC Power (20 marks)
Given. $v_s(t)=\sqrt2\cdot100\cos(\omega t)$ V (100 V rms); position 1: $L_1$ in series with $C_1$ in series with $R_1$ (a plain series R-L-C, confirmed against the source figure); position 2: the source drives the parallel tank $C_2\parallel L_2$ directly ($L_1,R_1,C_1$ are not in the circuit at all in this position).
Given data
Quantity
Value
$R_1$
$10\,\Omega$
$L_1,\,C_1$ (position 1)
$80\text{ mH},\ 100\,\mu\text{F}$
$L_2,\,C_2$ (position 2)
$0.5\text{ H},\ 0.5\,\mu\text{F}$
$V_s$ (rms)
$100\text{ V}$
[Figure not reproduced. See the official exam paper.]
Find. (a) $P,Q$ at 60 Hz, position 1; (b) resonant frequency, position 1; (c) $i_1(t)$, $P$, $Q$ at resonance; (d) frequency for $i_2(t)=0$, position 2.
Approach. Position 1 is a textbook series RLC loop, so use $\underline{Z}_1=R_1+j\omega L_1+\frac{1}{j\omega C_1}$ directly; resonance is where $\mathrm{Im}(\underline{Z}_1)=0$. Position 2 is a parallel LC tank, whose input current vanishes at its own (anti-)resonant frequency.
a) Active/reactive power at 60 Hz. With $\omega=2\pi(60)=376.99\text{ rad/s}$:
$$\underline{Z}_1=R_1+j\omega L_1+\frac{1}{j\omega C_1}=10+j(30.16-26.53)=10+j3.63\ \Omega$$
$$\underline{I}_1=\frac{100\angle0^\circ}{10+j3.63}=8.83-j3.21\text{ A}, \qquad \underline{S}=V_s\underline{I}_1^{*}$$
$$P=\boxed{883.4\text{ W}}, \qquad Q=\boxed{321.0\text{ var (inductive)}}$$
b) Resonant frequency. Series resonance occurs where the net reactance vanishes, $\omega_0L_1=\dfrac{1}{\omega_0C_1}$:
$$\omega_0=\frac{1}{\sqrt{L_1C_1}}=\frac{1}{\sqrt{(0.08)(100\times10^{-6})}}=353.55\text{ rad/s} \quad\Rightarrow\quad f_0=\boxed{56.27\text{ Hz}}$$
This is called the (series) resonant frequency: at $\omega_0$ the impedance is purely resistive ($\underline Z_1=R_1$) and minimal in magnitude, so the source current is maximal.
c) At resonance. $\underline{Z}_1(\omega_0)=R_1=10\,\Omega$ exactly, so
$$\underline{I}_1=\frac{100\angle0^\circ}{10}=10\angle0^\circ\text{ A} \quad\Rightarrow\quad i_1(t)=\sqrt2\cdot 10\cos(353.55\,t)\text{ A}$$
$$P=\boxed{1000\text{ W}}, \qquad Q=\boxed{0\text{ var}}$$
(purely resistive at resonance, as expected).
d) Position 2, $i_2(t)=0$. With S at 2 the source drives $C_2\parallel L_2$ directly; the input current is zero when the parallel-tank admittance is zero, i.e. at the tank's own (anti-)resonant frequency:
$$\omega_0'=\frac{1}{\sqrt{L_2C_2}}=\frac{1}{\sqrt{(0.5)(0.5\times10^{-6})}}=2000\text{ rad/s} \quad\Rightarrow\quad f_0'=\boxed{318.31\text{ Hz}}$$
