Question 3 of 7: AC Steady-State — Power, Branch Currents and Resonance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — December 2017
National Exams, 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; all seven are solved below as a complete study resource.
Reference texts
Sadiku, C.K. Alexander & M.N.O. Sadiku, Fundamentals of Electric Circuits — DC/AC circuit analysis, Thevenin theorem, magnetic circuits (Ch. 13).
S. Chapman, Electric Machinery Fundamentals — magnetic circuits and reluctance.
R. Boylestad, Electronic Devices and Circuit Theory — diode rectifiers and RC filters.
M.M. Mano, Digital Design — combinational logic design.
Question 3: AC Steady-State — Power, Branch Currents and Resonance (20 marks)
Given. $R$ in series with $L_1$ feeding a node where $C_1$ and $R_{LOAD}$ are in parallel; $v_s(t)=100\cos(\omega t)$ V (peak amplitude 100 V).
Given data
$R$
$L_1$
$C_1$
$V_m$
$f$ (case a)
$R_{LOAD}$ (case a)
10Ω
10 mH
10 μF
100 V
60 Hz
100Ω
Find. (a) $P,Q$ supplied by the source at 60 Hz; (b) $i_{2L}(t)$, $i_2(t)$; (c) the frequency of maximal source-current amplitude with $R_{LOAD}\to\infty$, and its name; (d) $i_1(t)$, $P$, $Q$ at that frequency.
Figure 3: Series R-L1 driving parallel C1 || RLOAD
Approach. Convert to phasors, form $Z_{total}=R+j\omega L_1 + (Z_{C_1}\parallel R_{LOAD})$, solve for $I_1$, then current-divide into $C_1$ and $R_{LOAD}$; for (c)/(d), removing $R_{LOAD}$ collapses the circuit to a simple series $R$-$L_1$-$C_1$ loop whose current is maximal at series resonance.
a) Impedances and source current at 60 Hz. $\omega=2\pi(60)=376.99$ rad/s, so $Z_L=j\omega L_1=j3.770\,\Omega$ and $Z_C=1/(j\omega C_1)=-j265.26\,\Omega$.
$$Z_{par}=Z_C\parallel R_{LOAD}=\frac{(-j265.26)(100)}{100-j265.26}=87.56-j33.01\,\Omega$$
$$Z_{tot}=R+Z_L+Z_{par}=97.56-j29.24\,\Omega = 101.84\angle{-16.68^\circ}\,\Omega$$
$$I_1=\frac{V_m}{Z_{tot}}=\frac{100\angle 0^\circ}{101.84\angle{-16.68^\circ}}=0.982\angle 16.68^\circ\text{ A}$$
Active/reactive power supplied (peak-phasor convention, $S=\tfrac12 V_mI_1^{*}$):
$$P = \boxed{47.03\text{ W}}\ ,\qquad Q = \boxed{-14.09\text{ VAR}}$$
(negative $Q$: the parallel $C_1$ branch makes the net load slightly capacitive at 60 Hz).
b) Branch currents. Node-B voltage $V_B=I_1Z_{par}=91.88\angle{-3.97^\circ}$ V, then current-divide into $R_{LOAD}$ and $C_1$:
$$I_{2L}=\frac{V_B}{R_{LOAD}}=0.919\angle{-3.97^\circ}\text{ A}\ \Rightarrow\ \boxed{i_{2L}(t)=0.919\cos(\omega t-3.97^\circ)\text{ A}}$$
$$I_2=\frac{V_B}{Z_{C_1}}=0.346\angle 86.03^\circ\text{ A}\ \Rightarrow\ \boxed{i_2(t)=0.346\cos(\omega t+86.03^\circ)\text{ A}}$$
c) Resonant frequency. With $R_{LOAD}\to\infty$ the parallel branch reduces to $C_1$ alone, so the circuit becomes a simple series $R$-$L_1$-$C_1$ loop. The source-current amplitude $I_1=V_m/|R+j\omega L_1+1/(j\omega C_1)|$ is maximal when the reactive part vanishes, i.e. $\omega L_1=1/(\omega C_1)$:
$$\omega_0=\frac{1}{\sqrt{L_1C_1}}=\frac{1}{\sqrt{(0.01)(10\times10^{-6})}}=3162.3\text{ rad/s}\ \Rightarrow\ f_0=\boxed{503.3\text{ Hz}}$$
This is the circuit's (series) resonant frequency.
d) At resonance. $Z_{tot}(\omega_0)=R=10\,\Omega$ (purely resistive — the inductive and capacitive reactances cancel exactly), so
$$I_1=\frac{V_m}{R}=\frac{100}{10}=10\text{ A}\ \Rightarrow\ \boxed{i_1(t)=10\cos(\omega_0 t)\text{ A}}$$
$$P=\tfrac12V_mI_1=\tfrac12(100)(10)=\boxed{500\text{ W}}\ ,\qquad Q=\boxed{0\text{ VAR}}$$
