Question 4 of 7: AC Steady-State Phasor Analysis — Series-Fed Parallel Resonant Load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — May 2018
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
C.K. Alexander & M.N.O. Sadiku, Fundamentals of Electric Circuits — DC/AC circuit analysis, Thévenin theorem, first-order transients.
M.N.O. Sadiku, Elements of Electromagnetics — magnetic circuits and reluctance.
R. Boylestad, Electronic Devices and Circuit Theory — diode rectifiers.
M.M. Mano, Digital Design — combinational logic design.
Given. Figure 4: the source $v_s(t)$ drives $R_1$ and $L_1$ in series, feeding a node where $R_2$, $C$, and $L_2$ are all in parallel back to the return rail; $v_s(t)=\sqrt2\,(10)\cos(100t)$ V is written in the standard RMS-phasor convention $v(t)=\sqrt2\,|V|\cos(\omega t+\angle V)$, so $\underline{V_s}=10\angle0^\circ$ V (RMS) at $\omega=100$ rad/s.
Given data
$L_1$
$L_2$
$R_1$
$R_2$
$C$
$\omega$
$V_s$ (RMS)
80 mH
20 mH
8 Ω
2 Ω
5 mF
100 rad/s
10 V∠0°
Find. $\underline{Z_{L1}},\underline{Z_{L2}},\underline{Z_C}$; the phasor $\underline{V_1}$ across the parallel group; the phasor $\underline{I_1}$ delivered by the source; and $\underline{Z_{eq}}$ seen by the source.
Figure 4: Series R1-L1 source feeding parallel R2 || C || L2 (ω=100 rad/s)
Approach. Convert each element to its impedance at $\omega=100$ rad/s, combine $R_2$, $L_2$, $C$ as a parallel admittance, add the series $R_1+j\omega L_1$, and apply Ohm's law/current-voltage division for the phasors.
a) Impedances.
$$\underline{Z_{L1}}=j\omega L_1=j(100)(0.080)=\boxed{j8\,\Omega}\ ,\quad \underline{Z_{L2}}=j\omega L_2=j(100)(0.020)=\boxed{j2\,\Omega}\ ,\quad \underline{Z_C}=\frac{1}{j\omega C}=\frac{1}{j(100)(0.005)}=\boxed{-j2\,\Omega}$$
Combine the parallel load. $L_2$ and $C$ are equal-and-opposite reactances at this $\omega$ (in fact $\omega=1/\sqrt{L_2C}$, the tank's own anti-resonance), so their admittances cancel exactly, leaving only $R_2$:
$$\underline{Z_{load}}=\left(\frac{1}{R_2}+\frac{1}{j2}+\frac{1}{-j2}\right)^{-1}=\left(\frac{1}{2}+j0.5-j0.5\right)^{-1}=\boxed{2\,\Omega\ (\text{purely resistive})}$$
d) Equivalent impedance seen by the source.
$$\underline{Z_{eq}}=(R_1+\underline{Z_{L1}})+\underline{Z_{load}}=(8+j8)+2=10+j8=\boxed{12.81\angle38.66^\circ\ \Omega}$$
c) Current phasor $\underline{I_1}$.
$$\underline{I_1}=\frac{\underline{V_s}}{\underline{Z_{eq}}}=\frac{10\angle0^\circ}{12.81\angle38.66^\circ}=\boxed{0.781\angle{-38.66^\circ}\text{ A}}$$
b) Voltage phasor $\underline{V_1}$. By Ohm's law across the (purely resistive) parallel load:
$$\underline{V_1}=\underline{I_1}\,\underline{Z_{load}}=(0.781\angle{-38.66^\circ})(2\angle0^\circ)=\boxed{1.562\angle{-38.66^\circ}\text{ V}}$$
Final results — Question 4 (all RMS phasors)
Quantity
Value
$\underline{Z_{L1}}$
$j8\,\Omega$
$\underline{Z_{L2}}$
$j2\,\Omega$
$\underline{Z_C}$
$-j2\,\Omega$
$\underline{Z_{load}}$
$2\,\Omega\angle0^\circ$
$\underline{V_1}$
1.562 V∠−38.66°
$\underline{I_1}$
0.781 A∠−38.66°
$\underline{Z_{eq}}$
12.81 Ω∠38.66° ($=10+j8\,\Omega$)
Check: $\omega=100$ rad/s exactly matches $1/\sqrt{L_2C}=1/\sqrt{(0.020)(0.005)}=100$ rad/s, the anti-resonant frequency of the $L_2\parallel C$ tank — this is why their parallel combination vanishes and the load reduces to pure $R_2$; it is a deliberate feature of the problem, not a coincidence to second-guess.
