AC Steady-State Phasor Analysis with a Resonant Tank: 04-BS-4, May 2017, Q4

Question 4: AC Steady-State Phasor Analysis with a Resonant Tank (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Figure 4: a series R₁–L₁ branch fed by v_s(t) drives node 1 (v₁(t), i₁(t) as marked), where R₂, C, and L₂ all sit in parallel to the return rail.

Given data
Quantity	Value
L₁	80 mH
L₂	20 mH
R₁	8 Ω
R₂	2 Ω
C	5 mF
v_s(t)	√2·10cos(100t) V

Find. Z_L1, Z_L2, Z_C; the phasors V₁ and I₁; and i_C(t) in the time domain.

Figure 4: series R₁L₁ source feeding the parallel R₂∥C∥L₂ node

Approach. Convert every element to its impedance at ω=100 rad/s, combine the three parallel branches into Z_p, then apply the voltage-divider and Ohm's law relations for phasors; convert i_C back to the time domain last.

Impedances (part a). With ω=100 rad/s: $$ Z_{L1}=j\omega L_1 = j(100)(0.080) = \boxed{j8\ \Omega}, \qquad Z_{L2}=j\omega L_2 = j(100)(0.020) = \boxed{j2\ \Omega}, $$ $$ Z_C = \frac{1}{j\omega C} = \frac{1}{j(100)(0.005)} = \boxed{-j2\ \Omega}. $$
Parallel node impedance. L₂ and C are exact opposites ($j2\ \Omega$ and $-j2\ \Omega$), so their admittances cancel — the tank is at anti-resonance at this frequency and only R₂ survives in the parallel combination: $$ Y_p = \frac{1}{R_2}+\frac{1}{Z_C}+\frac{1}{Z_{L2}} = 0.5 + j0.5 - j0.5 = 0.5\text{ S} \ \Rightarrow\ Z_p = \boxed{2\ \Omega\ (\text{purely resistive})}. $$
Voltage phasor V₁ (part b). Using rms phasor $V_s=10\angle0^\circ\text{ V}$ (since $v_s(t)=\sqrt2\,(10)\cos(100t)$) and the divider across Z_p: $$ V_1 = V_s\cdot\frac{Z_p}{R_1+Z_{L1}+Z_p} = 10\cdot\frac{2}{8+j8+2} = \frac{20}{10+j8} = \boxed{1.562\angle{-38.66^\circ}\text{ V}}. $$
Current phasor I₁ (part c). $$ I_1 = \frac{V_s}{R_1+Z_{L1}+Z_p} = \frac{10}{10+j8} = \boxed{0.781\angle{-38.66^\circ}\text{ A}}. $$ (Consistent with $I_1=V_1/Z_p$ since Z_p is real, both share the same angle.)
Capacitor current, time domain (part d). $$ I_C = \frac{V_1}{Z_C} = \frac{1.562\angle{-38.66^\circ}}{2\angle{-90^\circ}} = \boxed{0.781\angle{51.34^\circ}\text{ A}}. $$ Converting the rms phasor back to an instantaneous cosine (peak = $\sqrt2\times$rms): $$ i_C(t) = \sqrt2(0.781)\cos(100t+51.34^\circ) = \boxed{1.104\cos(100t+51.34^\circ)\text{ A}}. $$

Final results
Quantity	Value
Z_L1	j8 Ω
Z_L2	j2 Ω
Z_C	−j2 Ω
V₁	1.562∠−38.66° V
I₁	0.781∠−38.66° A
i_C(t)	1.104cos(100t+51.34°) A

Topic: AC steady-state phasor analysis with an LC tank at anti-resonance.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)$, phasor voltage divider, parallel admittance combination, anti-resonance $\omega_0=1/\sqrt{L_2C}$.

Why this works: at the exact frequency where $Z_{L2}=-Z_C$, the two reactive branches' admittances cancel identically, so the parallel combination collapses to the lone resistor — a deliberate design point in the problem, not a coincidence, and it makes the divider arithmetic purely real-valued in the denominator's reactive part.

Common pitfall: forgetting that the given source is already an rms-cosine amplitude ($\sqrt2\,V_{rms}\cos$), so the phasor magnitude is the 10 V coefficient directly, not $\sqrt2\times10$; the $\sqrt2$ only reappears when converting a computed rms phasor back to a time-domain peak amplitude.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 9–10 (sinusoids and phasors, AC steady-state analysis).

Question 4 of 7: AC Steady-State Phasor Analysis with a Resonant Tank

Question 4: AC Steady-State Phasor Analysis with a Resonant Tank (20 marks)