Question 4 of 7: AC Steady-State — Power and Resonance

Question 4: AC Steady-State — Power and Resonance (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. A sinusoidal source $V_s(t)=100\cos\omega t$ V (peak 100 V, i.e. 70.71 V rms) with series resistance R1 driving the node V2. Two branches hang off V2: an L1∥C1 tank (current I1) and a series R2–L2–C2 branch (current I2).

Find. (a) active/reactive power at 60 Hz; (b) the frequency making I2 in phase with V2 (and its name); (c) I1(t), I2(t) at that frequency; (d) reactive power there.

Approach. Work in phasors. Combine the two branch impedances in parallel at V2, add R1, and compute $S=V_{s,\text{rms}}I_s^{*}$ at 60 Hz. The "I2 in phase with V2" condition is the series resonance of the R2–L2–C2 branch, $\omega=1/\sqrt{L_2C_2}$; evaluate the branch currents and reactive power there.

Branch impedances: tank $Z_1(\omega)=\dfrac{1}{\,j\omega C_1 + 1/(j\omega L_1)\,}$ and series branch $Z_2(\omega)=R_2+j\omega L_2+\dfrac{1}{j\omega C_2}$. With $Z_p=Z_1\|Z_2$, the source delivers $I_s=V_s/(R_1+Z_p)$ and $V_2=V_s\,Z_p/(R_1+Z_p)$.

1. a) Power at 60 Hz ($\omega=2\pi(60)=377$ rad/s). Here $Z_1=+j3.82\ \Omega$ (inductive — the tank's self-resonance is 503 Hz, well above 60 Hz) while $Z_2\approx 10-j1.33\times10^{7}\ \Omega$ (C2 nearly open). The series branch draws almost nothing, so the source current is set by R1 and the tank: $$S=V_{s,\text{rms}}\,I_s^{*}=P+jQ.$$ $$\boxed{P \approx 631\text{ W},\qquad Q \approx +483\text{ VAR (inductive)}}$$ (equivalently $|S|=794$ VA at power factor 0.79 lagging).
2. b) Frequency for I2 in phase with V2. Current I2 flows through the series R2–L2–C2 branch; it is in phase with the driving node voltage V2 only when that branch is purely resistive, i.e. at its series resonance: $$\omega_0=\frac{1}{\sqrt{L_2 C_2}}=\frac{1}{\sqrt{(5)(200\times10^{-12})}}=3.162\times10^{4}\text{ rad/s},\qquad f_0=\frac{\omega_0}{2\pi}.$$ $$\boxed{f_0 = 5.03\text{ kHz — the (series) resonant frequency}}$$
3. c) Branch currents at resonance. At $\omega_0$ the series branch is purely resistive, $Z_2=R_2=10\ \Omega$, while the tank is capacitive, $Z_1=-j3.19\ \Omega$. Solving with peak phasors ($V_s=100\angle0^\circ$ V): $$i_1(t)=14.44\cos(\omega_0 t + 43.8^\circ)\text{ A},\qquad i_2(t)=4.61\cos(\omega_0 t - 46.2^\circ)\text{ A}.$$ $$\boxed{|I_1|=14.4\text{ A (pk)},\quad |I_2|=4.61\text{ A (pk)}}$$ I2 is exactly in phase with V2 (both at $-46.2^\circ$), confirming part (b).
4. d) Reactive power at resonance. Using rms values, $S=V_{s,\text{rms}}I_s^{*}=680.9 - j333.0$ VA, so $$\boxed{Q = -333\text{ VAR (net capacitive)}}$$ (the tank's dominant capacitive current at this frequency makes the source supply net negative reactive power).