AC nodal analysis and the capacitor current: 22-Elec-A1 Circuits, December 2016, Q4

Question 4: AC nodal analysis and the capacitor current [12 + 8]

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. $\omega=5\text{ rad/s}$. Source $10\sin(5t+25^\circ)\text{ V}$ fixes node $v_1$. Elements: $0.5\text{ F}$ in series with $1\text{ H}$ from $v_1$ to $v_3$ (top branch); $4\,\Omega$ from $v_1$ to $v_2$; $2\text{ H}$ from $v_2$ to $v_3$; $0.4\text{ F}$ from $v_2$ to ground (its current is $i$); and a $2\cos 5t\text{ A}$ source injecting into $v_3$.

Element	Impedance at $\omega=5$
$0.5\text{ F}$	$-j0.4\,\Omega$
$1\text{ H}$	$+j5\,\Omega$
Top branch $v_1$–$v_3$	$-j0.4+j5=+j4.6\,\Omega$
$4\,\Omega$	$4\,\Omega$
$2\text{ H}$	$+j10\,\Omega$
$0.4\text{ F}$	$-j0.5\,\Omega$

Find. the phasors $V_2,V_3$ and the time-domain capacitor current $i(t)$.

[Figure not reproduced: Figure 4 — redrawn. Using a cosine reference: $V_1=10\angle-65^\circ\text{ V}$ ($10\sin(5t+25^\circ)=10\cos(5t-65^\circ)$) and the current source is $2\angle0^\circ\text{ A}$. The reference node is the bottom rail below $v_2$. See the official exam paper.]

Approach. Convert to phasors at $\omega=5$, treat $v_1$ as a known source node, write KCL at $v_2$ and $v_3$ in admittance form, solve the $2\times2$ complex system, then $i=V_2/Z_{0.4\text{F}}$.

Phasor sources (cosine reference). $10\sin(5t+25^\circ)=10\cos(5t-65^\circ)\Rightarrow V_1=10\angle-65^\circ\text{ V}$; $2\cos5t\Rightarrow \mathbf{I}_s=2\angle0^\circ\text{ A}$ into $v_3$.
(a) KCL at node $v_2$. Currents leaving through $4\,\Omega$, $2\text{ H}$ and the $0.4\text{ F}$ to ground sum to zero: $$\frac{V_2-V_1}{4}+\frac{V_2-V_3}{j10}+\frac{V_2}{-j0.5}=0.$$
(a) KCL at node $v_3$. The current source feeds this node: $$\frac{V_3-V_1}{j4.6}+\frac{V_3-V_2}{j10}=2\angle0^\circ.$$ These two are the requested node-voltage equations.
(b) Assemble the admittance matrix. With $Y_{4}=0.25$, $Y_{2H}=-j0.1$, $Y_{0.4}=j2$, $Y_{top}=-j0.2174$: $$\begin{bmatrix}0.25+j1.9 & j0.1\\ j0.1 & -j0.3174\end{bmatrix}\!\begin{bmatrix}V_2\\V_3\end{bmatrix}=\begin{bmatrix}Y_4V_1\\ Y_{top}V_1+2\end{bmatrix}.$$
Solve. The complex solve gives $$\boxed{V_2=1.42\angle-149.98^\circ\text{ V},\quad V_3=2.51\angle-2.98^\circ\text{ V}.}$$
Capacitor current. Through the $0.4\text{ F}$ ($Z=-j0.5\,\Omega$), $$\mathbf{I}=\frac{V_2}{-j0.5}=j2\,V_2=2.84\angle-59.98^\circ\text{ A},$$ so $$\boxed{i(t)=2.84\cos(5t-59.98^\circ)\ \text{A}}$$ (equivalently $2.84\sin(5t+30.0^\circ)\text{ A}$).

Quantity	Value
$V_2$	$1.42\angle-149.98^\circ\text{ V}$
$V_3$	$2.51\angle-2.98^\circ\text{ V}$
Capacitor current $\mathbf{I}$	$2.84\angle-59.98^\circ\text{ A}$
$i(t)$	$2.84\cos(5t-59.98^\circ)\text{ A}$

Topic: Steady-state AC node analysis in the phasor domain.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)$; KCL with admittances $Y=1/Z$; time-domain recovery $x(t)=|\mathbf X|\cos(\omega t+\angle\mathbf X)$.

Why this works: At one frequency every element becomes a complex impedance, so KCL is a linear complex system. Node $v_1$ is pinned by the ideal source and drops out as an unknown, leaving a compact $2\times2$ solve. The two sinusoidal sources must share one reference (here cosine), which is why the sine source is rewritten as a $-65^\circ$ cosine before taking phasors.

Common pitfall: mixing sine and cosine references (a $90^\circ$ error), or using $1/(j\omega C)$ where $j\omega C$ is meant; the $0.4\text{ F}$ has the small impedance $-j0.5\,\Omega$, so its admittance $j2\text{ S}$ dominates node $v_2$.

Source: Sadiku, Fundamentals of Electric Circuits, §9.4–9.5 (impedance, phasors) and §10.3 (nodal analysis of AC circuits).

Question 4 of 6: AC nodal analysis and the capacitor current

Question 4: AC nodal analysis and the capacitor current [12 + 8]