AC steady-state mesh analysis (phasors): 22-Elec-A1 Circuits, May 2016, Q4

Question 4: AC steady-state mesh analysis (phasors) [10 + 5 + 5]

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. Angular frequency $\omega=5\text{ rad/s}$. Element impedances:

Element	Value	Impedance at $\omega=5$
$R_1$	$2\,\Omega$	$2\,\Omega$
$R_2$	$5\,\Omega$	$5\,\Omega$
$C=0.1\text{ F}$	—	$\dfrac{1}{j\omega C}=-j2\,\Omega$
$L=2\text{ H}$	—	$j\omega L=j10\,\Omega$

Source phasors (cosine reference): $\mathbf V_s=20\angle30^\circ\text{ V}$; and $i_s=15\sin(5t+20^\circ)=15\cos(5t-70^\circ)$ so $\mathbf I_s=15\angle{-70^\circ}\text{ A}$.

Find. The mesh equations, the mesh currents $I_1,I_2$, and $v_o(t)$ across $R_1$.

[Figure not reproduced: Figure 4 — redrawn. Mesh 1 (left): $v_s$, and the shared middle branch $R_1$ (with $V_0$) in series with $C$. Mesh 2 (right): the shared middle branch, $R_2$, $L$, and the current source $i_s$. Both mesh currents are clockwise. See the official exam paper.]

Approach. The current source sits in mesh 2's outer branch, so it fixes $I_2$ directly; the single KVL around mesh 1 then gives $I_1$, and $v_o=R_1(I_1-I_2)$.

(a) Mesh 1 KVL (clockwise). The only source is $v_s$; the shared branch $R_1+Z_C$ carries $(I_1-I_2)$: $$\mathbf V_s=(R_1+Z_C)(I_1-I_2)\;\Rightarrow\;(2-j2)\,I_1-(2-j2)\,I_2 = 20\angle30^\circ.$$
(a) Mesh 2 constraint. The current source $i_s$ (arrow up) lies only in mesh 2; a clockwise mesh current runs downward through that branch, opposing $i_s$, so $$\boxed{I_2=-\mathbf I_s=-15\angle{-70^\circ}=15\angle110^\circ\text{ A}}.$$ (The full mesh-2 KVL $(R_1+Z_C)(I_2-I_1)+(R_2+Z_L)I_2+\mathbf V_{x}=0$ is redundant — it only serves to find the voltage $\mathbf V_x$ across the current source, not $I_2$.)
(b) Shared-branch current. From the mesh-1 equation, $$I_1-I_2=\frac{\mathbf V_s}{R_1+Z_C}=\frac{20\angle30^\circ}{2-j2}=\frac{20\angle30^\circ}{2\sqrt2\,\angle{-45^\circ}}=7.071\angle75^\circ\text{ A}.$$
(b) Mesh currents. Adding $I_2$, $$I_1=I_2+7.071\angle75^\circ=15\angle110^\circ+7.071\angle75^\circ=(-3.30+j20.93)\text{ A}.$$ $$\boxed{I_1=21.18\angle98.96^\circ\text{ A},\qquad I_2=15\angle110^\circ\text{ A}}$$
(c) Output voltage. $V_0$ is across $R_1$, which carries the shared-branch current $(I_1-I_2)$: $$\mathbf V_0=R_1(I_1-I_2)=2\times7.071\angle75^\circ=14.14\angle75^\circ\text{ V}.$$ $$\boxed{v_o(t)=14.14\cos(5t+75^\circ)\text{ V}}$$ Note $v_o$ depends only on $(I_1-I_2)$, so it is independent of the sign convention chosen for the current source.

Quantity	Value
$I_2$ (set by $i_s$)	$15\angle110^\circ\text{ A}$
$I_1-I_2$	$7.07\angle75^\circ\text{ A}$
$I_1$	$21.18\angle98.96^\circ\text{ A}$
$v_o(t)$	$14.14\cos(5t+75^\circ)\text{ V}$

Topic: Phasor mesh analysis of an AC circuit containing a current source.

Key relations: $Z_L=j\omega L$, $Z_C=1/(j\omega C)$; $\sin\theta=\cos(\theta-90^\circ)$ for a common reference; a current source in a single mesh fixes that mesh current.

Why this works: Converting every source and element to a phasor/impedance turns the differential mesh equations into linear algebra. A current source shared by only one mesh removes an unknown outright, leaving one KVL equation for the remaining mesh.

Common pitfall: forgetting to convert the sine source to the cosine reference (a $-90^\circ$ shift), or mis-signing $I_2$ relative to $i_s$; note the graded output $v_o$ is sign-robust because it uses the difference $I_1-I_2$.

Source: Sadiku, §10.1–10.4 (sinusoidal steady-state nodal/mesh analysis).

Question 4 of 6: AC steady-state mesh analysis (phasors)

Question 4: AC steady-state mesh analysis (phasors) [10 + 5 + 5]