Question 4 of 6: AC mesh analysis in the phasor domain
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction, node/mesh analysis with dependent sources, first-order RC transients, AC phasor analysis, Thévenin equivalents, maximum-power transfer and Laplace (s-domain) analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of super-node bookkeeping and the initial-condition source models used in Q6.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here as a study resource. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are written as magnitude $\angle$ angle with angles in degrees; the cosine time reference is used throughout Q4 and Q5 (so $10\sin(377t+90^\circ)=10\cos 377t$ becomes $10\angle 0^\circ$), and the sources marked rms in Q5 are treated as rms.
Question 4: AC mesh analysis in the phasor domain [5 + 8 + 5 + 2]
Figure‑4 — Four‑mesh AC network (ω = 377 rad/s). All mesh currents taken clockwise.
Given. All sources oscillate at $\omega=377\text{ rad/s}$. Four clockwise mesh currents $I_1$–$I_4$ are defined as drawn. Element and source list:
Element
Value / impedance
Source
Phasor
$R_1$
$8\,\Omega$
$10\sin(377t+90^\circ)\text{ V}$
$10\angle0^\circ\text{ V}$
$R_2$
$6\,\Omega$
$4\cos 377t\text{ A}$
$4\angle0^\circ\text{ A}$
$C_1=2.7\text{ mF}$
$-j0.982\,\Omega$
$5\sin(377t+45^\circ)\text{ A}$
$5\angle{-45^\circ}\text{ A}$
$C_2=1.3\text{ mF}$
$-j2.040\,\Omega$
$L=13.3\text{ mH}$
$+j5.014\,\Omega$
Find. (a) the phasor circuit; (b) the mesh equations; (c) $V_o$ (taken across $R_2$) and $v(t)$.
Check: output node. The paper does not label $V_o$ on the network. It is taken here as the voltage across the output resistor $R_2$ (the branch carrying mesh current $I_4$ on the right edge), which is the conventional “output” of this ladder; if the marker intended a different element the phasor result scales accordingly, but the mesh set-up in parts (a)–(b) is unchanged.
Approach. Replace each element by its impedance and each source by its phasor (cosine reference); the two current sources fix $I_2$ and the difference $I_4-I_3$, leaving one super-mesh KVL and one ordinary mesh KVL to solve.
Phasor transform (part a). At $\omega=377$: $Z_{C1}=\dfrac{1}{j\omega C_1}=-j0.982\,\Omega$, $Z_{C2}=-j2.040\,\Omega$, $Z_L=j\omega L=+j5.014\,\Omega$, and $R_1=8$, $R_2=6\,\Omega$. Writing every source in cosine form, $10\sin(377t+90^\circ)=10\cos377t\Rightarrow10\angle0^\circ\text{ V}$, $4\cos377t\Rightarrow4\angle0^\circ\text{ A}$, and $5\sin(377t+45^\circ)=5\cos(377t-45^\circ)\Rightarrow5\angle{-45^\circ}\text{ A}$.
Current-source constraints (part b). The $5\angle{-45^\circ}$ source sits alone on the right edge of mesh $I_2$ (its clockwise edge points down, the arrow up), so $I_2=-5\angle{-45^\circ}=5\angle135^\circ\text{ A}$. The $4\angle0^\circ$ source on the shared middle branch gives the super-mesh constraint $I_4-I_3=4\angle0^\circ\text{ A}$.
Mesh KVL equations (part b). With all currents clockwise: $$\text{Mesh }I_1:\;(R_1+Z_{C2})I_1-R_1I_3-Z_{C2}I_2=10\angle0^\circ,$$ $$\text{Super-mesh }I_3\cup I_4:\;(Z_{C1}+R_1)I_3+(R_2+Z_L)I_4-R_1I_1-Z_LI_2=0,$$ closed by the constraint $I_4=I_3+4\angle0^\circ$ and the known $I_2$.
Solve the system (part c). Substituting the impedances and the known $I_2$ and solving the three complex equations gives $$I_1=3.19\angle{-136.5^\circ},\quad I_3=5.61\angle{-153.5^\circ},\quad I_4=2.71\angle{-112.2^\circ}\text{ A}.$$
Output voltage (part c). $V_o$ is the drop across $R_2$, carrying $I_4$: $$\boxed{V_o=I_4R_2=(2.71\angle{-112.2^\circ})(6)=16.24\angle{-112.2^\circ}\text{ V}.}$$ Returning to the time domain (cosine reference), $$\boxed{v(t)=16.24\cos(377t-112.2^\circ)\text{ V}.}$$
