Question 3 of 6: AC mesh analysis, power factor and source power
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Elec-A1 Circuits, May 2014. Closed-book, 3-hour paper; a Laplace-transform table is supplied in the appendix. Six questions of equal value; any five constitute a complete paper. Full worked solutions to all six questions are given below so the set can be used for study regardless of which five a candidate chose.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — nodal/mesh analysis, Thévenin’s theorem and maximum-power transfer, first-order transients, AC steady-state and resonance, and Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the s-domain and initial-condition models.
each network below is redrawn element-by-element from the original drawing, and every reference-node, mesh-direction and source-polarity choice is stated with its solution so a reader can reproduce every sign. Phasor answers are quoted in magnitude∡angle form with angles in degrees; all AC sources in Q3 and Q4 are rms.
Question 3: AC mesh analysis, power factor and source power [8 + 4 + 2 + 6]
Given. Three mesh currents as drawn: $I_1$ (right loop, CW), $I_2$ (bottom-left loop, CW), $I_3$ (top-left loop, CCW).
Branch
Element
Left branch (mesh $I_2$ only)
$I_s=10\angle{-}30^{\circ}$ A (arrow down)
Top-left
$5\ \Omega$
Shared $I_2$–$I_3$ (horizontal)
$j12\ \Omega$ (inductor)
Shared $I_3$–$I_1$ (upper centre)
$-j4\ \Omega$ (capacitor)
Shared $I_2$–$I_1$ (lower centre)
$-j3\ \Omega$ (capacitor)
Top-right
$10\ \Omega$
Right branch (mesh $I_1$ only)
$E_s=30\angle0^{\circ}$ V, + up
Find. The mesh equations, $I_1,I_2,I_3$, the source power factor, and $P$ supplied by $E_s$.
[Figure not reproduced: Figure 3 — Redrawn AC network. The current source occupies the outer-left branch of mesh $I_2$ alone, so it fixes $I_2$; $I_3$ is drawn counter-clockwise, so on every branch it shares with the clockwise meshes the two currents add . See the official exam paper.]
Approach. The current source sits on mesh $I_2$’s outer branch (arrow down, opposing the clockwise sense), so $I_2=-I_s$ outright. Two KVL equations for meshes $I_3$ and $I_1$ close the system; then the source current gives power factor and power.
Current-source constraint. The outer branch of mesh $I_2$ carries $I_s$ downward, opposite to the clockwise $I_2$, so $$I_2=-I_s=-10\angle{-}30^{\circ}=\boxed{10\angle150^{\circ}\ \text{A}}.$$
KVL, mesh $I_3$ (CCW). Its branches are $5\ \Omega$, $j12$ (shared with $I_2$) and $-j4$ (shared with $I_1$); because $I_3$ is CCW while $I_1,I_2$ are CW, the shared currents add: $$5\,I_3+j12\,(I_3+I_2)-j4\,(I_3+I_1)=0\;\Rightarrow\;(5+j8)\,I_3-j4\,I_1+j12\,I_2=0.$$
KVL, mesh $I_1$ (CW). Its branches are $10\ \Omega$, $E_s$, $-j3$ (shared with the CW $I_2$ → difference) and $-j4$ (shared with the CCW $I_3$ → sum): $$10\,I_1-j4\,(I_1+I_3)-j3\,(I_1-I_2)+E_s=0\;\Rightarrow\;-j4\,I_3+(10-j7)\,I_1+j3\,I_2=-E_s.$$
Solve. Substituting $I_2=10\angle150^{\circ}$ and $E_s=30\angle0^{\circ}$ into the pair and solving the $2\times2$ complex system, $$I_1=\boxed{5.705\angle140.07^{\circ}\ \text{A}},\qquad I_3=\boxed{10.346\angle4.32^{\circ}\ \text{A}}.$$
Source current. The current delivered by $E_s$ leaves its + terminal through the 10 Ω. By KCL that current is $$I_{E_s}=5.705\angle{-}39.93^{\circ}\ \text{A}\ (=-I_1),$$ lagging the $30\angle0^{\circ}$ V by $39.93^{\circ}$.
Power factor. $$\text{pf}=\cos(\theta_V-\theta_I)=\cos(39.93^{\circ})=\boxed{0.767\ \text{lagging}}.$$
Power supplied by $E_s$. With rms phasors, $$\mathbf{S}_{E_s}=E_s\,I_{E_s}^{*}=(30)(5.705\angle39.93^{\circ})=171.2\angle39.93^{\circ}\ \text{VA},$$ so $$P_{E_s}=|\mathbf{S}|\cos\theta=171.2(0.767)=\boxed{131.2\ \text{W}}\quad(Q=109.9\ \text{var}).$$
