Question 1 of 6: Equivalent resistance, source and branch currents
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 and Δ–Y reduction, mesh/nodal 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; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
[Figure not reproduced: Figure 1 — DC resistor network driven by the 30 V source (redrawn from the printed figure). See the official exam paper.]
Given. A $30\text{ V}$ source drives terminals a–b. Label the internal junctions as drawn: $C$ (top, just after the input $2\,\Omega$, where $I_T$ is read), $D$ (top-right, after the $3\,\Omega$ carrying $I_o$), $E$ (centre, below $C$), $F$ (right, below $D$) and $G$ (bottom, below $E$). The bottom-right corner $H$ carries no branch of its own.
Branch
Value
Branch
Value
a–$C$ (input)
$2\,\Omega$
$E$–$F$
$6\,\Omega$
$C$–$E$
$9\,\Omega$
$E$–$G$
$4\,\Omega$
$C$–$D$ ($I_o$)
$3\,\Omega$
$F$–$H$
$2\,\Omega$
$D$–$F$
$3\,\Omega$
$H$–$G$
$5\,\Omega$
$b$–$G$ (return)
$3\,\Omega$
Find. (a) $R_{ab}$; (b) $I_T$; (c) $I_o$.
Approach. Collapse the two pass-through nodes, convert the remaining $C$–$E$–$F$ delta to a wye to reduce the interior to a single $R_{CG}$, then apply Ohm’s law and one nodal back-substitution for $I_o$.
Collapse the pass-through nodes. $H$ joins only the $2\,\Omega$ and $5\,\Omega$, so those are in series: $R_{FG}=2+5=7\,\Omega$. Node $D$ joins only the two $3\,\Omega$ resistors, so the $C\!-\!D\!-\!F$ path is $3+3=6\,\Omega$ — but $I_o$ still flows in the $C\!-\!D$ segment, so $D$ is kept for part (c). The interior is now the four nodes $C,E,F,G$ with $R_{CE}=9$, $R_{CF}=6$, $R_{EF}=6$, $R_{EG}=4$, $R_{FG}=7$ (all $\Omega$).
Delta→wye on triangle $C\text{-}E\text{-}F$. With $\Sigma=R_{CE}+R_{EF}+R_{CF}=9+6+6=21\,\Omega$,
$$R_C=\frac{R_{CE}R_{CF}}{\Sigma}=\frac{9\cdot6}{21}=2.571\,\Omega,\quad R_E=\frac{R_{CE}R_{EF}}{\Sigma}=\frac{9\cdot6}{21}=2.571\,\Omega,\quad R_F=\frac{R_{CF}R_{EF}}{\Sigma}=\frac{6\cdot6}{21}=1.714\,\Omega.$$
The wye centre $n$ connects to $C$, $E$, $F$; the legs $E$ and $F$ still reach $G$ through $4\,\Omega$ and $7\,\Omega$.
Reduce to $R_{CG}$. From $n$ the two paths to $G$ are $(R_E+4)=6.571\,\Omega$ and $(R_F+7)=8.714\,\Omega$ in parallel, in series with $R_C$:
$$R_{CG}=R_C+\big[(R_E+4)\,\|\,(R_F+7)\big]=2.571+\frac{6.571\cdot8.714}{6.571+8.714}=2.571+3.747=6.318\,\Omega.$$
Equivalent resistance at a–b. Add the input and return resistors that sit in series outside the interior:
$$\boxed{R_{ab}=2+R_{CG}+3=2+6.318+3=11.32\,\Omega.}$$
Source current. The $30\text{ V}$ source sees $R_{ab}$, and $I_T$ is that same series current:
$$\boxed{I_T=\frac{30}{R_{ab}}=\frac{30}{11.32}=2.65\text{ A}.}$$
Branch current $I_o$. The node voltage at $C$ follows from the input drop, $V_C=30-I_T(2)=30-5.30=24.70\text{ V}$. Solving the interior nodal equations (done in the companion Python check) gives $V_D=20.31\text{ V}$, so the current down the top $3\,\Omega$ ($C\!\rightarrow\!D$) is
$$\boxed{I_o=\frac{V_C-V_D}{3}=\frac{24.70-20.31}{3}=1.46\text{ A}.}$$
