Question 1 of 7: DC Resistive Network — KCL, KVL, Branch Current and Source Power
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 04-BS-4 Electric Circuits and Power — undated sitting (internal
evidence places it at May 2019). 3 hours, closed book, one
double-sided aid sheet, approved Casio/Sharp calculator only. The exam instructs
"any five questions constitute a complete paper" — all seven are solved below as a
complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits
(7th ed.) — Ch. 2–4 (resistive circuits, node/mesh analysis, Thévenin/Norton,
maximum power transfer), Ch. 7 (first-order RC/RL transients), Ch. 9–10 (sinusoidal
steady state, phasors, AC power); Chapman, Electric Machinery Fundamentals
(5th ed.) — Ch. 1 (magnetic circuits, reluctance, fringing, force on an armature);
Sadiku, Elements of Electromagnetics — Ch. 5, 8 (magnetic energy and force);
Mano & Ciletti, Digital Design — Ch. 2–3 (Boolean algebra, combinational
logic design).
Question 1: DC Resistive Network — KCL, KVL, Branch Current and Source Power (20 marks)
Given. The resistive network of Figure 1, with a current source $I_s=10\text{ A}$ (arrow from node B to node A) and a voltage source $V_s=24\text{ V}$ (the "+" terminal on node C) bridging the two halves of the network.
Given data
Quantity
Value
$R_1$
$4\ \Omega$
$R_2$
$3\ \Omega$
$R_3$
$6\ \Omega$
$R_4$
$6\ \Omega$
$R_5$
$3\ \Omega$
$I_s$
$10\text{ A}$
$V_s$
$24\text{ V}$
Find. The four nodal KCL equations, the two loop KVL equations, the current through $R_1$, and the power delivered by $I_s$.
Figure 1 — DC circuit for Question 1 (nodes A, B, C, D as labelled).
Approach. Label the four node voltages, write KCL at each node and KVL around each named loop symbolically (parts a–b), then solve the network by nodal analysis (a supernode absorbs the floating source $V_s$) to get every node voltage and hence $I_{R_1}$ and the power leaving the current source (parts c–d).
Part (a) — KCL at nodes A, B, C, D. Take currents leaving each node as positive, with $I_{R_1},\dots,I_{R_5}$ the currents flowing away from A/B/C/D through the labelled resistors and $I_{V_s}$ the current flowing from C to D through the source branch:$$\text{Node A: } I_s = I_{R_1}+I_{R_2}+I_{R_4}$$$$\text{Node B: } I_{R_1}+I_{R_3}+I_{R_5} = I_s$$$$\text{Node C: } I_{R_2} = I_{R_3}+I_{V_s}$$$$\text{Node D: } I_{R_4}+I_{V_s} = I_{R_5}$$Node A and Node B are the same equation written from opposite ends of the $I_s$ branch — as expected, only three of the four KCL equations are independent.
Part (b) — KVL around loops ABCA and BCDB. Walking each loop clockwise and summing voltage drops to zero:$$\text{Loop ABCA: } -I_{R_1}R_1 + I_{R_2}R_2 + I_{R_3}R_3 = 0$$$$\text{Loop BCDB: } -I_{R_3}R_3 + V_s + I_{R_5}R_5 = 0$$The second equation is exactly the definition of $V_s$ once $I_{R_3}$ and $I_{R_5}$ are known, and is used below as a check.
Part (c) — solve the network for $I_{R_1}$. Ground node B ($V_B=0$) and use nodal analysis at A and at the supernode C–D (the ideal source $V_s$ ties $V_C-V_D=24\text{ V}$, so C and D cannot be nodal unknowns separately):$$\dfrac{V_A}{R_1}+\dfrac{V_A-V_C}{R_2}+\dfrac{V_A-V_D}{R_4}=I_s$$$$\dfrac{V_C-V_A}{R_2}+\dfrac{V_C}{R_3}+\dfrac{V_D-V_A}{R_4}+\dfrac{V_D}{R_5}=0,\qquad V_C-V_D=V_s$$ Substituting the given resistor and source values and solving the $3\times3$ linear system gives $V_A=24\text{ V}$, $V_C=24\text{ V}$, $V_D=0\text{ V}$. Hence $$I_{R_1}=\dfrac{V_A-V_B}{R_1}=\dfrac{24\text{ V}}{4\ \Omega}=\boxed{6\text{ A}}$$
Part (d) — power generated by $I_s$. The current source pumps current $I_s$ from the low-potential node (B, $0\text{ V}$) to the high-potential node (A, $V_A$), so it is delivering (not absorbing) power: $$P_{I_s}=V_A\cdot I_s = (24\text{ V})(10\text{ A})=\boxed{240\text{ W}}$$ As a check, $I_{R_2}=(V_A-V_C)/R_2=0$ (both ends sit at $24\text{ V}$), $I_{R_4}=(V_A-V_D)/R_4=4\text{ A}$, and $I_{R_1}+I_{R_2}+I_{R_4}=6+0+4=10\text{ A}=I_s$, matching the Node-A KCL equation from part (a).
