Question 1 of 7: DC Bridge Network — KCL/KVL, V ab and Power in R 4
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 — Electric Circuits and Power — National Exam, May 2014.
Closed book; one aid sheet permitted; Casio/Sharp approved calculator only.
Any five of the seven questions constitute a complete paper — every question is answered.
Reference texts: Sadiku & Alexander, Fundamentals of
Electric Circuits (DC/AC network analysis, Thevenin, resonance);
Chapman, Electric Machinery Fundamentals (magnetic circuits);
Boylestad, Electronic Devices and Circuit Theory (diode rectifiers);
Mano, Digital Design (combinational logic).
Question 1: DC Bridge Network — KCL/KVL, Vab and Power in R4(20 marks)
Given. The bridge-style network of Figure 1: node A is the common top
rail (tied directly to $V_s{+}$, and to the tops of $R_1$, $R_3$ and $R_5$); node C is the
common bottom (ground) rail; $R_1$ runs A–B, $R_2$ runs B–C, $R_3$ runs A–D,
$R_4$ runs D–C, $R_5$ runs A–C directly, and the current source $I_s$ bridges
B–D (arrow B→D).
Given data
Quantity
Value
$R_1$
10 Ω
$R_2$
10 Ω
$R_3$
5 Ω
$R_4$
10 Ω
$R_5$
100 Ω
$I_s$ (B→D)
9 A
$V_s$ (A to C, + at A)
10 V
Find. The KCL equations at B and D; the KVL equations for loops ACA,
ABCA, ADCA; the voltage $V_{ab}=V_A-V_B$; and the power dissipated in $R_4$.
[Figure not reproduced: Figure 1 — redrawn from the printed figure ($R_2$ runs B–C as a separate branch from $R_1$, and $V_s$ lands directly on node A). See the official exam paper.]
Check: in Figure 1, $R_2$ is its own branch from B to the ground rail C, and $V_s$ connects directly to the top rail (node A) with no series resistor (it is not in series with $V_s$ feeding node B). All values below use this topology.
Approach. Because $V_s$ sits directly on node A, $V_A=V_s$ is fixed;
write one KCL equation per floating node (B, D) and one KVL equation per independent
loop (ACA, ABCA, ADCA), then solve the resulting linear system for the branch currents.
(a) KCL at node B and node D. Current leaving a node sums to zero.
At B, three branches meet ($R_1$ to A, $R_2$ to C, $I_s$ to D):
$$\dfrac{V_B-V_A}{R_1}+\dfrac{V_B-V_C}{R_2}+I_s=0 \quad\Longrightarrow\quad
I_1+I_2=I_s$$
At D, three branches meet ($R_3$ to A, $R_4$ to C, $I_s$ arriving from B):
$$\dfrac{V_D-V_A}{R_3}+\dfrac{V_D-V_C}{R_4}-I_s=0 \quad\Longrightarrow\quad
I_3+I_s=I_4$$
where $I_1,I_2,I_3,I_4$ are the currents in $R_1,R_2,R_3,R_4$ (each defined flowing away
from node A / into node C).
(b) KVL for loops ACA, ABCA, ADCA. Because $R_5$ and $V_s$ are the
only two elements directly spanning A–C, loop ACA is simply
$$I_5R_5=V_s$$
(this reproduces $V_A=V_s$ via Ohm's law on $R_5$). Loop ABCA (A→B via $R_1$,
B→C via $R_2$, C→A via $V_s$) gives
$$I_1R_1+I_2R_2=V_s$$
Loop ADCA (A→D via $R_3$, D→C via $R_4$, C→A via $V_s$) gives
$$I_3R_3+I_4R_4=V_s$$
(c) Solve for $V_{ab}$. $I_5=V_s/R_5=10/100=0.1$ A confirms
$V_A=10$ V. Combining the ABCA loop with the KCL-at-B equation
($I_1=I_2+9$, $I_1R_1+I_2R_2=10$) gives two equations in $I_1,I_2$:
$$10I_1+10I_2=10,\qquad I_1-I_2=9$$
Solving: $I_1=5.0$ A, $I_2=-4.0$ A, so
$$V_B=I_2R_2=(-4.0)(10)=-40.0\text{ V}$$
$$\boxed{V_{ab}=V_A-V_B=10-(-40)=50.0\text{ V}}$$
(d) Power in $R_4$. Combine the ADCA loop with KCL-at-D
($I_4=I_3+9$, $I_3R_3+I_4R_4=10$):
$$5I_3+10I_4=10,\qquad I_4-I_3=9$$
Solving: $I_3=-16/3\approx-5.333$ A, $I_4=11/3\approx3.667$ A, so
$$\boxed{P_{R_4}=I_4^2R_4=(3.667)^2(10)=134.44\text{ W}}$$
