DC Bridge Network — KCL/KVL, Vab, Power in R4: 04-BS-4, December 2017, Q1

Question 1: DC Bridge Network — KCL/KVL, Vab, Power in R4 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bridge network of Figure 1: $V_s$ (A–C) drives resistors $R_1$(A–B), $R_2$(B–C), $R_3$(A–D), $R_4$(D–C) and $R_5$(A–C directly across the source), with an ideal current source $I_s$ connected B→D.

Given data
$R_1$	$R_2$	$R_3$	$R_4$	$R_5$	$I_s$	$V_s$
10 Ω	10 Ω	5 Ω	10 Ω	100 Ω	9 A	10 V

Find. The KCL equations at B, 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 1: DC bridge circuit for Question 1

Approach. Define branch currents $I_1$(A→B), $I_2$(B→C), $I_3$(A→D), $I_4$(D→C), $I_5$(A→C); apply KCL at B and D, then KVL around the three source-containing loops, and solve the resulting linear system.

a) KCL at nodes B and D. Current $I_1$ enters B from R1 and splits into $I_2$ (through R2) and the current source $I_s$ leaving toward D; at D, $I_3$ (through R3) and $I_s$ (entering from B) combine and leave through R4 as $I_4$. $$\text{Node B: } I_1 = I_2 + I_s \qquad \text{Node D: } I_3 + I_s = I_4$$
b) KVL for loops ACA, ABCA, ADCA. $R_5$ is wired directly in parallel with the ideal source $V_s$ (loop ACA), so its voltage is fixed regardless of the rest of the network; loops ABCA and ADCA each close through the source branch. $$\text{ACA: } V_s = I_5 R_5 \qquad \text{ABCA: } I_1R_1+I_2R_2=V_s \qquad \text{ADCA: } I_3R_3+I_4R_4=V_s$$
c) Solve for $V_{ab}$. Substitute $I_1=I_2+I_s$ into the ABCA equation: $(I_2+9)(10)+I_2(10)=10 \Rightarrow 20I_2=-80 \Rightarrow I_2=-4\text{ A}$, so $I_1=5\text{ A}$. Then $V_{ab}=V_A-V_B=I_1R_1$. $$V_{ab} = I_1 R_1 = (5\text{ A})(10\,\Omega) = \boxed{50\text{ V}}$$ A node-voltage cross-check (ground at C, $V_A=V_s=10$ V since $R_5$ floats across the ideal source) gives $V_B=-40$ V and $V_A-V_B=50$ V — the same result.
d) Power in $R_4$. Substitute $I_3=I_4-9$ into the ADCA equation: $(I_4-9)(5)+I_4(10)=10 \Rightarrow 15I_4=55 \Rightarrow I_4=\tfrac{11}{3}=3.667\text{ A}$. $$P_{R_4} = I_4^2 R_4 = (3.667\text{ A})^2(10\,\Omega) = \boxed{134.4\text{ W}}$$

Final results — Question 1
Quantity	Value
$I_1,\,I_2$ (R1, R2)	5.000 A, −4.000 A
$I_3,\,I_4$ (R3, R4)	−5.333 A, 3.667 A
$I_5$ (R5)	0.100 A
$V_{ab}$	50.0 V
$P_{R_4}$	134.4 W

Check: $I_2$ and $I_3$ come out negative — this simply means the true current in R2 and R3 flows opposite to the assumed reference direction (B→C, A→D); the magnitudes and $V_{ab}$/$P_{R_4}$ results are unaffected.

Topic: Kirchhoff's Current and Voltage Laws in a resistive bridge network.

Key relations: $\sum I_{in}=\sum I_{out}$ at every node (KCL); $\sum V_{drops}=0$ around every closed loop (KVL); $P=I^2R$.

Why this works: KCL is charge conservation at a node; KVL is energy conservation around a closed path. Any two independent node equations plus the loop equations for the branches carrying sources fully determine a network with two unknown nodes, without needing a full mesh or nodal solve of the whole bridge.

Common pitfall: forgetting that an ideal current source forces a *fixed* current regardless of the voltage across it — students often try to write a KVL loop straight through $I_s$ and get stuck; route the loop through the source-defined branches (ACA/ABCA/ADCA) instead, as done here.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, Ch. 2–3 (KCL/KVL, node/loop analysis).

Question 1 of 7: DC Bridge Network — KCL/KVL, Vab, Power in R4

Question 1: DC Bridge Network — KCL/KVL, Vab, Power in R4 (20 marks)