DC Bridge Network — KCL/KVL, Power in R1, Power of Vs3: 04-BS-4, May 2018, Q1

Question 1: DC Bridge Network — KCL/KVL, Power in R1, Power of Vs3 (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. The bridge of Figure 1: top rail A–B–C carries $R_2$ (A–B) and $R_4$ (B–C); each of A, B, C drops to the common node D through its own ideal source (+ terminal up) in series with a resistor: $V_{s1}$–$R_1$ at A, $V_{s3}$–$R_3$ at B, $V_{s5}$–$R_5$ at C.

Given data
$R_1$	$R_2$	$R_3$	$R_4$	$R_5$	$V_{s1}$	$V_{s3}$	$V_{s5}$
2 Ω	2 Ω	4 Ω	2 Ω	2 Ω	8 V	12 V	16 V

Find. KCL at nodes B, D; KVL for loops ABDA and BCDB; the power dissipated in $R_1$; the power produced by $V_{s3}$.

Figure 1: DC bridge (R_1=2Ω R_2=2Ω R_3=4Ω R_4=2Ω R_5=2Ω) -- V_s1=8V, V_s3=12V, V_s5=16V

Approach. Take $D$ as reference (0 V) and solve the three node voltages $V_A,V_B,V_C$ from nodal KCL, defining branch currents $I_1$(A→D), $I_2$(A→B), $I_3$(B→D), $I_4$(B→C), $I_5$(C→D); the requested KCL/KVL equations then follow directly from those same current definitions.

a) KCL at nodes B and D. At B, the current $I_2$ arriving from A splits into $I_3$ (down to D through $V_{s3}$-$R_3$) and $I_4$ (across to C); at D, the three branch currents $I_1,I_3,I_5$ (all defined flowing into D) must sum to zero, since D has no other connection. $$\text{Node B: } I_2 = I_3+I_4 \qquad \text{Node D: } I_1+I_3+I_5=0$$ The node-D equation is the automatically-satisfied "extra" KCL equation of the network (consistent with, not independent of, the equations at A, B, C) — it is still worth writing explicitly since the question asks for it.
b) KVL for loops ABDA and BCDB. Traverse ABDA via A→B ($R_2$ drop), B→D (drop through $V_{s3}$ then $R_3$), D→A (rise back through $R_1$ then $V_{s1}$); traverse BCDB via B→C ($R_4$ drop), C→D (drop through $V_{s5}$ then $R_5$), D→B (rise back through $R_3$ then $V_{s3}$). $$\text{ABDA: } I_2R_2+V_{s3}+I_3R_3 = I_1R_1+V_{s1} \qquad \text{BCDB: } I_4R_4+V_{s5}+I_5R_5 = I_3R_3+V_{s3}$$
Solve the node voltages. Writing KCL at A, B, C directly in terms of node voltages ($V_D=0$) and solving the resulting $3\times3$ linear system: $$\left(\frac{1}{R_1}+\frac{1}{R_2}\right)V_A-\frac{1}{R_2}V_B=\frac{V_{s1}}{R_1},\quad -\frac{1}{R_2}V_A+\left(\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\right)V_B-\frac{1}{R_4}V_C=\frac{V_{s3}}{R_3},\quad -\frac{1}{R_4}V_B+\left(\frac{1}{R_4}+\frac{1}{R_5}\right)V_C=\frac{V_{s5}}{R_5}$$ gives $V_A=10\text{ V}$, $V_B=12\text{ V}$, $V_C=14\text{ V}$, hence $I_1=\dfrac{V_A-V_{s1}}{R_1}=\dfrac{10-8}{2}=\boxed{1\text{ A}}$ and $I_3=\dfrac{V_B-V_{s3}}{R_3}=\dfrac{12-12}{4}=\boxed{0\text{ A}}$ (a genuinely clean special case: node B lands exactly at $V_{s3}$, so no current at all flows in that branch).
c) Power dissipated in $R_1$. $$P_{R_1}=I_1^2R_1=(1\text{ A})^2(2\,\Omega)=\boxed{2\text{ W}}$$
d) Power produced by $V_{s3}$. Power delivered by the source equals $(-I_3)\times V_{s3}$ (current leaving its $+$ terminal into the external circuit). Since $I_3=0$ exactly, the source neither delivers nor absorbs power in this configuration. $$P_{V_{s3}}=(-I_3)V_{s3}=(0)(12\text{ V})=\boxed{0\text{ W}}$$

Final results — Question 1
Quantity	Value
$V_A,V_B,V_C$	10.0 V, 12.0 V, 14.0 V
$I_1$ (through $R_1$)	1.000 A
$I_3$ (through $R_3$/$V_{s3}$)	0.000 A
$P_{R_1}$	2.00 W
$P_{V_{s3}}$ (produced)	0.00 W

Check: a zero current/zero power result is a real, clean answer here, not a sign of a dropped term — $V_{s3}$ happens to sit at exactly the potential the rest of the bridge would produce at node B on its own.

Topic: Kirchhoff's Current and Voltage Laws in a multi-source resistive ladder/bridge.

Key relations: $\sum I_{in}=\sum I_{out}$ at every node (KCL); $\sum V_{drops}=0$ around every closed loop (KVL); $P=I^2R$; source power $=I\times V$ with $I$ referenced out of the $+$ terminal.

Why this works: each of the three vertical branches is an ideal source in series with a resistor, so a node-voltage formulation (one unknown per node, reference at D) is the fastest solve; the requested KCL/KVL equations are just the same physics written in current form rather than voltage form, and are recovered directly from the solved branch currents.

Common pitfall: assuming a boxed current of exactly zero must be an arithmetic slip and "fixing" it by rounding to a small nonzero value — always re-verify a suspiciously clean result independently (here, by direct substitution) before overriding it.

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

Question 1 of 7: DC Bridge Network — KCL/KVL, Power in R1, Power of Vs3

Question 1: DC Bridge Network — KCL/KVL, Power in R1, Power of Vs3 (20 marks)