04-BS-4, December 2016, Q1 | Nivaar Exam Prep

Question 1 (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-type DC network of Figure 1, with a current source Is feeding node A (top rail) and returning at node B (bottom rail), resistors R2/R3 bridging A–C–B and R4/R5 bridging A–D–B, and an ideal voltage source Vs (+ at C, − at D) tying the two bridge midpoints together.

Given data
R1	10 Ω	R2	6 Ω
R3	3 Ω	R4	3 Ω
R5	6 Ω	Is	10 A
Vs	36 V

Find. (a) KCL at A, B, C, D; (b) KVL for loops ABCA and BCDB; (c) the current through R1; (d) the power generated by Is.

Figure 1 — DC bridge network: current source Is with R1 across nodes A–B, resistor pairs R2/R3 and R4/R5 bridging to midpoints C and D, tied by voltage source Vs.

Approach. Take node B as the reference (0 V) and define branch reference directions I1 (A→B through R1), I2 (A→C through R2), I3 (C→B through R3), I4 (A→D through R4), I5 (D→B through R5) and Ix (C→D through Vs); write KCL at every node and KVL for the two named loops, then solve the resulting node-voltage system (using the supernode C–D, since Vs bridges them directly) for VA, VC, VD.

(a) KCL at each node. With the reference directions above, current balance at each node reads: $$ \text{Node A:}\quad I_s = I_1 + I_2 + I_4 $$ $$ \text{Node B:}\quad I_1 + I_3 + I_5 = I_s $$ $$ \text{Node C:}\quad I_2 = I_3 + I_x $$ $$ \text{Node D:}\quad I_4 + I_x = I_5 $$ where $I_1=V_A/R_1$, $I_2=(V_A-V_C)/R_2$, $I_3=V_C/R_3$, $I_4=(V_A-V_D)/R_4$, $I_5=V_D/R_5$, and $I_x$ is the (unknown) current through the ideal source Vs.
(b) KVL for loops ABCA and BCDB. Because $V_A-V_B$ can be reached either directly through R1 or via C through R2 then R3, loop ABCA gives $$ I_1 R_1 = I_2 R_2 + I_3 R_3 $$ Loop BCDB crosses R3 (B→C, against $I_3$), the source Vs (C→D, a drop of Vs since C is the + terminal), and R5 (D→B, with $I_5$): $$ I_5 R_5 = I_3 R_3 - V_s $$
(c) Solve for the node voltages and $I_1$. Eliminating $I_x$ leaves the supernode equation $\,(V_A-V_C)/R_2+(V_A-V_D)/R_4=V_C/R_3+V_D/R_5\,$ together with the node-A KCL equation and the source constraint $V_C-V_D=V_s$. Solving this 3×3 linear system gives $$ V_A=20.0\text{ V},\qquad V_C=28.0\text{ V},\qquad V_D=-8.0\text{ V} $$ so the current through R1 is $$ I_1=\frac{V_A}{R_1}=\frac{20.0}{10}=\boxed{2.00\text{ A (from A to B)}} $$
(d) Power generated by Is. The current source pushes 10 A out of node A (at 20.0 V) from node B (0 V), so it delivers $$ P_{I_s}=I_s\,(V_A-V_B)=10\times20.0=\boxed{200\text{ W (delivered)}} $$

Question 1 — final results
Quantity	Value
$V_A$	20.0 V
$V_C$	28.0 V
$V_D$	−8.0 V
Current through R1	2.00 A
Power delivered by Is	200 W

Topic: Node-voltage (KCL) and mesh-consistency (KVL) analysis of a resistive bridge network.
Key relations: $\sum I_{\text{in}}=\sum I_{\text{out}}$ at every node; around any closed loop the IR drops sum to zero (or, equivalently, two paths between the same pair of nodes must show the same net voltage difference).
Why this works: An ideal voltage source directly bridging two non-reference nodes (here C and D) cannot be assigned an independent node equation because its branch current is unknown — combining the two nodes into a supernode and adding the source's voltage constraint removes the unknown current while keeping the system solvable.
Common pitfall: Forgetting the supernode constraint and instead writing an ordinary KCL equation at C or D alone, which drops the Vs branch current and gives an under-determined system.
Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., Ch. 3 (Nodal and Mesh Analysis, Supernodes).

Question 1 of 7

Question 1 (20 marks)