DC Bridge Network — KCL/KVL and an Unknown Source Current: 04-BS-4, May 2016, Q1

Question 1: DC Bridge Network — KCL/KVL and an Unknown Source Current (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-style DC network of Figure 1 has an ideal voltage source V₆ across node A and the ground rail, an unknown ideal current source I₅ between node B and ground, and four resistors as tabulated. A measurement shows I₄ = 2 A in the branch through R₄.

Given data
Quantity	Value
R1	10 Ω
R2	11 Ω
R3	2 Ω
R4	5 Ω
V6	50 V
I4 (measured)	2 A

Find. The KCL/KVL equations at the labelled nodes and loops, the unknown source current I₅, and the power dissipated in R₁.

Figure 1: DC bridge circuit (R1=10, R2=11, R3=2, R4=5, V6=50V, I4=2 A)

Approach. Node C connects to the ground rail only through R4, so the measured I4 fixes V_C directly; V6 fixes V_A directly since it is an ideal source; walking those two fixed potentials through R1 and R3 isolates I5 from a single KCL equation at node B.

KCL at nodes A, B, C (part a). Taking currents leaving each node as positive: $$ \text{Node A: } I_1 + I_2 = I_6 \qquad \text{Node B: } I_2 + I_5 = I_3 \qquad \text{Node C: } I_1+I_3 = I_4 $$ (I1 through R1, I2 through R2, I3 through R3, I4 through R4, I5 the unknown source, I6 through V6.)
KVL for the three loops (part b). Walking each loop and summing voltage drops to zero: $$ \text{Loop ACBA: } I_1R_1 - I_3R_3 - I_2R_2 = 0 $$ $$ \text{Loop ACDA: } I_1R_1 + I_4R_4 - V_6 = 0 $$ $$ \text{Loop ABCDA: } I_2R_2 + I_3R_3 + I_4R_4 - V_6 = 0 $$ (the third loop is the sum of the first two and is not independent, but the exam asks for all three.)
Fix the two known nodes. The ideal source gives $V_A = V_6 = 50\text{ V}$ directly. Node C reaches ground only through R4, so the measurement fixes $$ V_C = I_4R_4 = (2\text{ A})(5\ \Omega) = 10\text{ V}. $$ Then $ I_1 = \dfrac{V_A-V_C}{R_1} = \dfrac{50-10}{10} = \boxed{4\text{ A}}. $
KCL at C, then Ohm's law on R3. From node C, $I_3 = I_4-I_1 = 2-4 = -2\text{ A}$ (the physical current in R3 runs from C down to B, opposite the assumed reference). Since $I_3=(V_B-V_C)/R_3$, $$ V_B = V_C + I_3R_3 = 10 + (-2)(2) = 6\text{ V}. $$
Ohm's law on R2, then KCL at B for I5 (part c). $$ I_2 = \frac{V_A-V_B}{R_2} = \frac{50-6}{11} = 4\text{ A}. $$ From the node-B equation, $ I_5 = I_3 - I_2 = -2-4 = \boxed{-6\text{ A}}$ — magnitude 6 A, directed opposite to the assumed "into B" reference arrow (i.e. the source actually delivers 6 A from node B down to the ground rail).
Power in R1 (part d). $$ P_{R_1} = I_1^2R_1 = (4)^2(10) = \boxed{160\text{ W}}. $$

Final results
Quantity	Value
V_A	50 V
V_C	10 V
V_B	6 V
I₁	4 A
I₂	4 A
I₃	−2 A
I₄ (given)	2 A
I₅	−6 A
P_R1	160 W

Topic: Nodal/KCL-KVL analysis of a resistive bridge network containing a mixed ideal voltage source and an unknown ideal current source.

Key relations: Kirchhoff's current law ($\sum I_{in}=\sum I_{out}$ at each node), Kirchhoff's voltage law around any closed loop, Ohm's law $V=IR$.

Why this works: an ideal voltage source fixes the potential of its own node outright, and a resistor with only the ground rail beyond it turns a measured branch current directly into a node voltage. Chaining those two facts through the two intervening resistors collapses what looks like a 6-unknown network into two direct substitutions, leaving only the node-B KCL equation to solve for the unknown source.

Common pitfall: reporting I5 as a negative number and stopping there. A negative result relative to an assumed reference direction is not an error — it means the physical current flows opposite to the arrow drawn on the figure; always state the magnitude and the true direction.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 2–3 (Ohm's law, nodal analysis, Kirchhoff's laws).

Question 1 of 7: DC Bridge Network — KCL/KVL and an Unknown Source Current

Question 1: DC Bridge Network — KCL/KVL and an Unknown Source Current (20 marks)