DC Bridge Network — Kirchhoff's Laws and an Unknown Branch Resistance: 04-BS-4, May 2017, Q1

Question 1: DC Bridge Network — Kirchhoff's Laws and an Unknown Branch Resistance (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. A three-node DC network (Figure 1): the top rail is node A (top of R_o, R₁, R₂); node B ties the bottom of R₁, the top of R₃, and the R_o–V_s branch; node C is the ground rail (V_s−, bottom of R₃ and R₂). V_s sits directly between B and C with no series element, so it fixes V_B outright.

Given data
Quantity	Value
R₁	15 Ω
R₂	6 Ω
R₃	8 Ω
I₂	2 A
V_s	24 V

Find. V_AB, V_AC; I₁, I₃, I_s; the unknown R_o and I_o; and the power dissipated in R₂.

Figure 1: DC bridge network (R_o unknown, node A = top rail, node B = mid-tie, node C = ground)

Approach. V_s fixes V_B directly; the given I₂ through R₂ fixes V_A; Ohm's law on R₁ and R₃ then gives every branch current, and KCL at node A isolates the unknown I_o (hence R_o).

Fix the two known node voltages. Taking node C as ground (0 V), V_s connects directly from B to C with no intervening element, so $$ V_B = V_s = 24\text{ V}. $$ R₂ carries the given I₂ from A to C, so $$ V_A = I_2 R_2 = (2)(6) = \boxed{12\text{ V}}. $$
Voltages V_AB and V_AC (part a). $$ V_{AB} = V_A - V_B = 12 - 24 = \boxed{-12\text{ V}}, \qquad V_{AC} = V_A - V_C = 12 - 0 = \boxed{12\text{ V}}. $$
Currents I₁ and I₃ (part b, i). Ohm's law on R₁ (B→A) and R₃ (B→C): $$ I_1 = \frac{V_B-V_A}{R_1} = \frac{24-12}{15} = \boxed{0.8\text{ A}}, \qquad I_3 = \frac{V_B-V_C}{R_3} = \frac{24-0}{8} = \boxed{3\text{ A}}. $$
Current I_o and resistance R_o (part c). KCL at node A (currents I_o and I₁ arrive, I₂ leaves): $$ I_o = I_2 - I_1 = 2 - 0.8 = \boxed{1.2\text{ A}}. $$ Ohm's law on the R_o branch (B→A) then gives the unknown resistance: $$ R_o = \frac{V_B-V_A}{I_o} = \frac{24-12}{1.2} = \boxed{10\ \Omega}. $$
Current I_s (part b, ii). KCL at node B (I_s arrives; I_o, I₁, I₃ leave): $$ I_s = I_o + I_1 + I_3 = 1.2 + 0.8 + 3 = \boxed{5\text{ A}}. $$
Power dissipated in R₂ (part d). $$ P_{R_2} = I_2^2 R_2 = (2)^2(6) = \boxed{24\text{ W}}. $$

Final results
Quantity	Value
V_AB	−12 V
V_AC	12 V
I₁	0.8 A
I₃	3 A
I_s	5 A
R_o	10 Ω
I_o	1.2 A
P_R2	24 W

Topic: Node-voltage (KCL/KVL) analysis of a DC network containing an unknown branch resistance.

Key relations: Kirchhoff's current law at a node, Ohm's law $V=IR$, power $P=I^2R$.

Why this works: an ideal voltage source with no series element fixes its own node's potential outright, and a resistor carrying a stated current fixes the node beyond it. Once both A and B are known, every other branch current (and hence the unknown R_o) falls out of Ohm's law and a single KCL equation — no simultaneous system is needed.

Common pitfall: assuming R_o must be given data and hunting for it in the problem statement; the exam deliberately omits it so that KCL at node A is required to recover it as I_o, then R_o.

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 — Kirchhoff's Laws and an Unknown Branch Resistance

Question 1: DC Bridge Network — Kirchhoff's Laws and an Unknown Branch Resistance (20 marks)