Question 1 of 7: DC Bridge Network — KCL, KVL and Solving for an Unknown Resistor

Question 1: DC Bridge Network — KCL, KVL and Solving for an Unknown Resistor (equal value)

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.

[Figure not reproduced: Figure 1 — redrawn from the exam: R0 (unknown) and Vs sit in the left branch (node A); R4 bridges A to C; R1 bridges B to C; R3 bridges C to D; R2 bridges B to D; R5 returns D to the source's negative terminal. The top rail from R0's top to B carries I0, and continues as I2 into R2. See the official exam paper.]

Given. R1 = 10 Ω, R2 = 7 Ω, R3 = 10 Ω, R4 = 4 Ω, R5 = 1 Ω, Vs = 4 V (ideal source, + at node A). R0 is an unknown series resistor; the observed current I2 = 4 A supplies the extra equation needed to solve for it.

Find. (a) KCL at A, B, C; (b) KVL for loops ABCA, ACDA, BCDB; (c) R0; (d) I0 and the power dissipated in R0.

Approach. Because R0's top terminal ties to B through a plain wire, and Vs is ideal, the network has four true junctions (A, B, C, D); KCL at any three of them is independent (the fourth follows automatically), which is exactly why the exam asks for A, B, C only. Assign each branch current in the direction printed on the figure, write KCL at the junctions and KVL around the three named loops, then substitute the given resistor values and the observed I2 = 4 A to solve simultaneously for R0.

a) KCL. With I0 (A→B through R0), I4 (A→C through R4), I1 (C→B through R1), I3 (D→C through R3), I2 (B→D through R2) and I5 (D→E through R5) all taken in the direction printed on the figure:

1. Node A (current in from the source = current out through R0 and R4): $$I_5 = I_0 + I_4$$
2. Node B (current in from R0 and from R1 = current out through R2): $$I_0 + I_1 = I_2$$
3. Node C (current in from R4 and from R3 = current out through R1): $$I_4 + I_3 = I_1$$

(Node D gives $I_2 = I_3+I_5$, which is the sum of the other three and is not independent of them — this is why only three equations are requested for four junctions.)

b) KVL. Summing voltage drops with each resistor's drop written in the direction of its own defined current:

1. Loop ABCA (A→B via R0, B→C against I1, C→A against I4): $$I_0 R_0 - I_1 R_1 - I_4 R_4 = 0$$
2. Loop ACDA (A→C via I4, C→D against I3, D→A via I5 then the source): $$I_4 R_4 - I_3 R_3 + I_5 R_5 - V_s = 0$$
3. Loop BCDB (B→C against I1, C→D against I3, D→B against I2): $$I_1 R_1 + I_3 R_3 + I_2 R_2 = 0$$

c), d) Solve for R0. Vs is ideal, so node A sits at a fixed 4 V above the source's − terminal (node E, taken as 0 V) regardless of R0. Writing nodal equations at B, C, D in terms of conductances (with $G_0=1/R_0$ unknown) and substituting $I_2=(V_B-V_D)/R_2=4$ A together with the resistor values gives four linear equations in $V_B,V_C,V_D,G_0$:

1. Solve the linear system. $$V_B=\dfrac{3096}{95}=32.59\text{ V},\quad V_C=\dfrac{996}{95}=10.48\text{ V},\quad V_D=\dfrac{436}{95}=4.59\text{ V}$$ $$R_0 = \dfrac{V_A-V_B}{I_0} = -\dfrac{1358}{295}\ \Omega = -4.60\ \Omega$$
2. Recover the branch currents from these node voltages: $$I_0=\dfrac{V_A-V_B}{R_0}=6.21\text{ A},\quad I_1=\dfrac{V_C-V_B}{R_1}=-2.21\text{ A},\quad I_3=\dfrac{V_D-V_C}{R_3}=-0.59\text{ A}$$ $$I_4=\dfrac{V_A-V_C}{R_4}=-1.62\text{ A},\qquad I_5=\dfrac{V_D}{R_5}=4.59\text{ A}$$ Each of these satisfies the KCL equations of part (a) exactly (e.g. $I_0+I_4 = 6.21-1.62 = 4.59\text{ A}=I_5$✓).
3. Power in R0, using the magnitude of R0 since dissipated power is never negative: $$P_{R_0} = I_0^2\,|R_0| = (6.21)^2(4.60) = \boxed{177.6\text{ W}}$$