Question 1 of 7: DC Bridge Network — KCL, KVL, and the Unknown Source Current

Question 1: DC Bridge Network — KCL, KVL, and the 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. A four–node ladder (bridge) network driven by a voltage source Vs and an unknown current source Is. The voltage across R7 is measured as V7 = 1 V — this single observation, together with the source Vs, closes the problem for the unknown Is.

Find. (a) KCL at A, B, C; (b) KVL for the two named loops; (c) the power dissipated in R7; (d) the value of the current source Is.

[Figure not reproduced: Figure 1 — DC bridge/ladder network. Node F (bottom-right rail) is taken as the reference (0 V); the source fixes node E at +28 V. Branch-current reference arrows as printed. See the official exam paper.]

Approach. Choose node F as reference. The source Vs fixes node E; the observation V7 = 1 V fixes node D. Apply KCL at D, C, B, A in turn (each yields the next node voltage directly because the ladder is singly connected), then read Is from KCL at node B.

a) KCL equations (currents leaving each node = 0)

1. Node A — R1 and R2 drain downward to E while R3 delivers I3 from B: $$I_1 + I_2 = I_3$$
2. Node B — the source Is feeds the node; R3 and R4 carry it away: $$I_s = I_3 + I_4$$
3. Node C — R4 delivers current that splits into R5 and R6: $$I_4 = I_5 + I_6$$

b) KVL equations (sum of drops around each loop = 0)

1. Loop R1–R3–R4–R5–Vs — traversing E→A→B→C→F→E: $$V_s + I_1 R_1 + I_3 R_3 = I_4 R_4 + I_5 R_5$$
2. Loop R5–R6–R7 — the outer right mesh C→F→D→C: $$I_5 R_5 = I_6 (R_6 + R_7)$$

c) Power in R7 and d) the source current Is

With F = 0 V and E = +28 V, the observation gives node D directly, since V7 is the drop across R7 to the reference rail:

1. Node D from the observation. $V_D = V_7 = 1\text{ V}$, so the R7 current is $I_6 = V_7/R_7 = 1/2 = 0.5\text{ A}$.
2. Node C (KCL at D). $\dfrac{V_D-V_C}{R_6}+\dfrac{V_D}{R_7}=0 \Rightarrow V_C = V_D + R_6\dfrac{V_D}{R_7} = 1 + 34(0.5) = 18\text{ V}$.
3. Node B (KCL at C). $\dfrac{V_C-V_B}{R_4}+\dfrac{V_C}{R_5}+\dfrac{V_C-V_D}{R_6}=0 \Rightarrow V_B = 40\text{ V}$.
4. Node A (KCL at A). $V_A\!\left(\tfrac1{R_1}+\tfrac1{R_2}+\tfrac1{R_3}\right)=28\!\left(\tfrac1{R_1}+\tfrac1{R_2}\right)+\tfrac{V_B}{R_3}\Rightarrow V_A = 30\text{ V}$.
5. Branch currents. $I_1=\tfrac{V_A-28}{R_1}=0.667$, $I_2=0.333$, $I_3=\tfrac{V_B-V_A}{R_3}=1$, $I_4=\tfrac{V_B-V_C}{R_4}=2$, $I_5=\tfrac{V_C}{R_5}=1.5$, $I_6=0.5$ (all in A).
6. Power in R7 (part c). $P_{7}=\dfrac{V_7^{2}}{R_7}=\dfrac{1^2}{2}$, i.e. $$\boxed{P_{R_7}=0.5\text{ W}}$$
7. Source current (part d), KCL at B. $I_s=I_3+I_4=1+2$, so $$\boxed{I_s = 3\text{ A}}$$

Every node balances: at A, $I_3=I_1+I_2=1$ A; at B, $I_s=I_3+I_4=3$ A; at C, $I_4=I_5+I_6=2$ A — confirming the KCL set of part (a). The KVL loops close as well: $V_s+I_1R_1+I_3R_3=28+2+10=40=I_4R_4+I_5R_5=22+18$, and $I_5R_5=18=I_6(R_6+R_7)=0.5(36)$.