Question 1 of 7: DC Circuit — KCL, KVL, and Branch Quantities

Question 1: DC Circuit — KCL, KVL, and Branch Quantities (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 network of Figure 1: node A carries R₆ (to C), R₄ (to B), and the series R₁–V_s1 branch to the common ground rail D/E; node B carries R₄ (to A), R₂ (to ground), and the series R₅–V_s5 branch (to C); node C carries R₃ and the current source I_s to ground, besides R₆ and the R₅–V_s5 branch. All labelled currents follow the reference arrows printed on the figure.

Find. The KCL/KVL equations for the stated nodes and loops, then V_R2, I₂, and the power dissipated in R₂.

[Figure not reproduced: Figure 1 (redrawn): the bridge network, ground rail D–E common. See the official exam paper.]

Approach. Take the continuous bottom rail (D = E) as the 0 V reference, write nodal KCL at A, B, C using the figure's reference current directions, then solve the resulting 3×3 linear system for the node voltages.

1. (a) KCL at node A (I₁ leaves A down the R₁–V_s1 branch; I₄ enters A from B; I₆ enters A from C):
   $$I_1 = I_4 + I_6$$
2. KCL at node B (I₄ leaves B to A; I₂ leaves B down R₂; I₅ enters B from the V_s5–R₅ branch):
   $$I_4 + I_2 = I_5$$
3. KCL at node C (I₅, I₆, I₃ and I_s all leave C by their printed reference arrows):
   $$I_5 + I_6 + I_3 + I_s = 0$$
4. (b) KVL, loop ABDA (A→B via R₄, B→D via R₂, D→A via V_s1 then R₁):
   $$I_1 R_1 - I_2 R_2 + I_4 R_4 + V_{s1} = 0$$
5. KVL, loop ABCA (A→B via R₄, B→C via R₅ then V_s5, C→A via R₆):
   $$I_4 R_4 + I_5 R_5 - I_6 R_6 - V_{s5} = 0$$
6. Solve the nodal system. With D=E=0 V, each branch current is written from Ohm's law (the R₁–V_s1 series branch gives $I_1=(V_A-V_{s1})/R_1$; the R₅–V_s5 branch gives $I_5=(V_C+V_{s5}-V_B)/R_5$). Substituting into the three KCL equations and solving simultaneously gives:
   $$V_A = 25\text{ V}, \qquad V_B = 10\text{ V}, \qquad V_C = -5\text{ V}$$
7. (c) Voltage across R₂. R₂ runs directly from node B to ground, so $V_{R2}=V_B$:
   $$\boxed{V_{R2} = 10.0\text{ V}}$$$$I_2 = \frac{V_{R2}}{R_2} = \frac{10}{2} = \boxed{5.0\text{ A}}$$
8. (d) Current I₂ and power dissipated in R₂.
   $$P_{R2} = I_2^2 R_2 = (5.0)^2(2) = \boxed{50.0\text{ W}}$$Substituting back, the three KCL equations and both KVL loop equations all evaluate to exactly zero at this solution.