Question 1 of 7: DC Ladder Network — KCL, KVL and Source Current

Question 1: DC Ladder Network — KCL, KVL and 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 DC ladder with the seven resistors, source $V_s$, and the measured rung voltage $V_7$ below (node $F$ = reference, $V_E=V_s$).

Find. (a) KCL at $A,B,C$; (b) KVL for the two named loops; (c) the source current $I_s$; (d) the power dissipated in $R_1$.

1. Part (a) — KCL at the three nodes. Using the branch currents drawn in Figure 1 (with $I_3$ entering $A$, $I_s$ entering $B$, $I_4$ entering $C$):$$I_3=I_1+I_2,\qquad I_s=I_3+I_4,\qquad I_4=I_5+I_6.$$
2. Part (b) — KVL for the two loops. Around loop $R_1R_3R_4R_5V_s$ (path $E{\to}A{\to}B{\to}C{\to}F{\to}E$) and loop $R_5R_6R_7$ (path $C{\to}D{\to}F{\to}C$):$$I_4R_4+I_5R_5-I_1R_1-I_3R_3=V_s,\qquad I_6R_6+I_7R_7=I_5R_5.$$
3. Anchor on the measured rung. $V_D=V_7=1$ V, so $I_7=V_7/R_7=0.5$ A, and since $R_6$ is in series with $R_7$ at node $D$, $I_6=I_7=0.5$ A.
4. Walk left to node $C$. $V_C=V_D+I_6R_6=18$ V, hence $I_5=V_C/R_5=1.5$ A and, by KCL at $C$, $I_4=I_5+I_6=2$ A.
5. Reach node $A$ and solve for the node voltage. $V_B=V_C+I_4R_4=40$ V. Writing KCL at $A$ ($\,(V_B-V_A)/R_3=(V_A-V_E)/R_1+(V_A-V_E)/R_2\,$) gives $$V_A=30\text{ V},\qquad I_1=0.667\text{ A},\quad I_2=0.333\text{ A},\quad I_3=I_1+I_2=1\text{ A}.$$
6. Part (c) — source current. KCL at $B$ closes the problem:$$\boxed{I_s=I_3+I_4=3\text{ A}.}$$
7. Part (d) — power in $R_1$. $P_{R_1}=I_1^2R_1=(0.667)^2(3)=1.333$ W.