Given. The network of Figure 1: top rail nodes A-B-C-D joined by $R_3,R_4,R_6$; a parallel pair $R_1\parallel R_2$ drops from A to node E; the DC source $V_s$ sits in the bottom rail between E and F (positive at F); the current source $I_s$ bridges F up to B; $R_5$ drops from C to F; and $R_7$ (carrying the observed $V_7$) drops from D to F.
Given data
Quantity
Value
$R_1,R_2,R_3,R_4$
$3,\ 6,\ 10,\ 11\ \Omega$
$R_5,R_6,R_7$
$12,\ 34,\ 2\ \Omega$
$V_s$
$28\text{ V}$
$V_7$ (observed)
$1\text{ V}$
Find. The three node-A/B/C KCL equations, the two named-loop KVL equations, $P_7$, and the current-source value $I_s$.
Figure 1: DC network for Question 1 (R1=3Ω R2=6Ω R3=10Ω R4=11Ω R5=12Ω R6=34Ω R7=2Ω, Vs=28V, observed V7=1V)
Approach. $R_6$-$R_7$ is a series branch carrying the same current as the observed $V_7$, so $I_7$ is known immediately; propagate that current backward through the ladder (C, then B, then A) using Ohm's law and KCL at each node, then close with KCL at B to isolate $I_s$.
a) KCL at nodes A, B, C (currents defined per the arrows in Figure 1: $I_1,I_2$ down A→E, $I_3$ from B→A, $I_4$ from B→C, $I_5$ down C→F, $I_6=I_7$ through $R_6$-$R_7$ from C→D→F, and $I_s$ from F→B):$$\text{Node A: } I_3 = I_1+I_2 \qquad \text{Node B: } I_s = I_3+I_4 \qquad \text{Node C: } I_4=I_5+I_6$$These three equations are the answer to (a); they are also used numerically below.
b) KVL for loop $R_1R_3R_4R_5V_s$ (traversing E→A→B→C→F→E): summing drops against the assumed current directions gives $$V_s = I_1R_1 + I_3R_3 - I_4R_4 - I_5R_5$$ KVL for loop $R_5R_6R_7$ (a source-free loop between the two parallel C-F branches): $$I_5R_5 = I_6R_6 + I_7R_7$$ These hold for any valid current solution; part (d) below finds the numbers that satisfy them.
c) Power in $R_7$. $R_6$ and $R_7$ are in series (both carry $I_7$), and $V_7$ is given directly, so $$I_7 = \dfrac{V_7}{R_7} = \dfrac{1}{2} = 0.5\text{ A}, \qquad P_7 = V_7 I_7 = (1)(0.5) = \boxed{0.5\text{ W}}$$
d) Propagate to find $I_s$. Since $R_5$ and the $R_6$-$R_7$ series pair both span the same two nodes (C and F), they share the same voltage: $$V_{CF} = I_6(R_6+R_7) = (0.5)(34+2) = 18\text{ V} \;\Rightarrow\; I_5 = \dfrac{V_{CF}}{R_5} = \dfrac{18}{12}=1.5\text{ A}$$ Node C gives $I_4 = I_5+I_6 = 1.5+0.5 = 2.0\text{ A}$. Taking $V_E=0$ (ground) and $V_F=V_s=28\text{ V}$: $$V_C = V_F + V_{CF} = 28+18 = 46\text{ V}, \qquad V_B = V_C + I_4R_4 = 46+(2.0)(11) = 68\text{ V}$$ Node A has only $R_1,R_2,R_3$ attached, so $\dfrac{V_A}{R_1}+\dfrac{V_A}{R_2}+\dfrac{V_A-V_B}{R_3}=0$ (current leaving A sums to zero); solving, $$V_A = \dfrac{V_B}{1+R_3(1/R_1+1/R_2)} = \dfrac{68}{1+10(1/3+1/6)} = 11.33\text{ V}$$ so $I_1=V_A/R_1=3.78\text{ A}$, $I_2=V_A/R_2=1.89\text{ A}$, and $I_3=(V_B-V_A)/R_3=(68-11.33)/10=5.67\text{ A}$. Finally, node B's KCL from part (a) gives $$I_s = I_3+I_4 = 5.67+2.0 = \boxed{7.67\text{ A}}$$ (flowing F→B, i.e. in the direction the arrow in Figure 1 shows).
Final Results -- Question 1
Quantity
Value
Node-A KCL
$I_3=I_1+I_2$
Node-B KCL
$I_s=I_3+I_4$
Node-C KCL
$I_4=I_5+I_6$
Loop $R_1R_3R_4R_5V_s$
$V_s=I_1R_1+I_3R_3-I_4R_4-I_5R_5$
Loop $R_5R_6R_7$
$I_5R_5=I_6R_6+I_7R_7$
$P_7$
$0.5\text{ W}$
$I_s$
$7.67\text{ A}$ (F→B)
The recovered branch currents ($I_1=3.78\text{ A}$, $I_2=1.89\text{ A}$, $I_3=5.67\text{ A}$, $I_4=2.0\text{ A}$, $I_5=1.5\text{ A}$, $I_6=I_7=0.5\text{ A}$) satisfy every KCL node and both KVL loops simultaneously, confirming the observed $V_7=1\text{ V}$ is internally consistent with the given resistor network -- this cross-check is the main value of writing the symbolic equations in (a)/(b) before solving numerically in (d).
