Q6 (a) — [8 marks]

Nodal Analysis · Current through the 4 Ω resistor between A and B

Problem. Using nodal analysis, determine the current through the 4 Ω resistor connected between terminals A and B for the network shown.

Fig. 1 — Annotated network. V_m is across the TOP 4 Ω (between nodes C and D, + on left); V_n is across the 1 Ω. The 0.2 V_m source has + on left (B-side) and − on right (D-side).

Step 1 · Reference, Nodes, and Controlling Variables

Take the bottom rail as ground. Four essential node voltages: $V_C, V_A, V_B, V_D$.

The 24 V source fixes:

$$V_C = 24\ \text{V}$$

Controlling variables (+ terminals as marked in the figure):

$$V_n = V_C - V_A = 24 - V_A \qquad \text{(across the 1 Ω)}$$ $$V_m = V_C - V_D = 24 - V_D \qquad \text{(across the TOP 4 Ω)}$$

Step 2 · Voltage-Source Constraint (Supernode B–D)

The dependent voltage source 0.2 V_m sits between B and D with + on left (at B), − on right (at D):

$$V_B - V_D = 0.2\,V_m = 0.2\,(24 - V_D)$$ $$5(V_B - V_D) = 24 - V_D \;\Rightarrow\; 5V_B - 4V_D = 24$$ $$\boxed{\,V_B = \tfrac{4V_D + 24}{5}\,} \quad(\text{A})$$

Step 3 · KCL at Node A

Sum of currents leaving A (the 10 A source enters A from below):

$$\frac{V_A - 24}{1} + \frac{V_A - V_B}{4} - 10 = 0$$

Multiply by 4:

$$4V_A - 96 + V_A - V_B - 40 = 0$$ $$\boxed{\,5V_A - V_B = 136\,}\quad(\text{B})$$

Step 4 · KCL at Supernode {B, D}

Currents leaving the supernode:

From B back to A through middle 4 Ω: $\dfrac{V_B - V_A}{4}$
From B to ground through the 0.4 V_n current source (arrow UP into B): $-\,0.4\,V_n$
From D back to C through the top 4 Ω: $\dfrac{V_D - 24}{4}$
From D to ground through the 2 Ω: $\dfrac{V_D}{2}$

$$\frac{V_B - V_A}{4} - 0.4\,(24 - V_A) + \frac{V_D - 24}{4} + \frac{V_D}{2} = 0$$

Multiply by 20:

$$5(V_B - V_A) - 8(24 - V_A) + 5(V_D - 24) + 10V_D = 0$$ $$5V_B - 5V_A - 192 + 8V_A + 5V_D - 120 + 10V_D = 0$$ $$\boxed{\,3V_A + 5V_B + 15V_D = 312\,}\quad(\text{C})$$

Step 5 · Solve the System

From (B): $V_B = 5V_A - 136$. Substitute into (A):

$$5V_A - 136 = \frac{4V_D + 24}{5}$$ $$25V_A - 680 = 4V_D + 24$$ $$25V_A - 4V_D = 704 \quad(\text{i})$$

Substitute $V_B = 5V_A - 136$ into (C):

$$3V_A + 5(5V_A - 136) + 15V_D = 312$$ $$28V_A + 15V_D = 992 \quad(\text{ii})$$

From (i): $V_A = \dfrac{704 + 4V_D}{25}$. Substitute into (ii):

$$28 \cdot \frac{704 + 4V_D}{25} + 15V_D = 992$$ $$28(704 + 4V_D) + 375V_D = 24\,800$$ $$19\,712 + 112V_D + 375V_D = 24\,800$$ $$487\,V_D = 5\,088 \;\Rightarrow\; V_D = \frac{5088}{487} \approx 10.45\ \text{V}$$

Back-substitute:

$$V_A = \frac{704 + 4(5088/487)}{25} = \frac{14\,528}{487} \approx 29.83\ \text{V}$$ $$V_B = 5V_A - 136 = \frac{6408}{487} \approx 13.16\ \text{V}$$

Step 6 · Current through the 4 Ω between A and B

Take the reference direction from A → B:

$$I_{AB} = \frac{V_A - V_B}{4} = \frac{(14\,528 - 6\,408)/487}{4} = \frac{8120}{4 \cdot 487} = \frac{2030}{487}$$

Current through 4 Ω (A → B)

$$\boxed{\,I_{AB} \;=\; \dfrac{2030}{487} \;\approx\; 4.17\ \text{A}\,}$$

The positive sign confirms conventional current flows from A to B.

Step 7 · Verification

$V_n = 24 - V_A = \tfrac{11\,688 - 14\,528}{487} = -\tfrac{2840}{487} \approx -5.83$ V

$V_m = 24 - V_D = \tfrac{11\,688 - 5088}{487} = \tfrac{6600}{487} \approx 13.55$ V

Source constraint:

$$V_B - V_D = \frac{6408 - 5088}{487} = \frac{1320}{487}$$ $$0.2\,V_m = 0.2 \cdot \frac{6600}{487} = \frac{1320}{487} \;\checkmark$$

KCL at A:

$$5V_A - V_B = \frac{5 \cdot 14\,528 - 6408}{487} = \frac{66\,232}{487} = 136 \;\checkmark$$

KCL at supernode {B, D}:

$$3V_A + 5V_B + 15V_D = \frac{3 \cdot 14528 + 5 \cdot 6408 + 15 \cdot 5088}{487}$$ $$= \frac{43\,584 + 32\,040 + 76\,320}{487} = \frac{151\,944}{487} = 312 \;\checkmark$$

Summary of Node Voltages

Symbol	Meaning	Value
$V_C$	Top-left node (top of 24 V)	24 V
$V_A$	Node A	14528/487 ≈ 29.83 V
$V_B$	Node B	6408/487 ≈ 13.16 V
$V_D$	Top-right node (top of 2 Ω)	5088/487 ≈ 10.45 V
$V_n$	Voltage across 1 Ω	−2840/487 ≈ −5.83 V
$V_m$	Voltage across TOP 4 Ω	6600/487 ≈ 13.55 V
$I_{AB}$	Current through middle 4 Ω (A→B)	2030/487 ≈ 4.17 A

Required answer

I_AB ≈ 4.17 A

Controlling V_n

−5.83 V

Controlling V_m

13.55 V