Q5 (a)

Node-Voltage Analysis · Circuit with Dependent Sources

Problem. In the circuit shown below, find the value of v using the node-voltage method.

Fig. 1 — Annotated circuit with chosen nodes V₁ (top-left), V₂ (top-right), V₃ (middle-left), V₄ (middle-right). Ground = bottom rail.

Step 1 · Choose Reference and Define Node Voltages

Take the bottom rail as the reference (ground). Four essential node voltages remain:

Node	Location
$V_1$	Top-left rail (top of left 4 Ω, − terminal of 16 V source, left terminal of 8i_x source)
$V_2$	Top-right rail (right terminal of 8i_x source, right end of 2 Ω-with-$v_y$, top of dependent current source)
$V_3$	Middle-left (+ of 16 V, top of 2 Ω-with-$i_x$, left of 5 A source)
$V_4$	Middle-right (right of 5 A, top of 4 Ω-with-$v$, left of 2 Ω-with-$v_y$)

Controlling variables expressed in terms of node voltages:

$$i_x = \frac{0 - V_3}{2} = -\frac{V_3}{2} \qquad \text{(current flowing UP through left 2 Ω)}$$ $$v_y = V_4 - V_2 \qquad \text{(}+ \text{ on left of the 2 Ω-}v_y\text{ resistor)}$$ $$v = V_4 \qquad \text{(}+ \text{ on top of right 4 Ω, − at ground)}$$

Step 2 · Constraint Equations from Voltage Sources

16 V independent source (between V₃ and V₁)

+ at bottom (V₃), − at top (V₁):

$$V_3 - V_1 = 16 \;\;\Rightarrow\;\; V_1 = V_3 - 16 \quad(\text{A})$$

8i_x dependent voltage source (between V₁ and V₂)

− on left (V₁), + on right (V₂):

$$V_2 - V_1 = 8 i_x = 8\!\left(-\frac{V_3}{2}\right) = -4 V_3$$ $$V_2 = V_1 - 4V_3 = (V_3 - 16) - 4V_3 = -3V_3 - 16 \quad(\text{B})$$

Since V₁ and V₃ are tied by the 16 V source, and V₁ and V₂ are tied by the 8i_x source, the three nodes {V₁, V₂, V₃} form a single supernode. Write one KCL for the whole supernode plus the two constraint equations above.

Step 3 · KCL at Node V₄

Currents leaving V₄:

Through 4 Ω-with-$v$ to ground: $\dfrac{V_4}{4}$
Through 2 Ω-with-$v_y$ to V₂: $\dfrac{V_4 - V_2}{2}$
5 A current source enters V₄ from the left ⇒ contributes $-5$ to "leaving"

$$\frac{V_4}{4} + \frac{V_4 - V_2}{2} - 5 = 0$$

Multiply by 4:

$$V_4 + 2(V_4 - V_2) = 20 \;\;\Rightarrow\;\; 3V_4 - 2V_2 = 20 \quad(\text{C})$$

Step 4 · KCL for Supernode {V₁, V₂, V₃}

Sum of currents leaving the supernode through external branches:

From V₁ down through 4 Ω(left) to ground: $\dfrac{V_1}{4}$
From V₃ down through 2 Ω-with-$i_x$ to ground: $\dfrac{V_3}{2}$
From V₃ rightward through 5 A source to V₄: $+5$ (leaves V₃)
From V₂ leftward through 2 Ω-with-$v_y$ to V₄: $\dfrac{V_2 - V_4}{2}$
From V₂ down through (3/4)$v_y$ source to ground: $\dfrac{3}{4} v_y$

$$\frac{V_1}{4} + \frac{V_3}{2} + 5 + \frac{V_2 - V_4}{2} + \frac{3}{4}v_y = 0$$

Multiply by 4:

$$V_1 + 2V_3 + 20 + 2(V_2 - V_4) + 3v_y = 0$$

Substitute $v_y = V_4 - V_2$:

$$V_1 + 2V_3 + 20 + 2V_2 - 2V_4 + 3V_4 - 3V_2 = 0$$ $$V_1 + 2V_3 - V_2 + V_4 + 20 = 0 \quad(\text{D})$$

Step 5 · Solve the System

Substitute (A) and (B) into (D):

$$(V_3 - 16) + 2V_3 - (-3V_3 - 16) + V_4 + 20 = 0$$ $$V_3 - 16 + 2V_3 + 3V_3 + 16 + V_4 + 20 = 0$$ $$6V_3 + V_4 + 20 = 0 \;\;\Rightarrow\;\; V_4 = -6V_3 - 20 \quad(\text{E})$$

Substitute (B) into (C):

$$3V_4 - 2(-3V_3 - 16) = 20$$ $$3V_4 + 6V_3 + 32 = 20 \;\;\Rightarrow\;\; V_4 = \frac{-12 - 6V_3}{3} = -2V_3 - 4 \quad(\text{F})$$

Equate (E) and (F):

$$-6V_3 - 20 = -2V_3 - 4$$ $$-4V_3 = 16 \;\;\Rightarrow\;\; V_3 = -4\ \text{V}$$

Back-substitute:

$$V_4 = -2(-4) - 4 = 4\ \text{V}$$ $$V_1 = V_3 - 16 = -20\ \text{V}, \quad V_2 = -3(-4) - 16 = -4\ \text{V}$$ $$i_x = -V_3/2 = 2\ \text{A}, \quad v_y = V_4 - V_2 = 4 - (-4) = 8\ \text{V}$$

Required answer

$$\boxed{\,v \;=\; V_4 \;=\; 4\ \text{V}\,}$$

Step 6 · Verification

Constraint (B): $V_2 - V_1 = -4 - (-20) = 16,\ \ 8 i_x = 8(2) = 16$ ✓

Constraint (A): $V_3 - V_1 = -4 - (-20) = 16$ ✓

KCL at V₄:

$$\frac{V_4}{4} + \frac{V_4 - V_2}{2} - 5 = \frac{4}{4} + \frac{4-(-4)}{2} - 5 = 1 + 4 - 5 = 0 \;\;\checkmark$$

KCL at ground (total current into ground from all branches):

From 4 Ω(left): $V_1/4 = -5$ A into ground
From 2 Ω-$i_x$: $V_3/2 = -2$ A into ground
From 4 Ω-$v$: $V_4/4 = +1$ A into ground
From (3/4)$v_y$ source: $+\tfrac34(8) = +6$ A into ground

$$-5 -2 + 1 + 6 = 0 \;\;\checkmark$$

Summary of Node Voltages

Symbol	Meaning	Value
$V_1$	Top-left node	−20 V
$V_2$	Top-right node	−4 V
$V_3$	Middle-left node	−4 V
$V_4$	Middle-right node	4 V (= v)
$i_x$	Current through left 2 Ω (upward)	2 A
$v_y$	Voltage across right 2 Ω	8 V

Required quantity

v = 4 V

Controlling current

i_x = 2 A

Controlling voltage

v_y = 8 V