Concept Foundation — Why Nodal Analysis Is Ideal Here

Every element in Fig. Q4 is connected between the same top node and the same bottom reference node. That means every branch has the same voltage. Instead of writing several loop equations, define one node voltage V and express each branch current in terms of V.

The circuit also contains a dependent current source . A dependent source is not assigned an arbitrary current; its value must always be written from its controlling variable. Here the controlling current is i₀ through the 2 Ω resistor.

Step 1 — Define the Node Voltage and Controlling Current

Take the bottom conductor as reference (0 V) and the top conductor as V volts. The current i₀ is shown downward through the 2 Ω resistor, so by Ohm’s law:

i_0 = V/2

Step 2 — Express the Dependent Source

The dependent current source arrow is downward and its value is i₀/4. Therefore:

i_d = i_0/4 = (V/2)/4 = V/8

Step 3 — Express the 8 Ω Branch Current

i_8Ω = V/8

Step 4 — Apply KCL at the Top Node

The 6 A source injects current upward into the top node. The three right-hand branch currents leave the top node downward. Hence incoming current = outgoing current:

6 = V/2 + V/8 + V/8

6 = 4V/8 + V/8 + V/8

6 = 6V/8 = 3V/4

V = 8 V

Step 5 — Calculate Power in the 8 Ω Resistor

P = V²/R

P_8Ω = 8²/8 = 8 W

KCL Cross-Check

i_0 = 8/2 = 4 A

i_d = i_0/4 = 1 A

i_8Ω = 8/8 = 1 A

4 + 1 + 1 = 6 A ✓

The branch currents exactly equal the 6 A source current, confirming the node equation and the final answer.

Concept Extension

Dependent sources remain active during normal circuit analysis. Their controlling relationship is part of the circuit physics. The fastest safe workflow is: choose node voltage → write controlling variable → write dependent source → apply KCL → calculate the requested power.

Exam Strategy

When many branches share the same two nodes, first test whether a single-node KCL equation can solve the problem. It is often much faster than mesh analysis.

Quick Trick

Same two nodes → one V. i₀ = V/2, dependent = V/8, 8 Ω current = V/8. KCL → V = 8 V → P = 8 W.

Why Other Options Are Wrong

• B — 16 W: would require |V| = √(16×8) ≈ 11.31 V, which does not satisfy KCL.
• C — 32 W: would require |V| = 16 V, inconsistent with the 6 A source and branch currents.
• D — 64 W: would require |V| ≈ 22.63 V, also impossible from the KCL relation.

Common Mistake

Ignoring the arrow direction, forgetting i₀ = V/2, or using i₀/4 as a constant current independent of V.

Concept Foundation — Why Nodal Analysis Is Ideal Here

Every element in Fig. Q4 is connected between the same top node and the same bottom reference node. That means every branch has the same voltage. Instead of writing several loop equations, define one node voltage V and express each branch current in terms of V.

The circuit also contains a dependent current source . A dependent source is not assigned an arbitrary current; its value must always be written from its controlling variable. Here the controlling current is i₀ through the 2 Ω resistor.

Step 1 — Define the Node Voltage and Controlling Current

Take the bottom conductor as reference (0 V) and the top conductor as V volts. The current i₀ is shown downward through the 2 Ω resistor, so by Ohm’s law:

i_0 = V/2

Step 2 — Express the Dependent Source

The dependent current source arrow is downward and its value is i₀/4. Therefore:

i_d = i_0/4 = (V/2)/4 = V/8

Step 3 — Express the 8 Ω Branch Current

i_8Ω = V/8

Step 4 — Apply KCL at the Top Node

The 6 A source injects current upward into the top node. The three right-hand branch currents leave the top node downward. Hence incoming current = outgoing current:

6 = V/2 + V/8 + V/8

6 = 4V/8 + V/8 + V/8

6 = 6V/8 = 3V/4

V = 8 V

Step 5 — Calculate Power in the 8 Ω Resistor

P = V²/R

P_8Ω = 8²/8 = 8 W

KCL Cross-Check

i_0 = 8/2 = 4 A

i_d = i_0/4 = 1 A

i_8Ω = 8/8 = 1 A

4 + 1 + 1 = 6 A ✓

The branch currents exactly equal the 6 A source current, confirming the node equation and the final answer.

Concept Extension

Dependent sources remain active during normal circuit analysis. Their controlling relationship is part of the circuit physics. The fastest safe workflow is: choose node voltage → write controlling variable → write dependent source → apply KCL → calculate the requested power.

Exam Strategy

When many branches share the same two nodes, first test whether a single-node KCL equation can solve the problem. It is often much faster than mesh analysis.

Quick Trick

Same two nodes → one V. i₀ = V/2, dependent = V/8, 8 Ω current = V/8. KCL → V = 8 V → P = 8 W.

Why Other Options Are Wrong

• B — 16 W: would require |V| = √(16×8) ≈ 11.31 V, which does not satisfy KCL.
• C — 32 W: would require |V| = 16 V, inconsistent with the 6 A source and branch currents.
• D — 64 W: would require |V| ≈ 22.63 V, also impossible from the KCL relation.

Common Mistake

Ignoring the arrow direction, forgetting i₀ = V/2, or using i₀/4 as a constant current independent of V.

Concept Foundation — Why Nodal Analysis Is Ideal Here

Every element in Fig. Q4 is connected between the same top node and the same bottom reference node. That means every branch has the same voltage. Instead of writing several loop equations, define one node voltage V and express each branch current in terms of V.

The circuit also contains a dependent current source . A dependent source is not assigned an arbitrary current; its value must always be written from its controlling variable. Here the controlling current is i₀ through the 2 Ω resistor.

Step 1 — Define the Node Voltage and Controlling Current

Take the bottom conductor as reference (0 V) and the top conductor as V volts. The current i₀ is shown downward through the 2 Ω resistor, so by Ohm’s law:

i_0 = V/2

Step 2 — Express the Dependent Source

The dependent current source arrow is downward and its value is i₀/4. Therefore:

i_d = i_0/4 = (V/2)/4 = V/8

Step 3 — Express the 8 Ω Branch Current

i_8Ω = V/8

Step 4 — Apply KCL at the Top Node

The 6 A source injects current upward into the top node. The three right-hand branch currents leave the top node downward. Hence incoming current = outgoing current:

6 = V/2 + V/8 + V/8

6 = 4V/8 + V/8 + V/8

6 = 6V/8 = 3V/4

V = 8 V

Step 5 — Calculate Power in the 8 Ω Resistor

P = V²/R

P_8Ω = 8²/8 = 8 W

KCL Cross-Check

i_0 = 8/2 = 4 A

i_d = i_0/4 = 1 A

i_8Ω = 8/8 = 1 A

4 + 1 + 1 = 6 A ✓

The branch currents exactly equal the 6 A source current, confirming the node equation and the final answer.

Concept Extension

Dependent sources remain active during normal circuit analysis. Their controlling relationship is part of the circuit physics. The fastest safe workflow is: choose node voltage → write controlling variable → write dependent source → apply KCL → calculate the requested power.

Exam Strategy

When many branches share the same two nodes, first test whether a single-node KCL equation can solve the problem. It is often much faster than mesh analysis.

Quick Trick

Same two nodes → one V. i₀ = V/2, dependent = V/8, 8 Ω current = V/8. KCL → V = 8 V → P = 8 W.

Why Other Options Are Wrong

• B — 16 W: would require |V| = √(16×8) ≈ 11.31 V, which does not satisfy KCL.
• C — 32 W: would require |V| = 16 V, inconsistent with the 6 A source and branch currents.
• D — 64 W: would require |V| ≈ 22.63 V, also impossible from the KCL relation.

Common Mistake

Ignoring the arrow direction, forgetting i₀ = V/2, or using i₀/4 as a constant current independent of V.