Correct Answer
A — 8 WThe common node voltage is 8 V, so the 8 Ω resistor dissipates 8 W.
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:
Step 2 — Express the Dependent Source
The dependent current source arrow is downward and its value is i₀/4. Therefore:
id = i0/4 = (V/2)/4 = V/8
Step 3 — Express the 8 Ω Branch Current
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
KCL Cross-Check
i0 = 8/2 = 4 A
id = i0/4 = 1 A
i8Ω = 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.
