Find the value of i using the above circuit by making use of the superposition theorem.
1.5 A
Node analysis settles this quickly, and the dependent source is the reason superposition must be applied with care.
Step 1 — find V at node 2. Write KCL there, remembering that the 2 A source pulls a fixed current out towards node 3:
\(\dfrac{8-V}{3}=\dfrac{V}{3}+2\)
\(8-V=V+6\quad\Rightarrow\quad 2V=2\quad\Rightarrow\quad V=1\ \text{V}\)
Step 2 — the dependent source now has a value.
\(\dfrac{V}{2}=\dfrac{1}{2}=0.5\ \text{A}\)
Step 3 — apply KCL at node 3. The 2 A arrives from the independent source, the controlled source removes 0.5 A, and the remainder leaves through R3:
\(i=2-0.5=1.5\ \text{A}\)
which is option 2.
| Contribution at node 3 | Current |
|---|---|
| Independent 2 A source | +2 A |
| Dependent source V/2 | −0.5 A |
| Through R3 | 1.5 A |
The rule the question is really testing. Superposition may be applied only to the independent sources; a dependent source is never deactivated, because its value is not an input to the circuit but a response to it. Killing it would destroy the very constraint that defines the network. So the correct procedure is: consider the 8 V source alone with the 2 A source open-circuited, then the 2 A source alone with the 8 V source short-circuited, keeping the controlled source live in both passes, and add the two answers.
A check on the result. R3 then drops \(1.5\times2=3\ \text{V}\), a modest voltage consistent with a 1 V node and a half-amp controlled source — and the answer is comfortably less than the 2 A being injected, exactly as the subtracting controlled source requires. Had the controlled source been drawn feeding into the node instead, the answer would have been 2.5 A, which is not offered; the option set therefore confirms the sense of the arrow in the figure.
Why controlled sources matter : they are how transistors and op-amps enter linear analysis, and their presence also destroys reciprocity, so a network containing one cannot be checked with the reciprocity theorem.
Hence, the current is 1.5 A.
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