Topic: Network Theory
For a network shown in figure, Calculate current i2.
- A. 2/3 A
- B. 3/2 A
- C. 1/3 A
- D. None of the above
Answer: A. 2/3 A
Explanation: Applying Nodal Analysis on the right part of the circuit. Let the nodes connecting the left and right halves be $V_1$ (top) and $V_3$ (bottom, reference = 0V). Let $V_4$ be the node between the two $1\Omega$ resistors and $V_5$ be the node on the far right. KCL at $V_4$: $(V_4 - V_1)/1 + (V_4 - 0)/1 + i_2 = 0$. KCL at $V_5$: $(V_5 - V_1)/1 + (V_5 - 0)/1 - i_2 = 0$. Subtracting the two equations gives $2(V_4 - V_5) + 2i_2 = 0 \implies V_5 - V_4 = i_2$. The current $i_2$ flows from $V_4$ through a 2V source and a $2\Omega$ resistor to $V_5$, so $i_2 = (V_4 + 2 - V_5) / 2$. Substituting $V_5 - V_4 = i_2$ into the equation gives $2i_2 = -i_2 + 2 \implies 3i_2 = 2 \implies i_2 = 2/3$ A.