Calculate i0 for the circuit diagram given below : [Figure description: A 30 V voltage source (+ at top) on the left drives a circuit. A 2 Ω resistor and a 3 Ω resistor are in series in the middle vertical branch (2 Ω on top, 3 Ω below). On the right is another vertical branch with a 2 Ω resistor on top and a 6 Ω resistor below it in series. The junction between the 2 Ω/3 Ω and the junction between 2 Ω/6 Ω are connected by a horizontal wire carrying current i0 (arrow pointing from the middle branch to the right branch).]
- A. 5/3 A
- B. -5/3 A
- C. 10/3 A
- D. -10/3 A
Answer: B. -5/3 A
Explanation: The wire joins the two mid-points: top 2 Ω ∥ 2 Ω = 1 Ω, bottom 3 Ω ∥ 6 Ω = 2 Ω, so VN = 20 V. Current entering middle junction from top is 5 A, leaving at bottom is 20/3 A, so wire supplies i0 = 5 − 20/3 = −5/3 A.