Calculate ix in the circuit diagram given below: [Figure description: A 10 V voltage source (+ on top) at the left in series with a 1 Ω resistor carrying current ix (arrow to the right). At the node after the 1 Ω resistor, a 4 Ω resistor goes vertically down to the bottom rail. A dependent current source 2ix (arrow pointing right) is connected between that node and the right-hand node, in parallel with a 2 Ω resistor drawn above it (top branch). The right-hand node connects through a 2 Ω resistor down to the bottom rail.]
- A. 2 A
- B. 10 A
- C. 5 A
- D. None of the above
Answer: B. 10 A
Explanation: Applying KCL at the intermediate nodes with the dependent source 2ix reveals node voltage V1 = 0, giving ix = (10 − 0)/1 = 10 A.