The Correct Answer and Explanation is:
The correct answers are v = 10 V and ix = -2 A.
To determine the values of the voltage v and the current ix in the given circuit, we can utilize Kirchhoff’s Voltage Law (KVL). KVL states that the algebraic sum of all the voltages around any closed path, or loop, in a circuit is equal to zero. This circuit contains two distinct loops that can be analyzed independently to find the unknown variables.
First, let’s analyze the left loop, which consists of the 12 V independent voltage source, the 12 Ω resistor, and the central 2 V element. We will sum the voltages by traversing the loop in a clockwise direction, starting from the bottom-left corner.
- Moving upward through the 12 V source, we encounter a voltage rise, so we add +12 V.
- Moving from left to right across the 12 Ω resistor, we see a voltage drop v, as indicated by the polarity signs (+ to -). This is represented as -v.
- Moving downward through the central element, we go from the positive to the negative terminal, which is a voltage drop of –2 V.
Setting the sum of these voltages to zero according to KVL:
12 V – v – 2 V = 0
10 V – v = 0
Solving for v, we get:
v = 10 V
Next, we apply KVL to the right loop, which contains the central 2 V element, the 8 V source, and the current-controlled dependent voltage source 3ix. We will again traverse clockwise, starting from the bottom of the loop.
- Moving upward through the dependent source 3ix, we go from the negative to the positive terminal, resulting in a voltage rise of +3ix.
- Moving across the 8 V source from right to left, we go from negative to positive, which is a voltage rise of +8 V.
- Moving downward through the central 2 V element, we have a voltage drop of –2 V.
The KVL equation for the right loop is:
3ix + 8 V – 2 V = 0
3ix + 6 V = 0
Solving for ix:
3ix = -6 V
ix = -2 A
The negative result for ix signifies that the current’s actual direction is opposite to the one indicated by the arrow in the diagram. Thus, a current of 2 A flows upwards through the central branch.
