Determine the value of IX (the current source) in the circuit shown.

Question

Determine the value of IX (the current source) in the circuit shown.

BlackTom AI · Accepted Answer

Let me restate the problem and examine what is fixed vs what is variable in the circuit.
- The leftmost element is a current source Ix delivering current into the top node from the bottom node.
- The middle branch is a 5 Ω resistor directly between the top and bottom nodes.
- The rightmost branch is a 10 Ω resistor in series with a 30 V source (with the 30 V source oriented so that its positive terminal is toward the top of the branch and the bottom of the source connects to the bottom node). This means the node between the 10 Ω resistor and the 30 V source sits at +30 V with respect to the bottom node.
- Let the top node voltage be V (with respect to the bottom node).

For this network, the currents flowing from the top node down through each branch are:
- Through the 5 Ω resistor: I5 = V / 5
- Through the right branch (10 Ω in series with the 30 V source): the current is determined by the voltage across the 10 Ω resistor, which is (V − 30) (since the node after the 10 Ω is fixed at +30 V). So I10 = (V − 30) / 10
The current source Ix injects current into the top node, and by KCL the sum of currents leaving the top node through the passive branches equals the current supplied by Ix:
 Ix = I5 + I10 = V/5 + (V − 30)/10
This gives a relation between Ix and V, but it does not fix a unique value of Ix by itself because V can adjust.

Now, evaluate each answer option by computing the corresponding V and branch currents, to see if each option yields a consistent circuit:

Option A: Ix = 12 A
- Solve 12 = V/5 + (V − 30)/10 = (2V + V − 30)/10 = (3V − 30)/10
  => 120 = 3V − 30 => 3V = 150 => V = 50 V
- Then I5 = V/5 = 50/5 = 10 A downward
- I10 = (V − 30)/10 = (50 − 30)/10 = 20/10 = 2 A downward
- Check: I5 + I10 = 10 + 2 = 12 A, which matches Ix. This is a consistent solution with V > 30 V, so the right branch current is in the expected direction.

Option B: Ix = 10 A
- Solve 10 = V/5 + (V − 30)/10 = (3V − 30)/10
  => 100 = 3V − 30 => 3V = 130 => V ≈ 43.33 V
- I5 ≈ 43.33/5 ≈ 8.67 A
- I10 ≈ (43.33 − 30)/10 ≈ 1.33 A
- I5 + I10 ≈ 8.67 + 1.33 = 10 A, consistent.

Option C: Ix = 15 A
- Solve 15 = V/5 + (V − 30)/10 = (3V − 30)/10
  => 150 = 3V − 30 => 3V = 180 => V = 60 V
- I5 = 60/5 = 12 A
- I10 = (60 − 30)/10 = 30/10 = 3 A
- I5 + I10 = 12 + 3 = 15 A, consistent.

Option D: Ix = 6 A
- Solve 6 = V/5 + (V − 30)/10 = (3V − 30)/10
  => 60 = 3V − 30 => 3V = 90 => V = 30 V
- I5 = 30/5 = 6 A
- I10 = (30 − 30)/10 = 0 A
- I5 + I10 = 6 + 0 = 6 A, consistent.

What this analysis shows is that each listed current value (12 A, 10 A, 15 A, 6 A) yields a self-consistent solution for V and the branch currents. In other words, the circuit allows a family of solutions parameterized by Ix, with V adjusting accordingly to satisfy Ix = V/5 + (V − 30)/10.

Therefore, based on the circuit equations, all of the provided options produce valid, internally consistent results; there is no single unique Ix determined solely by the given information. The correct interpretation is that Ix can be 12 A, 10 A, 15 A, or 6 A, and each choice corresponds to a different top-node voltage V (50 V, 43.33 V, 60 V, and 30 V respectively) that satisfies KCL for that Ix.

Determine the value of IX (the current source) in the circuit shown.

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