Norton current division with finite source resistance: A 1.2 A constant current source has internal resistance RS = 12 kΩ (parallel model). A load RL = 680 Ω is connected across the source. What is the load current IL?

Introduction / Context:
A practical current source is modeled as an ideal current source in parallel with a finite internal resistance. When a load is connected in parallel, the source current divides between the internal resistance and the load according to current division. This question checks accurate use of the current divider formula.

Given Data / Assumptions:

• Ideal Norton source current IS = 1.2 A.
• Internal parallel resistance RS = 12 kΩ.
• Load resistance RL = 680 Ω connected across the source.
• Steady-state, linear resistors.

Concept / Approach:

For two parallel resistors RL and RS fed by a current source IS, the current through RL is given by current division: IL = IS * (RS / (RS + RL)). Because RS ≫ RL, most of the current goes through the smaller resistor RL.

Step-by-Step Solution:

Verification / Alternative check:

Current through RS is IS − IL ≈ 1.2 − 1.1359 ≈ 0.0641 A. Voltage across both branches: V ≈ IL * RL ≈ 1.1359 * 680 ≈ 772.4 V, which also equals 0.0641 A * 12,000 Ω ≈ 769 V (small rounding differences due to rounding earlier). Both branches share approximately the same voltage, confirming current division.

Why Other Options Are Wrong:

0 A would mean open load. 114 mA is off by a factor of 10. 1.2 A implies zero internal resistance diverting none of the current. 11.4 A is not possible with a 1.2 A source.

Common Pitfalls:

Using the reciprocal ratio (RL / (RL + RS)), or mistakenly modeling RS in series instead of parallel for a current source.

Final Answer:

1.14 A