Source transformation in basic circuit theory: A 12 mA ideal current source with internal resistance RS = 1.2 kΩ is to be converted to an equivalent Thevenin (voltage) source. What is the resulting source voltage?

Introduction / Context:Many circuit problems ask you to transform a Norton (current-source) model into a Thevenin (voltage-source) model or vice versa. This is a standard technique in electrical engineering used to simplify analysis and to match sources with loads.

Given Data / Assumptions:

• Ideal current source Is = 12 mA.
• Series/internal resistance RS = 1.2 kΩ.
• We assume linear, time-invariant, resistive conditions with no dependent sources.

Concept / Approach:The source transformation rule is direct: a current source Is in parallel with RS is equivalent to a voltage source VTH in series with RS, where VTH = Is * RS. The resistance value remains the same in both equivalent forms (RS = RTH).

Step-by-Step Solution:

Verification / Alternative check:Check dimensional consistency: amperes * ohms = volts, so the units are correct. Also, if you were to convert back to Norton, IN = VTH / RTH = 14.4 / 1,200 = 0.012 A = 12 mA, confirming equivalence.

Why Other Options Are Wrong:

• 144 V: Off by a factor of 10; would require Is = 120 mA or RS = 12 kΩ.
• 7.2 V: Exactly half the correct value; likely from halving either the current or resistance erroneously.
• 72 mV: Incorrect scaling of milli and kilo prefixes.

Common Pitfalls:

• Mishandling metric prefixes (milli vs. kilo) causing thousand-fold errors.
• Confusing series vs. parallel placement of RS when picturing the transformation.

Final Answer:14.4 V