Norton–Thevenin conversion: A circuit has Norton current I_N = 10 mA and Norton resistance R_N = 10 kΩ. What is the equivalent Thevenin source?

Introduction / Context:
Norton and Thevenin equivalents are dual representations: Norton uses a current source in parallel with a resistance, while Thevenin uses a voltage source in series with a resistance. Converting between them is routine in network analysis and simplifies load calculations.

Given Data / Assumptions:

• Norton current I_N = 10 mA.
• Norton resistance R_N = 10 kΩ.
• Ideal linear network, so R_N = R_TH.

Concept / Approach:
The conversion uses V_TH = I_N * R_N and R_TH = R_N. The topology changes from current source in parallel to voltage source in series, preserving terminal behavior for any load.

Step-by-Step Solution:
Compute V_TH = I_N * R_N = 0.01 A * 10,000 Ω = 100 V.Set R_TH = R_N = 10 kΩ.Therefore Thevenin equivalent is 100 V in series with 10 kΩ.

Verification / Alternative check:
Open-circuit voltage equals 100 V, and short-circuit current equals 10 mA in both forms, confirming equivalence.

Why Other Options Are Wrong:
(a) 1000 V is a decade too high. (b) and (d) incorrectly keep the resistance in parallel in the Thevenin form.

Common Pitfalls:
Mixing series/parallel placement of the resistance; miscomputing V_TH by forgetting to convert mA to A.

Final Answer:
100 V with series resistance 10 kΩ