Questions: Thévenin and Norton Circuit Equivalents

Question 1

A circuit contains 12 resistors, 4 voltage sources, and 2 current sources. An engineer wants to test how different loads will behave when connected to two output terminals. What does the Thévenin equivalent tell her?

• A. She must re-solve the full circuit for every different load value
• B. The circuit looks like a single voltage source in series with a single resistor to any load connected at those terminals
• C. The circuit can be simplified only if the sources are all identical
• D. The equivalent works for this circuit only because it has more resistors than sources

Answer: (B)

Thévenin's theorem states that ANY linear circuit — regardless of complexity — appears to an external load as just V_th in series with R_th. The engineer computes these two values once, then uses the simple voltage-divider formula V_load = V_th × R_load / (R_th + R_load) for every different load. The theorem's power is precisely that it eliminates re-solving the full circuit for each load.

Question 2

A Norton equivalent has I_N = 4 A and R_N = 10 Ω. What is the Thévenin equivalent voltage V_th?

• A. 0.4 V — divide current by resistance
• B. 14 V — add current and resistance
• C. 40 V — multiply current by resistance
• D. Cannot be determined without knowing the original circuit topology

Answer: (C)

Thévenin and Norton are dual representations of the same terminal behavior. The conversion is V_th = I_N × R_N = 4 A × 10 Ω = 40 V, with R_th = R_N = 10 Ω. No knowledge of the original circuit is needed once you have the Norton equivalent — that is the entire point of circuit equivalents. Option D is the most tempting misconception: students think the original topology matters, but all terminal information is already captured in I_N and R_N.

Question 3

Thévenin and Norton equivalents always produce identical terminal behavior — they are two representations of the same underlying circuit, not two different approximations.

• T. True
• F. False

Answer: True

Thévenin (V_th in series with R_th) and Norton (I_N in parallel with R_N) are mathematically dual representations related by source transformation: I_N = V_th / R_th and R_N = R_th. Any voltage or current measured at the terminals is identical for both equivalents. The choice between them is purely one of analytical convenience.

Question 4

R_th in a Thévenin equivalent is the actual internal resistance of the voltage source that exists inside the original circuit.

• T. True
• F. False

Answer: False

This is a common and important misconception. R_th is the equivalent resistance seen looking back into the circuit from the terminals, computed by turning off all independent sources (replacing voltage sources with short circuits and current sources with open circuits) and finding the equivalent resistance of the remaining resistor network. An original circuit may contain no actual source resistance at all, yet still yield a non-zero R_th because of how resistors are arranged around the terminals.

Question 5

Why does maximum power transfer to a load occur when R_load = R_th, and why does this matter for engineering applications?

Model answer: With R_load = R_th, the load receives exactly half of V_th as voltage (by the voltage divider), which maximizes the power product V × I delivered to the load. For smaller R_load, more current flows but the voltage across the load is too small; for larger R_load, the voltage is close to V_th but too little current flows. The product P = V²/R is maximized at R_load = R_th.

The Thévenin equivalent makes this optimization transparent: without it, you would need to optimize over the full complex circuit. With it, you reduce the problem to a one-parameter comparison. This principle governs impedance matching in amplifier output stages, transmission line design, and sensor circuits — wherever transferring maximum power to a load is the design goal.