Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context:
Thevenin’s theorem lets you replace a complicated linear network with a single ideal voltage source in series with a resistance as seen from two terminals. The core quantity is the Thevenin voltage, which many learners confuse. This question asks whether V_th equals the open-circuit voltage at the terminals.
Given Data / Assumptions:
Concept / Approach:
By definition, the Thevenin voltage V_th is the terminal voltage when the output is open-circuited (no load current). With no load current, internal drops through the Thevenin resistance are zero, so the terminal voltage equals the ideal source voltage in the Thevenin model. Therefore, V_th = V_open-circuit is not merely a rule of thumb but the formal definition in Thevenin analysis.
Step-by-Step Solution:
Verification / Alternative check:
Build a resistive divider: with the output open, the node voltage is V_oc from the divider equation. In the Thevenin model, the same node is driven by V_th in series with R_th; with I_load = 0, drops across R_th are zero, so the terminal voltage equals V_th, matching V_oc.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing V_th with the loaded output; forgetting that R_th affects the voltage only when load current flows; mixing Norton parameters with Thevenin parameters.
Final Answer:
Correct — V_th equals the open-circuit terminal voltage.
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