Question 1

Practical current source: In real-world circuits, a current source cannot be ideal; it always has a finite internal resistance (typically large but not infinite). State whether this statement is true or false.

Accepted Answer

Introduction / Context:A current source is a two-terminal element intended to deliver a specified current regardless of the load. Ideal sources are useful abstractions, but every physical implementation deviates from ideal behavior. This question tests understanding of the difference between ideal and practical current sources and the role of internal resistance (compliance).Given Data / Assumptions:The device is a practical (real) current source, not an ideal model.It must operate over a finite compliance voltage range.Internal elements can be represented by an equivalent resistance in parallel with the current source (Norton form).Concept / Approach:In Norton representation, a practical current source is modeled as an ideal current source in parallel with a finite internal resistance Rint. A very large Rint makes the source closer to ideal, but it can never be infinite in practice. This resistance limits how perfectly the source holds current as load voltage varies and defines the compliance range where the source behaves approximately constant-current.Step-by-Step Solution:Model the practical current source as In in parallel with Rint.For a load RL, the delivered current deviates from In depending on Rint and RL: Iload = In * (Rint / (Rint + RL)).As Rint → ∞ (ideal case), Iload → In independent of RL; with finite Rint, some variation occurs.Therefore, the existence of finite Rint is inherent to practical sources.Verification / Alternative check:Measure current into two different load resistances. If the source were ideal (infinite internal resistance), current would be identical. Any measurable change indicates finite internal resistance and confirms practicality.Why Other Options Are Wrong:“False” implies infinite internal resistance is achievable in real hardware, which contradicts device physics and component limitations.Common Pitfalls:Confusing series vs. parallel placement in models: practical current sources use a parallel internal resistance (Norton), whereas practical voltage sources use a series internal resistance (Thevenin). Also, mixing up high but finite with truly infinite resistance.Final Answer:True

Question 2

Thevenin voltage definition: The Thevenin equivalent voltage VTH equals the open-circuit terminal voltage (not the short-circuit voltage). True or false?

Accepted Answer

Introduction / Context:Thevenin and Norton theorems are dual tools for simplifying linear networks. Correctly identifying VTH and IN is essential to avoid modeling errors. This statement tests whether you can distinguish open-circuit from short-circuit measurements in forming equivalents.Given Data / Assumptions:We want the Thevenin equivalent at a pair of terminals.All components are linear; dependent sources may be present.Phasor impedances may be used for AC, but the definitions of VTH and IN remain the same.Concept / Approach:Thevenin equivalent voltage VTH is the open-circuit terminal voltage Voc with the load removed. Norton equivalent current IN is the short-circuit current Isc across the terminals. The Thevenin/Norton impedances are identical and satisfy ZTH = Voc / Isc if both quantities are defined and linearity holds.Step-by-Step Solution:Remove the load to measure Voc ⇒ VTH = Voc.Short the terminals to measure Isc ⇒ IN = Isc.Find ZTH by deactivating independent sources or using ZTH = Voc / Isc.Construct Thevenin: VTH in series with ZTH; Norton: IN in parallel with ZN where ZN = ZTH.Verification / Alternative check:Compute the terminal V–I relation with the derived VTH and ZTH and compare with the original circuit for two distinct loads. Matching results verify that VTH was correctly taken as the open-circuit voltage, not the short-circuit value.Why Other Options Are Wrong:Choosing “True” confuses the short-circuit metric with VTH. The short-circuit quantity corresponds to Norton current, not Thevenin voltage.Common Pitfalls:Accidentally deactivating dependent sources when finding ZTH; use a test source method if needed. Also, forgetting that power cannot be directly superposed when computing equivalents.Final Answer:False

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