Read the following statements regarding Thevenin’s equivalent circuit : (a) The Thevenin’s voltage is calculated across the short circuit terminals. (b) The Thevenin’s voltage is calculated at the open circuit terminals. (c) The connection in the circuit is open if any voltage source is present. (d) The connection in the circuit is shorted if any voltage source is present. Which of the above statements are incorrect ?
(a) and (c)
Read the question carefully — it asks which statements are incorrect, so the true ones must be rejected. The statements come in two contradictory pairs, so exactly one of each pair is wrong.
Pair 1 — how VTh is found. The Thevenin voltage is the voltage that appears across the terminals when the load is removed, i.e. the open-circuit voltage:
\(V_{Th}=V_{OC}\)
So (b) is correct, and (a) is incorrect. Measuring across short-circuited terminals would always give zero volts — what a short circuit gives you is the Norton current, \(I_N=I_{SC}\), and the two are linked by \(R_{Th}=V_{OC}/I_{SC}\).
Pair 2 — how sources are treated when finding RTh. Every independent source is replaced by its internal resistance:
| Source | Replaced by | Because its ideal internal resistance is |
|---|---|---|
| Voltage source | short circuit | zero |
| Current source | open circuit | infinite |
So (d) is correct and (c) is incorrect — it applies the current-source rule to a voltage source.
The incorrect statements are therefore (a) and (c).
Why the rules are what they are. Deactivating a source means forcing its contribution to zero while leaving its resistance in place. Setting a voltage source to zero volts is exactly what a short circuit does; setting a current source to zero amps is exactly what an open circuit does. Getting this backwards is the single most common error in Thevenin problems.
One caution. These substitutions apply only to independent sources. A dependent source must be left in the circuit; RTh is then found by applying a 1 V test source at the terminals and computing \(R_{Th}=V_{test}/I_{test}\).
Hence, the incorrect statements are (a) and (c).
Similar Questions
Read the passage and answer the questions that follow based on your understanding of the passage :
General methods of n/w analysis become laborious and time consuming for large and complex network. For such situations the solution is network theorems. Besides, the other features of n/w theorems are (A) they are applicable to a useful and fairly wide class of networks, (B) their conclusions are simple and (C) they sometimes provide good physical insight into the problems.
The maximum power transfer implies that the load impedance must be the complex conjugate of the source impedance. The superposition theorem is valid for any linear, time invariant or time varying. It is useful in circuit analysis when the n/w has large number of sources. Thevenin's or Norton's theorem is applicable to any number of time invariant or time varying n/w. It is useful when only one part of the n/w is varying, while the other part remains constant. Thevenin's equivalent ckt is the voltage source equivalent at the terminals concerned. Millman's theorem is the extension of Thevenin's or Norton's theorem for a number of Current or Voltage sources respectively. The substitution theorem is applicable to any network and can be applied to a branch which is not coupled to other branches of the network. Tellegen's theorem is applicable to any lumped n/w regardless of the type of elements, which may be linear or non-linear, time varying or time invariant.
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