Thevenin's Theorem : Reason behind 'nullifying' or 'suppressing' sources?

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Thevenin's theorem requires voltage sources to be short-circuited and current sources to be open-circuited to accurately nullify their effects in circuit analysis. An inactive voltage source is equivalent to zero volts, making it behave like a short circuit, while an inactive current source has infinite impedance, functioning as an open circuit. If a voltage source were replaced with an open circuit, a voltage difference could still exist across its terminals, contradicting its inactive status. Similarly, replacing a current source with a short circuit would allow current to flow, which is not permissible for an inactive source. Understanding these principles is crucial for correct application of Thevenin's theorem in circuit analysis.
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WHY aren't voltage sources open circuited and current sources short circuited?

Thevenin's theorem (and even Superposition Theorem) states that we need to 'suppress' or 'nullify' the effect of all sources. We do this by short circuiting voltage sources and open circuiting current sources. WHY exactly can't we do this the other way round by open circuiting voltage sources and short circuiting current sources? What is wrong about doing that?
 
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An inactive voltage source is equivalent to zero volts between its terminals, independent of the current through it. This would be a "wire", or a short circuit.

An inactive current source is a current source in which no current can flow independent of the voltage across, which is represented by infinite impedance, and hence an open circuit.
 
Thanks a lot! You are awesome :)
 
If you were to replace the voltage source with an open circuit rather than a short circuit, there may exist a voltage difference across its terminals (dependent on the rest of the network) and hence it would not be considered an "inactive" voltage source. Similar arguments would apply to the current source.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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