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Why do you short voltage source and open current source while using superposition theorem?
You short a voltage source when doing this analysis because you do not know how much current will flow through the voltage source - consider it an undefined value. For the same reason, you open a current source since you know how much current will be flowing through it.
This is a simple explanation; I'm sure a more exhaustive, technical one could be made if this is not sufficient.
What else can I help you with?
How the superposition theorem is used to find the currents in a circuit supply by more than one voltage source?
1.Put a short circuit instead of voltage source 1 and find what you want with taking direction of current in that element(ris.ind.cap.) 2.puta short circuit instead of voltage source 2 and find what you want with taking direction of current in that element(ris.ind.cap.) 3.add current 1 and 2 for any element.
Theory of superposition theorem?
Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman's certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all "superimposed" on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active.
Can we apply Thevenin's theorem in circuit having multi source?
Yes, Thevenin's theorem can be applied to circuits with multiple sources. To do this, you can use techniques like superposition to analyze the effect of each source independently, then combine the results to find the overall response. Once you have the equivalent circuit for the portion of interest, you can simplify it to a single voltage source in series with a single resistor, which represents the Thevenin equivalent.
How do you change current into voltage?
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
How will you convert a open current circuit to a open voltage circuit?
The first thing you need to know is the internal resistance of the current source, the voltage source will have the same internal resistance. Then compute the open circuit voltage of the current source, this will be the voltage of the voltage source. You are now done.
What is the superposition?
This theorem is used to determine the value of current in specific branch of a multi voltage source circuit .
How can make a voltage source inactive in superposition theorem?
To make a voltage source inactive in the superposition theorem, you replace it with a short circuit. This means that you eliminate the voltage across the terminals of the source, allowing current to flow as if the voltage source were not present. Once the analysis is completed for all other sources, you can then reintroduce the effects of the voltage source by considering its contribution to the circuit.
What is the use of superposition theorem?
superposition can find the voltage and current effect of each source to a particular branch of the circuit and we can calculate the total effect of the sources to know the effect of the total sources to that branch
Can you calculate power using superposition theorem?
Yes, you can calculate power using the superposition theorem, but indirectly. The superposition theorem states that in a linear circuit with multiple independent sources, the total response (voltage or current) at any point can be found by considering one source at a time while replacing all other independent sources with their internal resistances. After determining the individual voltages or currents due to each source, you can then calculate the power for each case and sum them to find the total power absorbed or delivered by the circuit.
What is the Superposition Theorem for an Electrical Network?
superposition therorem states that in linear network containning more than one source of emf the resultant current in any branch is the algebraic sum of the current that would have been produced by each source of emf .taken sepertely with all other sources of emf replace by their internal resistance ........... that is called superposition theorem ..
How the superposition theorem is used to find the currents in a circuit supply by more than one voltage source?
1.Put a short circuit instead of voltage source 1 and find what you want with taking direction of current in that element(ris.ind.cap.) 2.puta short circuit instead of voltage source 2 and find what you want with taking direction of current in that element(ris.ind.cap.) 3.add current 1 and 2 for any element.
Why superposition theorem is not valid for power?
we cant consider two source at a time in superposition theorem....but power =v*i.so we cant calculate power.
How super position theorem solved?
The Superposition Theorem is used in linear circuit analysis to determine the contribution of each independent source to the overall circuit response. To apply it, you disable all but one independent source at a time: replace voltage sources with short circuits and current sources with open circuits. You then analyze the circuit to find the response (voltage or current) due to the active source. Finally, you sum all individual contributions to get the total response in the circuit.
How do you treat dependent source in superposition theorem or any kind of network theorem?
For a dependent source the main problem occurs while finding the equivalent impedance of the circuit . For this case : 1. Keep the dependent source as it is . 2. Apply a dc voltage across the o/p terminal . let it be Vdc 3. Let the current for this Vdc voltage source is Idc . 4. Find Zeq=Vdc / Idc. [Zeq is the equivalent impedance] [ Specially for superposition: we perhaps need not required the 2nd,3rd& 4th step . As we just have to calculate the current through the o/p . And , i think the dependent source should not be switched off any time . ]
Theory of superposition theorem?
Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman's certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all "superimposed" on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active.
What is the difference between the thevenin's theorem and norton's theorem?
Thevenin's theorem and Norton's theorem are both techniques used to simplify complex electrical circuits. Thevenin's theorem states that any linear circuit can be replaced by a single voltage source (Thevenin voltage) in series with a resistance (Thevenin resistance). In contrast, Norton's theorem states that the same circuit can be simplified to a single current source (Norton current) in parallel with a resistance (Norton resistance). While they are mathematically interchangeable, Thevenin's focuses on voltage sources, while Norton's emphasizes current sources.
Why do you replace voltage sources with short circuits and current sources with open circuits in thevenin's theorem?
The equivalent of an inactive Thevenin voltage source is a source with zero voltage between its terminals regardless of the current through it, best represented by a zero resistance, i.e. a short-circuit. The equivalent of an inactive Norton current source is a source through which no current can flow regardless of the voltage across it, best represented by an infinite resistance, i.e. an open circuit.
