Consider a
circuit,

Step 1

The solution
of the given problem is not done by the actual given circuit, we have to
find-out the value of any unknown (i.e, V,I or P) through any resistance as per
question is said we design a new circuit and the circuit is known as
“Thevenin’s equivalent circuit”.

Thevenin’s equivalent circuit has containing one voltage source in series with two resistances ‘R

Thevenin’s equivalent circuit has containing one voltage source in series with two resistances ‘R

_{L}’ & ‘R_{th}’. As shown in fig. below:-
Here,

V

_{th}= Thevenin’s equivalent voltage (“It is that equivalent voltage from required unknown through any resistance is removed”)
R

_{th}= Thevenin’s equivalent resistance (“It is that equivalent resistance from required unknown through any resistance is removed”) &
R

_{L}= Called “Load-Resistance” (The resistance in which unknown is required).

*Method Of Solving “Thevenin’s Theorem”*

*:-*__For finding R__Suppose that from above given circuit, asked to find-out the value of current through ‘R

_{L}: -_{4}’ resistance. Then the resistance along which any unknown is asked to find-out that resistance becomes “Load-resistance”.

Here, R

_{L}=R_{4}_{}__For finding R__For finding R

_{th}: -_{th}, the process is same as we discussed in “Method of solving grouping of resistance”. Just find-out equivalent b/w removed load-resistance. Like as:-

Step 1

^{st}:- Firstly removed the load-resistance from the given circuit

Step 3

^{rd}:- Now, R_{th}is actually the equivalent resistance b/w that two ends from load-resistance are removed of fig. Step 2^{nd}:-__For finding V__For finding V

_{th}: -_{th}in given circuit diagram, we have to follow the given steps:-

Step 2

^{nd}:- Solve the given circuit of step 1^{st}and find-out the value of current each resistance connected in b/w pt. ‘A’ & pt. ‘B’ for which we have to finding voltage. “V_{th}” is now actually the voltage difference b/w that two ends from where load-resistance is removed.
For finding
V

V

_{AB }we go from pt. ‘A’ to pt. ‘B’ using kvl for two open ends.V

_{th}=V_{A}-V_{B}=V_{AB}

__Now, the Thevenin’s equivalent circuit is,__
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