Monday, September 3, 2012

Thevenin’s Theorem And Method Of Solving ‘Thevenin’s Theorem’

Consider a circuit,

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 ‘RL’ & ‘Rth’. As shown in fig. below:-


Vth= Thevenin’s equivalent voltage (“It is that equivalent voltage from required unknown through any resistance is removed”)
Rth= Thevenin’s equivalent resistance (“It is that equivalent resistance from required unknown through any resistance is removed”) &
RL= Called “Load-Resistance” (The resistance in which unknown is required).

Method Of Solving “Thevenin’s Theorem”:-

For finding RL: - Suppose that from above given circuit, asked to find-out the value of current through ‘R4’ resistance. Then the resistance along which any unknown is asked to find-out that resistance becomes “Load-resistance”.

Here, RL=R4

For finding Rth: - For finding Rth, 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 1st:- Firstly removed the load-resistance from the given circuit
Step 2nd:- Deactivating all the energy sources of the fig. Step 1st:-

Step 3rd:- Now, Rth is actually the equivalent resistance b/w that two ends from load-resistance are removed of fig. Step 2nd:-

For finding Vth: - For finding Vth in given circuit diagram, we have to follow the given steps:-
Step 1st:- Removed load-resistance from the given problem circuit.

Step 2nd:- Solve the given circuit of step 1st and find-out the value of current each resistance connected in b/w pt. ‘A’ & pt. ‘B’ for which we have to finding voltage. “Vth” is now actually the voltage difference b/w that two ends from where load-resistance is removed.

For finding VAB we go from pt. ‘A’ to pt. ‘B’ using kvl for two open ends.
Now, the Thevenin’s equivalent circuit is,

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