Monday 15 October 2012

Norton’s Theorem Statement And Norton’s Theorem solved Method

Norton’s Theorem Statement And Norton’s Theorem solved Method

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 “Norton’s equivalent circuit”.

Norton’s equivalent circuit has containing one current source in parallel with two resistances ‘R_L’ & ‘R_N’. As shown in fig. below:-

Here,

I_N/I_S.C= Norton’s equivalent current (“It is that Current from required unknown through any resistance is removed and short-circuited.”)

R_N =Norton’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 “Norton’s Theorem”:-

For finding R_L: - 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, R_L=R₄

For finding R_N: - For finding R_N, Just find-out equivalent b/w removed load-resistance. Like as:-

Step 1^st:- Firstly removed the load-resistance from the given circuit

Step 2^nd:- Deactivating all the energy sources of the fig. Step 1^st:-

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

For finding I_N/I_S.C: - For finding I_N/I_S.C in given circuit diagram, we have to follow the given steps:-

Step 1^st:- Removed load-resistance from the given problem circuit.

Step 2^nd:- Short-circuit the open part from where the load-resistance is removed.

The current flow through short-circuited path is the required current I_N/I_S.C which we can find-out any of the method we have studied earlier in this website.(Easy to use Mesh-analysis)

Now, the Norton’s equivalent circuit is,