nortons theorem.pptx

nortons

1. NORTON’S THEOREM PRESENTED BY: NAME:P.SNEHA BRANCH:ECE-A ROLL NO:22R21A0446

2. NORTON’S THEOREM STATEMENT: Any linear, active, bilateral dc network having a number of voltage sources and/or current sources with resistances can be replaced by a simple equivalent circuit having single current source (I N ) in parallel with a single resistance (R N ). Where (I N ) is the known as Norton’s equivalent current through the terminal a-b. (R N ) is the Norton’s equivalent resistance viewed back into the network from terminal a-b. Note: independent voltage sources are short circuited and independent current sources are open circuited. Dependent sources will remain in the circuit for the calculation of Norton’s equivalent resistance.

3. NORTON’S THEOREM Procedure for converting any circuit into Norton's equivalent circuit Calculate Norton Current Step 1: remove the load resistance R L (through which current is required) and short circuit it. Let terminals of load are labelled as a-b. Therefore a-b is the short circuited. Step 2: Find the current through the terminal a-b by applying KCL, KVL, Ohm’s law or Superposition principle. This current is the short circuit current and it is known as Nortons equivalent current (I N ). Calculate Norton Resistance (equal to Thevinin resistance) Step 3: Set all Independent voltage Sources as short circuit and Current Sources open circuit. Dependent sources will not be changed Step 4: Calculate the resistance as “seen” through the terminals a-b into the network. This resistance is known as Norton’s equivalent resistance (R N ). Draw Equivalent Circuit Step 5: Replace the entire network by Nortons equivalent current (I N ) in parallel with Norton’s equivalent resistance (R N ) and connect the load resistance R L .

4. NORTON’S THEOREM Linea r , Active, Bilateral Network R L a b I L b Norton’s Equivalent Network R L a I L R N I N

5. NORTON’S THEOREM Example: Find the current through 3 ohm resistor by Norton’s Theorem for the network shown in fig.1a SOLUTION: STEP 1: Calculation of R N (calculation is same as R th ). Redraw the circuit by removing the 3 ohm resistor and short circuit the voltage sources as shown in fig. 1b Fig. 1a 6 ohm 1 ohm 12V 24V 3 ohm R 3 R 1 a R 2 b

6. NORTON’S THEOREM 6 ohm 1 ohm R 2 a b R 2 b 24V 12V a I N R 1 R N Step2: Calculation of Norton’s Current I N : Short circuit the terminals a-b and the current flow through a-b is I N Fig. 1c Fig. 1b R 1 and R 2 are in parallel 6 1 N 1 2 I  I  I  2 4  1 2  1 6 A  6  1  0.857  R  R 6  1 R  R 1 R 2 2 1 N I 1 I 2

7. NORTON’S THEOREM Step2: Draw the Norton’s Equivalent Circuit: I N 16A 0.857 ohm R N N I =16A R N = 0.857 ohm R3 = 3 Ohm Step3: Calculation of Current through R3, Reconnect R3 to Norton’s Equivalent Circuit (Fig. 1e) Apply Current divider rule  3.55 A 0.85 7  3 0.85 7 L I  16 I L R R N  R L I L  I N N

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