Thevenin’s equivalent circuit is a practical voltage source. In contrast, Nortorn’s equivalent circuit is a practical current source. This can be formally stated as
” Any two-terminal, linear circuit, of resistor and sources, can be replaced by a single current source in parallel with a resistor.”
To determine Norton’s equivalent circuit, Norton current, IN, and Norton resistance, RN, are required. The following steps outline the procedure required.
For the circuit diagram depicted in Figure, use Norton’s theorem to determine the current through the 5 Ohm resistance.
As mentioned above, The Norton’s resistance is the same as the Thevenin’s resistance, i.e
RN=21.67Ohm
To find Norton’s current, the terminals A and B in Fig (b) are short circuited. Then the current through the short circuited terminals is Norton current, IN or Isc
There are a number of ways to solve the above circuit such as KVL, KCL however, circuit reduction techniques have been chosen here.
Firstly, total current supplied by the voltage source can be found by adding the resistances using series/parallel combination.
Req=(15||20)+10 = (15x20)/(15+20)+10 = 18.57 Ohm
IS =VS / Req =10/18.57 = 0.54A
Since I =I10Ohm, ISC can be found by employing current divider rules (CDR) between the 15 Ohm and 20 Ohm resistances i.e.
ISC =ISx20/(15+20) = 0.54x20/35 = 0.31A
The Norton’s equivalent circuit can be drawn as shown in Figure
The current through the 5Ohm resistance can be determined by the application of CDR between RL and RN ,thus
” Any two-terminal, linear circuit, of resistor and sources, can be replaced by a single current source in parallel with a resistor.”
To determine Norton’s equivalent circuit, Norton current, IN, and Norton resistance, RN, are required. The following steps outline the procedure required.
- Remove the load resistance, RL
- IN is the SC current through the load terminals and
- RN is the resistance across the load terminals withal source replaced by their resistances. Clearly RN= RTH
For the circuit diagram depicted in Figure, use Norton’s theorem to determine the current through the 5 Ohm resistance.
As mentioned above, The Norton’s resistance is the same as the Thevenin’s resistance, i.e
RN=21.67Ohm
To find Norton’s current, the terminals A and B in Fig (b) are short circuited. Then the current through the short circuited terminals is Norton current, IN or Isc
Firstly, total current supplied by the voltage source can be found by adding the resistances using series/parallel combination.
Req=(15||20)+10 = (15x20)/(15+20)+10 = 18.57 Ohm
IS =VS / Req =10/18.57 = 0.54A
Since I =I10Ohm, ISC can be found by employing current divider rules (CDR) between the 15 Ohm and 20 Ohm resistances i.e.
ISC =ISx20/(15+20) = 0.54x20/35 = 0.31A
The Norton’s equivalent circuit can be drawn as shown in Figure
I5Ohm =0.31x21.67/(21.67+5) = 0.25A
Please read more: Thevenin’s theorem and Voltage and Current Laws part1
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