Norton theorem
Norton’s theorem statement
In Thevenin theorem, we see that in any complex circuit sources are converted in single voltage source Vth with the Series resistor Rth. Norton’s theorem is also a method to convert any complex network into a simple circuit.
Norton Theorem, any twoterminal complicated network circuit can be converted in single Current source IN in parallel with a single resistor RN.
Actually in norton we convert any complex or complicted circuit into a simple circuit and I get current value across a Load
According to Norton theorem, Any linear bilateral network irrespective of its complexities can be reduced into Norton’s equivalent circuit having a Norton short circuit current “Ir” in parallel with Norton’s equivalent resistance Rn in parallel with load resistor RL.
What is Load resistor – Across which resistor asked to find the current that resistor is called Load resistor.
Steps for solving the Norton theorem problems
Step1. Identify the load resistor RL.
Step 2.Replace RL by a short circuit at this place. The current flowing through this short circuit branch will be Norton’s current IN. This current also is known as shortcircuited current Isc.
Step 3. Remove the load resistance and replace all the active sources with their internal resistance.
Step 4. The equivalent resistance across the two open will be Norton’s resistance RN.
Step 5. Draw Norton’s equivalent circuit.
Step 6. Calculate IL using the identity IL=IN.RL/RN+RL
Example
Q.1 Find the value of current I in the given circuit using Norton’s theorem
Solution –
Here RL= 10Ω
Now I Replace the Load resistor with short circuit and Apply the KVL
I1=2A
40I172I1+12I3=120
Putting the value of I1
8072I2+12I3=120
72I1+12I3=40 ———Eq.1
And, 12I3+12I2=0 ———Eq.2
by solving the eq 1 & 2
I2=0.666A
Because of 10Ω resistor is shortcircuited then 12Ω resistor also should be shorted, and then we don’t need to find the current at IN.
Current I2 is passed across IN.
So,IN=I2=0.666A
Now we Remove the Load resistor RLK and repace all sources with their internal resistance.
RN=(20+40)//12
RN=60*12/60+12
RN=10Ω
by using Norton Identity
IL=IN.RL/RN+RL
IL=0.333A (Ans)
Q. 2 In the given network draw Norton’s equivalent circuit and find the current through 18.75Ω resistor.
Solution.–
Step 1Short circuit the load resistance (RL)
Step 2Calculate the short circuit current IN or ISC.
I1=30A
I3= ISC or IN
(in mesh 2) 10I1+20I25I3=50 ————eq.1
10*30+20I25I3=50
20I25I3=250
(In mesh 3) 5I2+5I3=50 ———eq. 2
by solving eq 1 & eq 2
I2=20A
I3=30A
I3=IN=30A
Step 3.Remove all sources with internal resistance and Remove Load resistance RL
RN=15*5/15+5 = 3.75
Step 4.Draw Norton equivalent circuit and calculate the IL using Identity IL=IN.RN/RN+RL
30*3.75/3.75+18.75=5A
S0, IL=5A (Ans)
Also read

Thevenin theorem

Kirchoff’s law (KCL, KVL)

Logic gates

Zener diode working

What is Optocoupler/optoIsolator