Solved problems on Norton’s theorem
Q1) Find the Norton’s equivalent circuit across A-B terminals for the circuit shown in figure 14.1
First we remove the 10Ω resistor and short circuit the terminals A&B. see figure 14.2
The current flowing through the short circuited terminals is called the Norton’s curren IN.
To find the IN we apply nodal equation for point C
(V – 30)/5 + V/10 + V/3 = 0
V = 180/19
Ohms law to 3Ω resistance
I = V/R
IN = (180/19) / 3
IN = 60/19
This is the Norton’s current.
Now we are about to find the Norton’s resistance. Note that this is equal to the thevenin’s resistance also.
See figure 14.3
RN = (10//5) + 3
RN = 19/3Ω
Q2) Find the Norton’s equivalent circuit across A-B terminals for the circuit shown in figure 14.4
First we short circuit the two terminals A&B and calculate the current flowing
See figure 14.5
When we short circuit A-B terminals it results in shorting out 12Ω resistor as shown in figure 14.6
Now we can simply find the current IN by applying Ohms law to 4Ω resistor,
I = V/R
IN = 24/4
IN = 6A
Then we find the RN . See figure 14.7
RN = 12//4
RN = 3Ω
Q3) Using Norton’s theorem calculates the current flowing through 12Ω resistor. You may refer figure 14.8
Wait for my next post to see the solution. Till then, try this yourself. It is very similar to Q1.