Electrical circuits lecture notes b. Tech

I
SC
= I
N
 = 2A. 
Fig(3) 
All voltage & current sources replaced by their internal impedances (i.e.
 
ideal voltage sources 
short circuited and ideal current sources open circuited)
and Open Load Resistor. as shown in 
fig.(4) 
Fig(4) 
Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R
N
) We have 
Reduced the 12V DC source to zero is equivalent to replace it with a short circuit as shown in 
fig(4), We can see that 3Ω resistor is in series with a parallel combination of 6Ω resistor and 2Ω 
resistor. i.e.: 
3Ω + (6Ω || 2Ω) ….. (|| = in parallel with) 
R
N
= 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)] 
R
N
= 3Ω + 1.5Ω 
R
N
 = 4.5Ω 
Fig(5) 

---

Connect the R
N
in Parallel with Current Source I
N 
and re-connect the load resistor. This is 
shown in fig (6) i.e. Norton Equivalent circuit with load resistor. 
Fig(6) 
Now apply the Ohm’s Law and calculate the load current through Load resistance across the 
terminals A&B. Load Current through Load Resistor is 
I
L
= I
N
x [R
N
/ (R
N
+ R
L
)] 
I
L=
2A x (4.5Ω /4.5Ω +1.5kΩ)
I
L
= 1.5A 
I
L
 = 1. 5A
Maximum Power Transfer Theorem: 
In many practical situations, a circuit is designed to provide power to a load. 
While for electric utilities, minimizing power losses in the process of transmission and 
distribution is critical for Efficiency and economic reasons, there are other applications in areas 
such as communications where it is desirable to maximize the power delivered to a load.
electrical applications with electrical loads such as Loud speakers, antennas, motors etc. it would 
be required to find out the condition under which maximum power would be transferred from
the 
circuit to the load.