Electrical circuits lecture notes b. Tech

V
Th
in series with 
R
Th
and the concerned branch resistance
 (or) 
load resistance across the load terminals(A&B) as shown in below fig.
Example: Find V
TH
, R
TH
 and the load current and load voltage flowing through R
L 
resistor 
as shown in fig. by using Thevenin’s Theorem?
Fig.(a) 
Solution: 
The resistance
R
L
 
is removed and the terminals of the resistance
R
L
 
are marked as
A & B 
as
 
shown in the fig. (1)
Fig.(1) 

---

Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V
TH
). We have 
already removed the load resistor from fig.(a), so the circuit became an open circuit as shown in 
fig (1). Now we have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 
12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor 
as it is open. So 12V (3mA x 4kΩ) will appear across the 4kΩ resistor. We also know that 
current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in 
parallel with 4k resistor. So the same voltage (i.e. 12V) will appear across the 8kΩ resistor as 
4kΩ resistor. Therefore 12V will appear across the AB terminals. So,
V
TH
 = 12V
Fig(2) 
All voltage & current sources replaced by their internal impedances (i.e.
 
ideal voltage sources 
short circuited and ideal current sources open circuited) as shown in fig.(3)
Fig(3) 
Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R
TH
)We have 
Reduced the 48V DC source to zero is equivalent to replace it with a short circuit as shown in 
figure (3) We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 
12k Ω resistor. i.e.: 
8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with) 
R
TH
= 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)] 
R
TH
= 8kΩ + 3kΩ