Thevenin's Theorem for DC Network

1. CIRCUIT THEORY
THEVENIN’S THEOREM
SYED HASAN SAEED
Wednesday, July 24, 2019 1syed hasan saeed

2. THEVENIN’S THEOREM D.C. NETWORK
STATEMENT: Any linear, active, bilateral dc network having a number of
voltage sources and/or current sources with resistances can be replaced
by a simple equivalent circuit having single voltage source (Vth) in series
with a single resistance (Rth).
Where (Vth) is the open circuit voltage known as Thevenin’s equivalent
voltage across the terminal a-b.
(Rth) is the Thevenin’s equivalent resistance viewed back into the network
from terminal a-b.
Note: independent voltage sources are short circuited and independent
current sources are open circuited. Dependent sources will remain in the
circuit.
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3. PROCEDURE FOR CONVERTING ANY CIRCUIT INTO
THEVENIN’S EQUIVALENT CIRCUIT AND ITS
CALCULATIONS:
Step 1: Identify and temporarily remove the resistance (load resistance RL) through
which current is required. Suppose these terminals are label as a-b. Therefore a-
b is the open circuited.
Step 2: Find the voltage across the terminal a-b by applying KCL, KVL, Ohm’s
law or Superposition principle. This voltage is the open circuit voltage (Voc) and
it is known as Thevenin’s equivalent voltage (Vth).
Step 3: Set all voltage Sources short circuited and Current Sources open circuited
Step 4: Calculate the resistance as “seen” through the terminals a-b into the
network. This resistance is known as Thevenin’s equivalent resistance (Rth).
Step 5: Replace the entire network by Thevenin’s equivalent voltage (Vth) in series
with Thevenin’s equivalent resistance (Rth).
Step 6: Reconnect the previously removed resistance (load resistance RL) to this
circuit.
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THEVENIN’S THEOREM D.C. NETWORK

4. Step 7: Calculate the Current through load resistance RL
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THEVENIN’S THEOREM D.C. NETWORK
Vth
Rth
RL
Linear, Active,
Bilateral
Network
RL
a
b
b
a
IL
IL
Thevenin’s Equivalent
Network
Fig. (1a)
Fig. (1b)

5. EXAMPLE: Find the current, voltage drop and power loss across the 5 ohm resistor
by applying the Thevenin’s theorem (as shown in fig. 2)
SOLUTION: STEP 1: Calculation of Vth
Remove 5 ohm resistor
Vth
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THEVENIN’S THEOREM D.C. NETWORK
12 V
2 ohm 3 ohm
4 ohm 5 ohm
Fig. 2
5 V
12 V
2 ohm 3 ohm
4 ohm
5 V
a
b
I
Fig. 2a

6. Apply KVL in LHS mesh:
12 - 2I - 4I = 0
I = 2A
Apply KVL in RHS mesh :
- Vth + 4I -3I -5 = 0
Put I = 2A
Vth = -3V
STEP 2: Calculation of Rth
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THEVENIN’S THEOREM D.C. NETWORK
2 ohm 3 ohm
4 ohm Rth
a
b
2 and 4 ohm resistance are in parallel and this combination is in series with 3 ohm
resistor.


 33.43
42
4*2
Rth
Fig. 2b

7. THEVENIN’S THEOREM D.C. NETWORK
STEP 3: Draw the Thevenin’s Equivalent Circuit
STEP 4: Connect 5 ohm resistor in the circuit as shown in fig.2d and calculate “ i ”
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Vth = 3
Rth = 4.33 ohm
Fig. 2c
Vth = 3
Rth = 4.33 ohm
RL = 5 ohm
i
: Thevenin’s Equivalent circuit
Fig.2d
0.322Ai
A322.0
33.9
3
533.4
3
RR
V
i
Lth
th







8. STEP 4: Calculation of Voltage drop across 5 ohm resistor :
STEP 5: Calculation of Power loss across 5 ohm resistor :
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THEVENIN’S THEOREM D.C. NETWORK
1.62V5*0.322R*iV5ohm 
0.518w5*(0.322)R*(i)LossPower 22


9. THANK YOU
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