elect principles 2 thevenin theorem
TRANSCRIPT
DC Networks – Thevenin Theorem
An active network having two terminals A and B can be replaced by a constant voltage source having an emf, E and an internal resistance, r.
The value of E is the open-circuit voltage between A and B and r is the equivalent resistance of the circuit with load removed and the sources of
emf replaced by their internal resistances.
Thevenin theorem can be used to understand the effect of output resistance on a voltage source under load conditions.
E
R1
A
B
R2
rint RTH
VTH
A
B
= V
DC Networks – Thevenin Theorem
Use the following procedure to obtain the Thevenin equivalent for the circuit shown, determine the terminal voltage VT and current through the load R3.
1. remove the load resistance from the circuit,
2. determine the open circuit voltage, VTH, across the break,
AR2 = 4
B
E = 20V
R1= 8 R3 = 15rint = 2VT
(use the potential divider formula)
R1= 8
A
B
R2 = 4
VTH
rint = 2
E = 20V
DC Networks – Thevenin Theorem
3. remove each source of emf and replace them with their internal resistances
4. dermine the Thevenin equivalent resistance, RTH ‘looking in’ at the break,
5. replace the load to the Thevenin equivalent circuit and determine the network parameters.
A
B
rint = 2 R1= 8
R2 = 4
RTH
B
A
RTH
RLOAD
VTH I
V
Activity
1. Use Thevenin theorem to derive the terminal voltage VT and load current.
DC Networks – Thevenin Theorem
AR2 = 2
R=20rint = 1Ω
E = 48V
R1=10 VT
B
2. Use Thevenin theorem to derive the terminal voltage VT and load current.
RL= 5VB = 24V
3
12 V
A
B
6 VT
Obtain the Thevenin equivalent circuit for the network shown and determine the value of load resistor required for a current of 0.5A to flow between terminals AB.
DC Networks – Thevenin Theorem
-
+
R1 = 4Ω
B
A
R2 = 6Ω
pd across R1 and R2 = 5 – 2 = 3V
-
+
1 2 Using the potential divider theorem
pd across R2 = 3 x
pd across R2 = 3 x 6
4 + 6
R2
R1 + R2
= 1.8V
pd across R1 = 3 – 1.8 = 1.2V
Determine the voltage across R1 and R2
R1 = 4Ω
E1 = 5V
E1 = 5V E2 = 2V
B
AR2 = 6Ω
E2 = 2V
Redraw the circuit indicating the pds and their polarity ( + ve side to highest source).
DC Networks – Thevenin Theorem3
path 1 = 5 – 1.2 = 3.8V
path 2 = 2 + 1.8 = 3.8V
ETH = 3.8V
-
+
4Ω
B
A
6Ω
VR1 = 1.2V
5V-
+
2V
VR2 = 1.8V
+ + --
The sum of the pd’s must be the same along both paths indicated,
tracing each path gives;
path1
path2
Replace each voltage source with its internal resistance (zero in this case). 4
Looking into the circuit from the terminals AB, the 6Ω and 4Ω then appear in parallel.
4 x 6
4 + 6 = 2.4 ΩRTH =
DC Networks – Thevenin Theorem
4Ω
B
A
6Ω
Replace each voltage source with its internal resistance (zero in this case). 4
Looking into the circuit from the terminals AB, the 6Ω and 4Ω then appear in parallel.
4 x 6
4 + 6 = 2.4 ΩRTH =
4
Draw the Thevenin equivalent circuit. 5 For a load current IL of 0.5A to flow. 6
RTOTAL = ETH
IL
=
3.8
0.5 = 7.6 Ω
RL = RTOTAL – RTH = 7.6 – 2.4 = 5.2Ω
-
+
RTH = 2.4ΩA
B
ETH 3.8V