Thevenin’s theorem provides a useful tool when solving complex and large electric circuit by reducing them to a single voltage source in series with a resistor. It is particularly advantageous where a single resistor or load in a circuit is subject to change.
Formally, the Thevenin’s theorem can be stated as
"Any two terminal linear electric circuit consisting of resistors sources, can be replaced by an equivant circuit containing a single voltage source in series with a resistor connected across the load."
In the circuit diagram shows in Figure, the current IL through the load resistance RL is same. Hence the circuits are equivalent as far as the load resistor RL is concerned.
The following steps outline the procedure to simplify an electric circuit using Thevenin’s theorem where Vth and Rth are the Thevenon’s voltage and Thevenin’s resistance respectively.
To evaluate RTh, remove the load resistance of 5 Ohm and replace the 10V voltage source by its internal resistance as depicted in Fig(a).
So we determine of Thevenin’s voltage,VTH and Thevenin’s resistance, RTH.
RTH is then given by
RTH =(10||20)+15 =(10x20)/(10+20)+15 = 21.67 Ohm
To determine VTH or V,sub>OC across the OC load terminals, note in Fig(b) that the voltage across the 20 Ohm resistance is the same as VOC since the right loop is open and no voltage drop occurs across the 15 Ohm resistance (I15 Ohm =0)
The 10 Ohm and 20 Ohm resistors are in series therefore VDR (Voltage Divider Rules) can be employed to determine VOC
VTH = VOC =10x20/(10+20) = 6.67V
The Thevenin’s equivalent circuit can now be drawn as show in below Fig. The load resistance of 5 Ohm is inserted back between the terminals A and B and the load current can be found as follows
Please read more: Norton’s theorem and Voltage and Current Laws part1
"Any two terminal linear electric circuit consisting of resistors sources, can be replaced by an equivant circuit containing a single voltage source in series with a resistor connected across the load."
In the circuit diagram shows in Figure, the current IL through the load resistance RL is same. Hence the circuits are equivalent as far as the load resistor RL is concerned.
- Remove the load resistance RL
- VTH is the open circuit (OC) voltage across the terminals and
- RTH is the resistance across the load terminal with all source replaced by their internal resistances
VTH=VOC and RTH =(VOC/ISC)
Example:Use Thevenin’s theorem to find the current through the 5 Ohm resistance in the circuit diagram as below.
RTH is then given by
RTH =(10||20)+15 =(10x20)/(10+20)+15 = 21.67 Ohm
To determine VTH or V,sub>OC across the OC load terminals, note in Fig(b) that the voltage across the 20 Ohm resistance is the same as VOC since the right loop is open and no voltage drop occurs across the 15 Ohm resistance (I15 Ohm =0)
The 10 Ohm and 20 Ohm resistors are in series therefore VDR (Voltage Divider Rules) can be employed to determine VOC
VTH = VOC =10x20/(10+20) = 6.67V
The Thevenin’s equivalent circuit can now be drawn as show in below Fig. The load resistance of 5 Ohm is inserted back between the terminals A and B and the load current can be found as follows
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