Thevenizing a Circuit
As an example, refer to Fig.1-a, where we want to find the voltage VL
across the 2Ω RL and its current IL. To use Thevenin’s theorem, mentally
disconnect RL. The two open ends then become terminals a and b.
Now we find the Thevenin equivalent of the remainder of the circuit
that is still connected to a and b. In general, open the part of the circuit
to be analyzed and “thevenise” the remainder of the circuit connected
to the two open terminals.
R1 = 3Ω
a
V
RL = 2Ω
R2 = 6Ω
b
Our only problem now is to find the value of the open-circuit voltage VTH
across ab and the equivalent resistance RTH. The Thevenin equivalent
always consists of a single voltage source in series with a single
resistance, as in Fig 1-d.
The effect of opening RL is shown in Fig 1-b. As a result, the 3Ω R1 and
6Ω R2 form a series voltage divider, without RL.
36V
Fig1(a)
R1 = 3Ω
a
V
Furthermore, the voltage across R2 now is the same as the open-circuit
voltage across terminals a and b. Therefore VR2 with RL open is Vab. This
is the VTH we need for the Thevenin equivalent circuit. Using the voltage
divider formula,
VR2 = 6 /9 x 36 V = 24V
VR2 = Vab = VTH = 24V
36V
Vab = 24V
R2 = 6Ω
b
Fig1(b)
R1 = 3 Ω
a
This voltage is positive at terminal a.
To find RTH , the 2Ω RL is still disconnected. However, now the
source V is short-circuited. So the circuit looks like Fig 1-c.
The 3Ω R1 is now in parallel with the 6Ω R2 , as they are both
connected across the same two points. This combined resistance is
RTH = 18/9 = 2Ω
As shown in Fig 1-d, the Thevenin circuit to the left of terminals a and b
then consists of the equivalent voltage VTH , equal to 24V, in series with
the equivalent series resistance RTH , equal to 2 Ω. This Thevenin
equivalent applies for any value of RL because RL was disconnected. We
are actually thevenizing the circuit that feeds the open ab terminals.
To find VL and IL, we can finally reconnect RL to terminals a and b of the
Thevenin equivalent circuit, as shown in Fig 1-e. Then RL is in series with
RTH and VTH. Using the voltage divider formula for the 2Ω RTH and 2Ω RL,
VL = ½ x 24V = 12V. To find IL as VL / RL, the value is 12V / 2Ω, which
is equal to 6A.
These answers of 6A for IL and 12V for VL apply to RL in both the original
circuit in Fig 1-a and the equivalent circuit in Fig 1-e. Note that the 6A IL
also flows through RTH.
The same answer could be obtained by solving the series-parallel circuit
in Fig 1-a, using Ohm’s Law. However, the advantage of Thevenizing
the circuit is that the effect of different values of RL can be calculated
easily. Suppose that RL were changed to 4Ω. In the Thevenin circuit,
the new value of VL would be 4/6 x 24V = 16V.
The new IL would be 16 V/ 4 Ω, which equals 4A.
In the original circuit, a complete new solution would be required
each time RL was changed.
Short
across V
Rab = 2Ω
R2 = 6 Ω
b
Fig1(c)
RTH = 2Ω
a
VTH = 24V
b
Fig1(d)
RTH = 2Ω
VTH = 24V
a
VL = 12V
Fig1(e)
b
{
RL = 2Ω