Calculate the Thevenin equivalent as seen
between terminals A and B for the following
circuit:
6kΩ
15kΩ
12V
A
3kΩ
B
•
We want to replace the whole circuit with a simpler
one which looks like this, and find the values of Vth
and Rth:
Rth=?
A
Vth=?
B
•
We use the fact that both the original and the
equivalent circuits should behave identically when
viewed at terminals A and B.
ELEC166 Worked Examples Week 3
1
•
First consider the open-circuit voltage between A
and B for both circuits.
In the original circuit, the 6kΩ||15kΩ pair and the
3kΩ resistors form a potential divider, so the voltage
between A and B is
3kΩ
12V ×
= 4.94V
3kΩ + 6kΩ || 15kΩ
In the equivalent circuit, since no current flows, the
voltage across Rth is zero. So the voltage between
A and B is just Vth. Hence Vth = 4.94V.
•
We still need to solve for Rth. If we now connect A
directly to B with a wire, identical currents should
flow in this wire in both circuits.
In the original circuit, this short-circuit current will be
I SC =
12V
12V
=
= 2.8mA
6kΩ in parallel with 15kΩ 4.286kΩ
(note that the 3kΩ resistor is shorted out).
In the equivalent circuit, this short-circuit current is
just equal to Vth/Rth. Hence
ELEC166 Worked Examples Week 3
2
Vth 4.94V
=
= 2.8 ×10 −3 A
Rth
Rth
and solving this gives Rth = 1765Ω.
•
Thus the final Thevenin equivalent circuit is:
1.765kΩ
A
4.94V
B
ELEC166 Worked Examples Week 3
3