# 1270 EXAM #2 prob 2 solution EX: Find the Thevenin equivalent

```1270
EXAM #2 prob 2 solution
S 11
EX:
Find the Thevenin equivalent circuit at terminals a and b. vx must not appear in your
solution. Note: α ≠ 0.
SOL'N:
The Thevenin equivalent voltage is vab across the a and b terminals with
nothing connected at a and b, with the + sign at a. We observe that, if
nothing is connected across a and b, no current flows in the left side of the
circuit, and the voltage drop, vx, across R 1 equals zero volts. This in turn
means the dependent source turns into an open circuit. Since no current
flows in the dependent source and in R1, no current flows in R 2 and the
voltage across R2 is zero. Thus, the voltage across the a and b
terminals,(i.e., the Thevenin equivalent voltage, vTh), is vs.
vTh = vs
Now we short out the a and b terminals to find current isc flowing from a
to b. We use the node-voltage method, defining a node at the junction of
the R's and dependent source.
We first define vx in terms of the node voltage.
vx = −vs − v1
Now we sum currents out of the node.
v1 − −vs
v
+ α (−vs − v1 ) + 1 = 0 A
R1
R2
or
⎛ 1
⎛
1 ⎞
v
1⎞
v1 ⎜ − α +
= − s + α vs = vs ⎜ α − ⎟
⎟
R2 ⎠
R1
R1 ⎠
⎝ R1
⎝
or
α−
v1 = vs
1
R1
1
1
−α +
R1
R2
= vs
α R1 − 1
R
1 − α R1 + 1
R2
Now we observe that isc is the same as the current through R1.
⎛
⎞
⎜
⎟
v − −vs vs
α R1 − 1
isc = 1
=
+ 1⎟
⎜
R1
R1 ⎜ 1 − α R + R1
⎟
1
R2
⎝
⎠
or
⎛
1 − α R1 +
v1 − −vs vs ⎜ α R1 − 1
isc =
=
+
⎜
R1
R1 ⎜ 1 − α R + R1 1 − α R +
1
1
R2
⎝
R1 ⎞
R2 ⎟
R1 ⎟⎟
R2 ⎠
or
R1
⎛
v ⎜
R2
isc = s ⎜
R1 ⎜ 1 − α R +
1
⎝
1
⎞
⎛
⎟
⎜
R2
= vs ⎜
⎟
R1 ⎟
⎜ 1 − α R1 +
R2 ⎠
⎝
⎞
⎟
⎛
⎞
1
= vs ⎜
⎟
R1 ⎟
⎝ R2 − α R1R2 + R1 ⎟⎠
R2 ⎠
or
⎛
⎞
1
isc = vs ⎜
⎝ R1 + R2 − α R1R2 ⎟⎠
Now we use vTh and isc to find the Thevenin equivalent resistance, RTh.
v
RTh = s =
isc
vs
= R1 + R2 − α R1R2
⎛
⎞
1
vs ⎜
⎝ R1 + R2 − α R1R2 ⎟⎠
Thevenin equivalent:
NOTE:
An alternative approach is to use a current divider to find an
equivalent resistance for the dependent source. The current
through R1 splits into two parts, and we can equate the current
in the dependent source with the current that would flow in an
equivalent resistance, Req.
vx
R2
= α vx
R1 R2 + Req
Solving for Req, we have a value that is free of vx.
⎛ 1
⎞
Req = R2 ⎜
− 1⎟
⎝ α R1 ⎠
We may then replace the dependent source with R eq and
calculate RTh as the resistance seen looking into the circuit
from a and b.
⎡ ⎛ 1
⎞⎤
RTh = R1 + R2 || Req = R1 + R2 || ⎢ R2 ⎜
− 1⎟ ⎥
⎣ ⎝ α R1 ⎠ ⎦
or
1
−1
⎛ ⎛ 1
⎞⎞
α R1
RTh = R1 + R2 ⎜ 1 || ⎜
− 1 ⎟ = R1 + R2
1
⎝ ⎝ α R1 ⎟⎠ ⎠
1+
−1
α R1
or
1
−1
α R1
RTh = R1 + R2
1
α R1
or
RTh = R1 + R2 (1 − α R1 )
```