Download 1250 LECTURE #3-5 correction EX: The following are corrected

1250
EX:
LECTURE #3-5 correction
The following are corrected equations for the example problem in video lecture
http://youtu.be/fUqX62tbrxE on finding Thevenin equivalent resistance using an
external voltage source. Here is the circuit:
SOL'N:
We turn off the independent source, 68 V, which becomes a wire, but we
keep the dependent source turned on. We then attach a 1 V source to the
output and determine the current flowing into the a terminal, as shown in
the following diagram:
Note that a wire should be drawn through the 68 V source. We now use
the node-voltage method to find Va and Vc, and from those voltages we
will determine Iex and then RTh.
Because of the 1 V source, we have Va = 1 V. So we only need an
equation for Vc. We sum the currents measured flowing out of Vc. Note
how we write Ix in terms of node voltages.
⎛ V − 1V ⎞
Vc − 4 ⎜ c
⎝ 6Ω ⎟⎠ Vc − 1V
Vc − 0V
+
+
= 0A
6Ω
2Ω
6Ω
Now we collect the terms multiplying Vc and put constants on the right
side of the equation.
⎛ 1
1
4
1 ⎞ −4 V 1V
Vc ⎜
+
−
+
=
+
⎝ 6 Ω 2 Ω 12 Ω 6 Ω ⎟⎠ 12 Ω 6 Ω
To clear the denominators, we multiply by the common denominator.
⎛ 1
⎛ −4 V 1V ⎞
1
4
1 ⎞
Vc ⎜
+
−
+
⋅12 Ω = ⎜
+
⋅12 Ω
⎟
⎝ 6 Ω 2 Ω 12 Ω 6 Ω ⎠
⎝ 12 Ω 6 Ω ⎟⎠
or
Vc ( 2 + 6 − 4 + 2 ) = −4 V+2V
or
6Vc = −2V
or
1
Vc = − V
3
Now we use Kirchhoff's current law to find find Iex. We sum the currents
out of the Va node.
1V − Vc 1V
+
− I ex = 0 A
6Ω
4Ω
or
1
1V − − V 1V
3 +
= I ex
6Ω
4Ω
or
4 1
+ A = I ex
18 4
or
9⎞
17
⎛ 8
I ex = ⎜
+ ⎟A=
A
⎝ 36 36 ⎠
36
Now we find RTh by dividing the voltage of the external source by the
current Iex.
RTh =
1V
36
=
Ω ≈ 2.12 Ω
17
17
A
36