Circuit Theory 1 · Thevenin and Norton

#29 Thevenin and Norton #29 — Dependent sources only

Uses a 1 A test source to find R_Th in two networks that contain dependent sources only.

Question

Apply a 1 A test source to two a-b port networks containing dependent sources only; find each R_Th and explain V_Th.

Written solution and narration transcript(shows the full solution)

Below are all the lines written in the notebook together with the full narration transcript.

1. Why a test source?
Example 1 uses a dependent voltage source; example 2 uses a dependent current source.
Why a test source?
Dependent sources only
No independent excitation
$\displaystyle V_{T}h=0 V$
I_t=1 A, direction b→a
$\displaystyle V_{t}=V_{\mathrm{ab}}$
$\displaystyle R_{T}h=\frac{V_{t}}{I_{t}}$
Narration transcript
Here the circuit contains only dependent sources. That changes the question. There is no independent source that can create a definite open circuit voltage by itself, so we do not start by hunting for V Thevenin. Instead, we measure the resistance seen from terminals a and b. The clean way is to attach a test source at the port. In these two examples we will inject 1 ampere from b to a, compute the resulting terminal voltage, and then use R Thevenin equals V test divided by I test.
2. Example 1 — circuit
V_1 is measured across the right 6 Ω resistor from the middle node toward terminal a.
Example 1 — circuit
Dependent voltage source: 1.5V₁
Resistors: 4 Ω, 6 Ω, 6 Ω
V₁: + at middle node, − at a
Test source: 1 A
Direction: b→a
Port voltage: V_y
Narration transcript
Example one has a dependent voltage source equal to 1.5 V 1. The voltage V 1 is across the right 6 ohm resistor, with plus on the left and minus on the terminal a side. The port is still a and b. Since the network has no independent source, the target is only R Thevenin seen at a b. We connect a 1 ampere test source upward, from b to a, and call the port voltage V y.
3. Example 1 — variables
V_x=V_1+V_y and V_y=3 V, so R_Th=3 Ω.
Example 1 — variables
Middle node: V_x
Port node:
$\displaystyle V_{y}=V_{\mathrm{ab}}$
$\displaystyle V_{1}=V_{x}-V_{y}$
$\displaystyle V_{x}=V_{1}+V_{y}$
Left node: 1.5V₁
Dependent source stays active
Narration transcript
Now slow down and define the node voltages. Let the middle top node be V x. Because the 6 ohm resistor on the right has voltage V 1 from left to right, and the port node is V y, we have V 1 equals V x minus V y. The same statement can be written as V x equals V 1 plus V y. That substitution keeps the dependent source and every resistor current consistent.
4. Example 1 — solution
V_x=V_1+V_y and V_y=3 V, so R_Th=3 Ω.
Example 1 — solution
$\displaystyle \frac{V_{x}-1.5V_{1}}{4}+\frac{V_{x}}{6}+\frac{V_{x}-V_{y}}{6}=0$
$\displaystyle 5V_{y}+2.5V_{1}=0$
$\displaystyle V_{y}=-0.5V_{1}$
$\displaystyle \frac{V_{y}-V_{x}}{6}-1=0$
$\displaystyle V_{1}=-6 V, V_{y}=3 V$
$\displaystyle R_{T}h=3 \Omega$
Narration transcript
Write KCL at the middle top node first. The current through 4 ohms is V x minus 1.5 V 1, all over 4. The current down through the middle 6 ohms is V x over 6. The current through the right 6 ohms is V x minus V y over 6, which is V 1 over 6. Substitute V x equals V 1 plus V y. After multiplying by 12, the equation becomes 5 V y plus 2.5 V 1 equals zero, so V y equals negative one half V 1. Now write KCL at terminal a. Current from a back through the right 6 ohm resistor is V y minus V x over 6, and the 1 amp test current enters the node, so V y minus V x over 6 minus 1 equals zero. That gives V 1 equal to negative 6 volts. Therefore V y is 3 volts. Since the test current is 1 ampere, R Thevenin is 3 ohms.
5. Example 2 — circuit
$Second a-b port network with a leftward 2V_{ab} dependent current source, 4 Ω, and 2 Ω resistors.$
V_{ab} is the voltage of terminal a relative to b.
Example 2 — circuit
Dependent current source: 2V_ab
Arrow points left from node a
Port branch: 2 Ω
Return branch: 4 Ω
I_t=1 A, direction b→a
$\displaystyle V_{x}=V_{\mathrm{ab}}$
Narration transcript
Example two looks different, but the idea is the same. The dependent source is now a current source of value 2 V a b, and its arrow points to the left. The port voltage V a b is positive at a and negative at b. Again, there is no independent source, so we attach the same 1 ampere test source from b to a and solve for the voltage that appears at the port.
6. Example 2 — solution
$Second network with a 1 A test source from b to a and port voltage V_x=V_{ab}.$
KCL gives V_x=0.4 V and R_Th=0.4 Ω.
Example 2 — solution
KCL at node a
$\displaystyle \frac{V_{x}}{2}+2V_{\mathrm{ab}}-1=0$
$\displaystyle V_{x}=V_{\mathrm{ab}}$
$\displaystyle 2.5V_{x}=1$
$\displaystyle V_{x}=0.4 V$
$\displaystyle R_{T}h=0.4 \Omega$
Narration transcript
Use the top right node, which is terminal a. Let V x equal V a b. Current down through the 2 ohm resistor is V x over 2. The dependent current source sends 2 V a b away from this node toward the left. The 1 ampere test source enters the node. KCL is therefore V x over 2 plus 2 V a b minus 1 equals zero. Since V x is the same as V a b, we have V x over 2 plus 2 V x equals 1. That is 2.5 V x equals 1, so V x equals 0.4 volts. With a 1 ampere test current, R Thevenin is 0.4 ohms.
7. Method summary
Both networks have V_Th=0 V and use a test source to measure R_Th.
Method summary
Source-free linear network:
$\displaystyle V_{T}h=0$
Keep dependent sources active
Attach a test source at the port
$\displaystyle R_{T}h=\frac{V_{t}}{I_{t}}$
Example 1: 3 Ω
Example 2: 0.4 Ω
Narration transcript
The takeaway is simple. When a port network has dependent sources but no independent source, do not invent a Thevenin voltage. Measure the port resistance. Keep every dependent source active, apply a test source, write KCL with the source direction included, and divide the measured port voltage by the test current. In example one the resistance is 3 ohms. In example two it is 0.4 ohms.