Circuit Theory 1 · Thevenin and Norton

#27 Thevenin and Norton #27 — 1 V test source and a second dependent-source example

Uses a 1 V test source and a short circuit to find the Thevenin resistance of two dependent-source circuits.

Question

First circuit with a 20 V independent source, 100 Ω and 10 Ω resistors, and a 9i_b dependent current source controlled by i_b. — V_Th=10 V is already known; R_Th will now be found with a 1 V test source.

Verify R_Th in the first circuit with a 1 V test source. For the second circuit, find V_Th, I_sc, R_Th, and the Norton equivalent seen by load R_3.

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. Example 1 recap
V_Th=10 V is already known; R_Th will now be found with a 1 V test source.
Example 1: where we left off
$\displaystyle V_{T}h=10 V$
Dependent source: 9i_b
Earlier methods: I_sc and 1 A
New method: 1 V test voltage
Turn off the 20 V source
Keep the dependent source
Narration transcript
In the previous lesson we found the Thevenin equivalent of this dependent-source circuit using short-circuit current and a one amp test source. Here we keep the same circuit, but use the third resistance method: add a one volt test voltage source at terminals a and b.
2. 1 V test source
i_b=−0.01 A and i_x=0.2 A, so R_Th=5 Ω.
1 V test source
20 V → short circuit
V_ab=1 V (+ at a)
$\displaystyle i_{b}=\frac{0-1}{100}$
$\displaystyle i_{b}=-0.01 A$
$\displaystyle 10i_{b}+i_{x}=\frac{1}{10}$
Test current: i_x
Narration transcript
First, the open-circuit voltage is still V Thevenin equal to 10 volts. To find R Thevenin with the voltage-source method, turn off the independent 20 volt source, so it becomes a short circuit. Then connect a 1 volt source from b to a, with the positive terminal at a.
3. Test-voltage result
i_b=−0.01 A and i_x=0.2 A, so R_Th=5 Ω.
Test-voltage result
$\displaystyle 10i_{b}+i_{x}=\frac{1}{10}$
$\displaystyle 10\left(-0.01\right)+i_{x}=0.1$
$\displaystyle i_{x}=0.2 A$
$\displaystyle R_{T}h=\frac{V_{t}}{i_{x}}$
$\displaystyle R_{T}h=\frac{1}{0.2}$
$\displaystyle R_{T}h=5 \Omega$
Narration transcript
The terminal node is now at 1 volt. The control current is i b equals zero minus one over 100, so i b is negative 0.01 ampere. KCL at the top node gives 10 i b plus i x equals one over 10. Substituting i b gives i x equal to 0.2 ampere. Therefore R Thevenin is one volt divided by 0.2 ampere, or 5 ohms.
4. Example 2
V_x is the voltage of node a relative to b and the load voltage across R_3.
Example 2: dependent voltage source
Load: R₃
Controlling voltage:
$\displaystyle V_{x}=V_{\mathrm{ab}}$
Dependent source: 2V_x
Resistors: 15 Ω and 5 Ω
Independent source: 2 A
Required: V_Th and R_Th
Narration transcript
Now use a second example. We want the Thevenin equivalent seen by the load resistor R 3. The circuit has a dependent voltage source of value 2 V x, a 15 ohm resistor, a 2 amp current source, and a 5 ohm branch. The voltage V x is the terminal voltage across R 3, from a to b.
5. Open circuit
The node equation gives V_x=V_Th=15 V.
Open-circuit voltage
Remove R₃
$\displaystyle V_{x}=V_{T}h$
$\displaystyle \frac{V_{x}-2V_{x}}{15}+\frac{V_{x}}{5}-2=0$
$\displaystyle \frac{2V_{x}}{15}=2$
$\displaystyle V_{x}=15 V$
$\displaystyle V_{T}h=15 V$
Narration transcript
Remove the load R 3 and find the open-circuit voltage. Since V x is the same as the terminal voltage, write KCL at the top node: V x minus 2 V x over 15, plus V x over 5, minus 2 equals zero. This simplifies to 2 V x over 15 equals 2, so V x equals 15 volts. Thus V Thevenin is 15 volts.
6. Short circuit
The full 2 A source current becomes the short-circuit current from a to b.
Short-circuit method
Short terminals a-b
$\displaystyle V_{x}=0$
2V_x=0 V → wire
No current in the 5 Ω branch
$\displaystyle I_{\mathrm{sc}}=2 A$
Current direction: a→b
Narration transcript
For the short-wire method, short terminals a and b. The terminal voltage V x becomes zero. That makes the dependent voltage source 2 V x equal to zero volts, so it acts like a short circuit, and the 5 ohm branch also has zero volts. The 2 amp current source now sends 2 amperes through the short.
7. Equivalent circuit
The Norton equivalent of example 2 is 2 A in parallel with 7.5 Ω.
Example 2 equivalent
$\displaystyle R_{T}h=\frac{V_{T}h}{I_{\mathrm{sc}}}$
$\displaystyle R_{T}h=\frac{15}{2}$
$\displaystyle R_{T}h=7.5 \Omega$
Thevenin: 15 V in series with 7.5 Ω
$\displaystyle I_{N}=\frac{15}{7.5}=2 A$
Norton: 2 A in parallel with 7.5 Ω
Narration transcript
The short-circuit current is 2 amperes. Therefore R Thevenin is V Thevenin divided by I short, equal to 15 volts divided by 2 amperes. The result is 7.5 ohms. The Thevenin model is a 15 volt source in series with 7.5 ohms, feeding the load R 3.
8. Method summary
The Norton equivalent of example 2 is 2 A in parallel with 7.5 Ω.
Method summary
Turn off independent sources
Keep dependent sources active
Test voltage:
$\displaystyle R=\frac{V_{t}}{I_{t}}$
Test current:
$\displaystyle R=\frac{V_{t}}{I_{t}}$
Short circuit:
$\displaystyle R=\frac{V_{T}h}{I_{\mathrm{sc}}}$
All three methods give the same R_Th
Narration transcript
The pattern is now complete. With dependent sources, do not turn off the dependent source by itself. For a test source, turn off only the independent sources and keep the controlling variable in the equations. For a short-circuit method, use V Thevenin divided by I short. Both approaches give the same resistance when the signs are tracked consistently.