#24 Thevenin and Norton equivalents #24 — independent sources

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

1. 1. Terminal equivalence

   

   Look from terminals a-b

   Complex linear network → simple equivalent

   Thevenin: V_Th in series with R_Th

   Norton: I_N in parallel with R_N

   $\displaystyle R_{N}=R_{T}h$

   $\displaystyle I_{N}=V_{T}h/R_{T}h$

   The load sees the same v-i behavior

   Narration transcript

   We now begin Thevenin and Norton equivalents. The goal is to replace everything connected to two terminals, a and b, by a simpler circuit that behaves the same from the outside. If the equivalent is one voltage source in series with one resistance, we call it a Thevenin equivalent. If it is one current source in parallel with one resistance, we call it a Norton equivalent. In both cases the load sees the same terminal voltage-current behavior.

2. 2. Independent-source procedure

   

   Independent-source procedure

   1) Remove the load; leave a-b open

   2) Voltage source → short circuit

   3) Current source → open circuit

   4) Find R_Th looking into the terminals

   $\displaystyle 5) V_{T}h=V_{o}c$

   6) Reconnect the load to the equivalent

   Narration transcript

   For circuits with only independent sources, the procedure is direct. First, remove the load. To find R Thevenin, turn off the independent sources: ideal voltage sources become short circuits, and ideal current sources become open circuits. Then look into terminals a and b. Second, find V Thevenin as the open-circuit voltage across a and b. Third, draw V Thevenin in series with R Thevenin. If a Norton form is needed, use I Norton equals V Thevenin divided by R Thevenin, with the same resistance in parallel.

3. 3. Example 1: R_Th

   

   Example 1: R_Th

   12 V → short circuit

   1 A → open circuit

   $\displaystyle 3 \Omega ∥ 6 \Omega =2 \Omega$

   8 Ω is in series with this pair

   $\displaystyle R_{T}h=8+2$

   $\displaystyle R_{T}h=10 \Omega$

   Narration transcript

   Example one has a 12 volt source, a 3 ohm resistor on the top branch, a 6 ohm resistor to the bottom node, a 1 amp current source pointing upward, and an 8 ohm resistor leading to terminal a. The load is removed from terminals a and b. For R Thevenin, turn off the sources. The 12 volt source becomes a short circuit, and the 1 amp source becomes an open circuit. Looking from a to b, the 8 ohm resistor is in series with 3 ohms in parallel with 6 ohms. So R Thevenin is 8 plus 2, equal to 10 ohms.

4. 4. Example 1: V_Th

   

   Example 1: V_Th

   Open load ⇒ i_8Ω=0

   $\displaystyle V_{a}=V_{T}h$

   $\displaystyle \frac{V_{T}h-12}{3}+\frac{V_{T}h}{6}-1=0$

   $\displaystyle 2V_{T}h-24+V_{T}h-6=0$

   $\displaystyle 3V_{T}h=30$

   $\displaystyle V_{T}h=10 V$

   Narration transcript

   Now find V Thevenin for example one. With the load open, no current flows through the 8 ohm resistor, so terminal a has the same voltage as the top middle node. Call that voltage V Thevenin. KCL at that node gives V Thevenin minus 12 over 3, plus V Thevenin over 6, minus 1 amp, equal to zero. Multiplying by 6 gives 2 V Thevenin minus 24 plus V Thevenin minus 6 equals zero. Therefore 3 V Thevenin equals 30, and V Thevenin is 10 volts.

5. 5. Thevenin and Norton forms

   

   Example 1 equivalents

   Thevenin: 10 V in series with 10 Ω

   $\displaystyle I_{N}=V_{T}h/R_{T}h$

   $\displaystyle I_{N}=\frac{10}{10}=1 A$

   Norton: 1 A in parallel with 10 Ω

   The same terminals a-b are preserved

   $\displaystyle V_{T}h=10 V, R_{T}h=10 \Omega , I_{N}=1 A$

   Narration transcript

   The Thevenin equivalent of example one is a 10 volt source in series with 10 ohms, seen at terminals a and b. The Norton equivalent is obtained by dividing V Thevenin by R Thevenin. 10 volts divided by 10 ohms gives 1 amp. So the Norton form is a 1 amp source in parallel with 10 ohms. These two forms are equivalent at the same terminals.

6. 6. Example 2: load current

   

   Example 2: current through 6 Ω

   $\displaystyle R_{T}h=4+\left(6 \Omega ∥ 12 \Omega \right)=8 \Omega$

   $\displaystyle \frac{V}{6}+\frac{V-12}{12}+\frac{V-V_{T}h}{4}=0$

   $\displaystyle \frac{V_{T}h-V}{4}-3=0$

   $\displaystyle V=16 V$

   $\displaystyle V_{T}h=28 V$

   $\displaystyle i=\frac{28}{8+6}=2 A ↓$

   Narration transcript

   The second source page uses the same idea to find the current through a 6 ohm load. After removing the load, the resistance seen from a and b is 4 ohms in series with 6 ohms parallel 12 ohms. The parallel part is 4 ohms, so R Thevenin is 8 ohms. For the open-circuit voltage, use two node equations. At the left top node V: V over 6 plus V minus 12 over 12 plus V minus V Thevenin over 4 equals zero. At the right top node: V Thevenin minus V over 4 minus 3 equals zero. Solving gives V equals 16 volts and V Thevenin equals 28 volts. With the 6 ohm load reconnected, the current is 28 divided by 8 plus 6, equal to 2 amperes downward.

7. 7. Method summary

   

   Thevenin-Norton checklist

   Viewpoint: the load terminals

   R_Th: turn off independent sources

   V_Th: open-circuit terminal voltage

   $\displaystyle I_{N}=V_{T}h/R_{T}h$

   Preserve directions and polarities

   Examples: 10 V/10 Ω and 28 V/8 Ω

   Narration transcript

   The main habit is to treat the load terminals as the viewpoint. Remove the load, find R Thevenin by turning off independent sources, find V Thevenin as the open-circuit terminal voltage, and then reconnect the load to the equivalent. For example one, R Thevenin is 10 ohms and V Thevenin is 10 volts, so the Norton current is 1 amp. For the load-current example, R Thevenin is 8 ohms, V Thevenin is 28 volts, and the load current is 2 amperes.

Source video: Circuit Theory #24 | Thevenin and Norton Equivalents - Introduction and Example (4:45)