EE 2115-01 - Lec 08 - 2026_09_11
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Overview
Hiren Trada finishes the source-transformation material by showing that Thevenin and Norton equivalents must preserve the same terminal voltage-current relationship, then applies series/parallel reduction and voltage division to circuit examples. Worked results include a 1.5 kΩ load dissipating about 2.48 mW, a Thevenin equivalent of about 1.9 V with 5.45 kΩ, and a final example illustrating how source polarity determines whether currents cancel or reinforce.
Key takeaways
- A source transformation is valid only when the source includes its associated resistance: series for a voltage source and parallel for a current source.
- Two networks are equivalent at their terminals when they produce the same voltage-current relationship for the same external test, regardless of their internal components.
- A junction does not make components parallel by itself; current must have an actual closed path to split, so open branches can leave components in series.
- For a series string, voltage division gives each resistor a share of the supply proportional to its resistance, such as VR1 = VS × R1/(R1 + R2).
- When load resistance is known, calculate its power by finding just its current or voltage, then applying P = I²R or P = V²/R; the 1.5 kΩ example yields about 2.48 mW.
- Source polarity controls whether transformed sources cancel or reinforce, so current directions and voltage polarities must remain consistent through every reduction.
Chapters
- A Thevenin equivalent represents a network with one voltage source and a series resistance; a Norton equivalent uses one current source and a parallel resistance.
- Source transformation requires the associated resistance: voltage sources need series resistance, while current sources need parallel resistance.
- Thevenin and Norton source values follow Ohm's law, V = IR, while the resistance remains the same in either equivalent.
- In the example with a voltage source and R1 and R2, the output voltage VAB is measured across R1, not directly across the supply.
- R1 and R2 carry the same current because the apparent connection beside R1 provides no closed path for current to branch into.
- Circuit elements are in series when all current through one must also pass through the other, even if extra wires or open branches meet at their junction.
- Transforming the original voltage source and its series resistance produces a current source with a parallel resistance, allowing parallel resistors to be combined.
- For series resistors, voltage division gives VR1 = VS × R1/(R1 + R2); the voltage across each resistor is proportional to its resistance.
- Thevenin and Norton forms provide equivalent ways to simplify a circuit while retaining the same terminal behavior.
- The example combines a 1 mA source with 1 kΩ and a 2 mA source with 1.2 kΩ by first transforming each source-resistor pair into a voltage source and series resistor.
- The transformed voltages, 1 V and 2.4 V, add to 3.4 V, while the series resistances add to 2.2 kΩ.
- Transforming the combined 3.4 V and 2.2 kΩ network back gives about 1.54 mA, with 2.2 kΩ in parallel.
- For the mixed-source example, the 1.5 kΩ resistor is treated as the load, and the requested quantity is its power.
- Power can be found from P = VI, P = I²R, or P = V²/R; because the load resistance is known, finding either its current or voltage is sufficient.
- Hiren Trada frames choosing the calculation route as an engineering efficiency decision: use the method that requires the fewest unknowns and remains easy to verify.
- The 5 V source and its 2 kΩ series resistance transform to a 2.5 mA source in parallel with 2 kΩ.
- That 2 kΩ branch combines in parallel with another 2 kΩ resistance to make 1 kΩ; alternatively, voltage division reduces the original branch to a 2.5 V source.
- The circuit is redrawn with the transformed branch connected at the same two breakpoints, preserving the original network's external connections.
- After reducing the branches, the example forms a 4.5 V source with 2 kΩ of series resistance, connected to the 1.5 kΩ load.
- The total resistance is 3.5 kΩ, so the circuit current is about 1.29 mA and voltage division gives about 1.93 V across the load.
- Using P = I²R or P = V²/R gives approximately 2.48 mW, illustrating how a second method can check a result.
- The network has 2 V with 10 kΩ and 3 V with 20 kΩ; source transformation gives 0.2 mA and 0.15 mA, respectively.
- The parallel current sources combine to 0.35 mA, and the worked reduction combines the 10 kΩ, 20 kΩ, and 30 kΩ resistor branches to about 5.45 kΩ.
- Transforming the Norton form back gives a Thevenin voltage of about 1.9 V; stopping at the current-source stage would instead provide the Norton equivalent.
- In the final example, transforming a 10 V source with 5 Ω gives a 2 A current source; its direction relative to the existing 2 A source determines whether the sources cancel or reinforce.
- With equal 2 A sources directed oppositely, the worked analysis concludes that the 3 Ω load current is zero.
- The example also considers current division between equal 5 Ω branches and a voltage-source transformation, emphasizing that consistent polarity and reference directions are essential.
Transcript excerpt
0:00 Good afternoon everyone. Uh Friday lecture 8. We still have a few slides of lecture seven to complete. So we'll get that uh done first and then we'll start lecture eight. Um seven and eight are essentially the same topic. Seven and uh and non equivalent circuits. Uh in lecture eight slides we solve uh some examples based on the theory that we learned in lecture seven. [clears throat] So let's bring this up here and we had stopped here. Any questions from anyone regarding this topic before I proceed. Okay. All right. So [clears throat and cough]
0:47 the last uh point I had made in the theory was we have a large circuit large electrical system and we can make a equivalent circuit um where we can describe the large circuit as a smaller equivalent circuit. That equivalent circuit is known as thean equivalent circuit where we represent the entire s all the sources in the entire circuit as one single source and all the resistors in the circuit as one single resistance and then few decades later came another uh theory was Norton's theorem and Norton's theorem stated that we could represent everything as a single current source and then the and norton are equivalent to each other so we had the concept of source transformation from there so So in source transformation what we are
1:35 doing is we can replace a voltage source and a current source and vice versa. But we cannot substitute a ideal voltage source directly with a ideal current source or a current source with a voltage source. There must be a resistance in there. Okay. So real life sources yes we can transform we can perform theoretical calculations and analysis. If we have a resistor in series with a voltage source and a
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Related notes
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Hiren Trada.