Helping me to understand the difference between pd of a source and a component

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SUMMARY

The discussion focuses on the distinction between the potential difference (pd) of a source, such as a battery, and the pd across individual components in a circuit. The user correctly identifies that the pd of a circuit is proportional to its current, as described by Ohm's Law (V=IR). The confusion arises when considering the relationship between the total pd supplied by the battery and the pd drops across components in a series circuit. The key takeaway is that the sum of the voltage drops across all components in a closed loop must equal the supply voltage, which clarifies the user's misunderstanding about energy distribution in the circuit.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Basic knowledge of electrical circuits and components
  • Familiarity with the concept of voltage and potential difference
  • Concept of energy transfer in electrical systems
NEXT STEPS
  • Study Kirchhoff's Voltage Law to understand voltage distribution in circuits
  • Explore series and parallel circuit configurations for deeper insights
  • Learn about energy conservation in electrical circuits
  • Investigate the role of resistance in determining current flow
USEFUL FOR

Students studying electrical engineering, hobbyists building circuits, and anyone seeking to deepen their understanding of electrical potential difference and circuit behavior.

Soul Glo
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Hi, I've recently been reading up on circuit electricity and as the title states have found difficulty understanding the difference between pd of a source and a component from the textbooks I've been reading.

So far I am to believe that the pd of a circuit is directly proportional to its current, fine - I can understand this as higher pd = more energetic electrons so they move around the circuit faster hence higher current... the pd of the battery can be seen as a sort of 'driving' force on the electrons.

However, when defining the pd I am told it is how much energy per unit charge used in passing through a certain component. - The battery supplies the pd and the electrons 'spend' it passing through the components. So the pd of the component can be seen as just how difficult it is, or how much energy is required for a charge to pass through it.

My misunderstanding started when imagining a series circuit in which the pd of the battery was doubled. Since I know the equation V=IR and understand that the pd of the battery is like its driving force I can see that the total current of the circuit would double also.

What I want to know is why the sum of the pds across the components must now add up the the new total pd when the resistance has not increased so the electrons would have no more difficulty in passing through them. Wouldn't the electrons not just be supplied this new doubled amount of energy but then use the same amount as before with half left over? Why does the increase of current seemingly make the electrons need more energy to pass through the same components that they already passed through for half the energy?!

I am sorry if these questions are stupid or I am misunderstanding on some fundamental level but please understand I don't have a teacher I am purely studying this topic for the first time on my own.

Thanks

*EDIT: after some thought I have been able to narrow down my problem to the following statement : 'the sum of the voltage drops across each component of one complete loop of the circuit is equal to the supply voltage.'

If I accept the above as fact it all makes sense, however I still feel that without explanation I can't really understand why this is.
 
Last edited:
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When studying circuits, use voltage which is potential difference between two points, then forget absolute potential. You are using the language of P when talking about PD, and that confuses you.
 
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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