Why is the internal resistance of the source zero in an ideal voltage source? Why does a practical voltage source have finite internal resistance?
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$\begingroup$ 1. Because it's easier to calculate (“ideal” is defined by this property) 2. Because the electrons need to flow through a material that's not a superconductor. Every material has resistance because electrons get scattered by the nuclei. $\endgroup$– Lukas BernsCommented Feb 4, 2018 at 15:40
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$\begingroup$ @LukasBerns - Electrons in conductors don't get scattered by the nuclei. The are scattered by deviations from the ideal crystal lattice periodicity. The most important are crystal vibrations (phonons) and dopants. $\endgroup$– freecharlyCommented Feb 4, 2018 at 15:53
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$\begingroup$ @freecharly Agreed. Controversially, perhaps, what is in essence Drude's theory is sometimes taught at A-level in the UK. I think that's alright as long as the health warnings are stressed – which they may not always be. $\endgroup$– Philip WoodCommented Feb 4, 2018 at 23:14
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3 Answers
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Because you don't want the pd of the source to change when you connect a load across the source. If it's supposed to be a 12 V source, that's what you want it to be.
A real voltage source can be modelled as a fixed emf (energy supplied per coulomb passing) in series with an unavoidable internal resistance. When you connect a load across the source, you cause a current: charge flows through the source of emf and the internal resistance (r) as well as the load itself – they're all in series! The current through the internal resistance causes a voltage drop (Ir) across it, which means that the source voltage (as measured across the terminals of the source) is (the emf – Ir), that is less than the emf !
Ideally, then, r should be zero. Then there'll be nothing to subtract from the emf, even when there's a load causing a current.
answered Feb 4, 2018 at 15:40
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A practical voltage source contains a nonvanishing internal resistance because it contains real-world (nonideal) components like resistors, transformers, inductors and capacitors. To take one example from this list, consider the transformer (a popular component in voltage supplies). Its job is to "gear up" or "gear down" the AC mains voltage to a value which upon rectification and filtering will produce the desired DC voltage. It contains windings of copper wire which, although copper is a conductor, exhibit nonzero resistance for the reasons cited by others above. Pulling large amounts of current through a transformer thus causes the windings to warm up, which increases the wire's resistance, leading to significant voltage drop. The result: the power supply is incapable of furnishing rated voltage for low resistance loads, and behaves like an ideal voltage source in series with an internal resistance.
answered Feb 4, 2018 at 19:14
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Why is the internal resistance of the source zero in an ideal voltage source?
For a voltage source with finite internal resistance $r$, the voltage $v_s$ across the terminals of the source is, by KVL, just
$$v_S = V_0 - i_S\cdot r$$
where $V_0$ is the open-circuit voltage across the voltage source and $i_S$ is the current out of the voltage source.
An ideal voltage source is defined as the circuit element that has a voltage $v_S$ across that is independent of the current $i_S$ through. Clearly, this can only be the case when the internal resistance $r = 0$.
Why does a practical voltage source have finite internal resistance?
An ideal voltage source can supply arbitrarily large current and thus, arbitrarily large power. No physical voltage source can supply arbitrarily large current and power; an ideal voltage source would be an essentially unlimited energy source.
answered Feb 5, 2018 at 3:18