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Internal resistance and practical source

Short answer

Internal resistance is the factor. For a voltage source, terminal voltage V = E − I·Rs; as load current I rises, the internal IR drop increases, causing V to fall linearly from the open-circuit value E.

What is the factor responsible for the deviation of the practical sources from their ideal behavior? Explain the effect of this factor on the terminal characteristics of the voltage source.

The factor is the internal resistance of the source (Rs for a voltage source, Rp for a current source). An ideal voltage source has Rs = 0 and an ideal current source has Rp = ∞; real sources do not.

Effect on a practical voltage source

A practical voltage source is an ideal source E in series with Rs:

V = E − I·Rs

  • At no load (I = 0), V = E.
  • As the load current increases, the internal drop I·Rs increases and the terminal voltage falls linearly.
  • At short circuit, V = 0 and I = E/Rs (the maximum current).
  • The V-I characteristic is a straight line with slope −Rs. The larger Rs is, the steeper the fall.
  V
  E |-------------------  ideal (horizontal line at E)
    |  \
    |     \               practical (slope = -Rs)
    |        \
    +-----------\------- I
               E/Rs

Voltage regulation

Regulation = (V_no load − V_full load)/V_full load. It is small when Rs is small, so a good voltage source has a very small internal resistance.

Example: E = 24 V, Rs = 2 Ω. At 3 A the terminal voltage is 24 − 6 = 18 V, not 24 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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