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Voltage and current sources, terminal voltage

Short answer

A practical voltage source has internal series resistance Rs; terminal voltage V = E − I·Rs. As load current increases, the internal drop I·Rs increases, so V decreases. A practical current source has internal parallel resistance.

Discuss voltage and current sources in brief. Also justify the statement "Terminal voltage goes on decreasing on increasing load current."

Voltage source

A voltage source supplies a given voltage to a load. An ideal one keeps the voltage constant for any load current (internal resistance zero). A practical voltage source is an ideal source E in series with an internal resistance Rs.

Current source

A current source supplies a given current to a load. An ideal one keeps the current constant for any load voltage (internal resistance infinite). A practical current source is an ideal source Is in parallel with a resistance Rp.

 Practical V source        Practical I source
  o--[Rs]--+--o             o--+--------+--o
   (E)    Load               (Is)       [Rp]  Load
  o--------+--o             o--+--------+--o

Conversion: Is = E/Rs and Rp = Rs (and E = Is·Rs).

Why terminal voltage falls as load current increases

For a practical voltage source:

V_terminal = E − I·Rs

E and Rs are fixed. When the load current I rises, the drop I·Rs inside the source rises, so the terminal voltage falls. The V-I characteristic is a straight line sloping downward from E (at no load) to zero (at short circuit, where I = E/Rs).

  V
  E |\
    |  \
    |    \      slope = -Rs
    |      \
    +--------\------- I
            E/Rs

Example: E = 12 V, Rs = 0.5 Ω. At 2 A, V = 12 − 1 = 11 V; at 10 A, V = 12 − 5 = 7 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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