- 2071 Chaitra · 6 marks
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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