Consider an ac source. Let VS be its open-circuit voltage (i.e., the voltage which exists across its terminals when nothing is connected to it), and ZS be its internal impedance. Let it be connected to a load impedance ZL, whose value can be varied, as shown in Fig. 1.
Now, suppose ZL, is infinite. It means that the terminals AB of the source are open-circuited. Under this condition, no current can flow. The terminal voltage VT is obviously the same as the emf Vs, since there is no voltage drop across ZS. Let us now connect a finite variable load impedance ZL, and then go on reducing its value. As we do this,the current in the circuit goes on increasing. The voltage drop across ZS also goes on increasing. As a result, the terminal voltage VT goes on decreasing.
For a given value of ZL, the current in the circuit is given as
𝐼= 𝑉𝑆 / (𝑍𝑆+𝑍𝐿)
Therefore, the terminal voltage of the source, which is the same as the voltage across the load, is
𝑉𝑇= 𝐼×𝑍𝐿=𝑉𝑆 /(𝑍𝑆+𝑍𝐿) × 𝑍𝐿= 𝑉𝑆 / (1+𝑍𝑆/𝑍𝐿)
From the above equation, we find that if the ratio 𝑍𝑆𝑍𝐿 is small compared to unity, the terminal voltage 𝑉𝑇 remains almost the same as the voltage VS. Under this condition, the source behaves as a good voltage source. Even if the load impedance changes, the terminal voltage of the source remains practically constant (provided the ratio 𝑍𝑆𝑍𝐿 is quite small). Such a source can then be said to be a "good (hut not ideal) voltage source"
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