When a dc supply such as the generator, battery is used, you initially assume that it will provide the desired voltage for
any resistive load no matter what load you may apply. Unfortunately,
this is not always the case. For instance, if we apply a 1 kΩ resistor
to a dc laboratory supply, it is fairly easy to set the voltage across
the resistor to 20 V. However, if we remove the 1 kΩ resistor and
replace it with a 100Ω resistor and don't touch the controls on the supply
at all, we may find that the voltage has dropped to 19.14 V. Change
the load to a 68 Ω resistor, and the terminal voltage drops to 18.72 V.
We discover that the load applied affects the terminal voltage of the supply.
The reason the terminal voltage drops with changes in load (current
demand) is that
The resistance level depends on the type of supply,
but it is always present. Every year new supplies come out that are less
sensitive to the load applied, but even so, some sensitivity still remains.
Due to the internal resistance of the supply, the ideal internal supply must be set to some value above the fixed value. The internal resistance will capture extra value of the applied voltage. Lets explain the idea of internal resistance and ideal voltage value with the following example.
**Fig.no.1: **Ideal and practical voltage sources.
The current in the circuit is determined considering the ideal voltage in Fig.no.1(a). by simply looking at the load and using Ohm's law; that is,
$ I_L = {20V \over 1000Ω} =20mA$, which is a relatively low current.
But there is not practically available any ideal voltage source, again every voltage source has some internal resistance as shown in Fig.no.1(b). The internal resistance of the battery in the figure is assumed to be 5Ω. To calculate current flow in the circuit, now we have to count the value of internal resistance as well to identify the effect of internal resistance of voltage source in the circuit.
$$ I_L = {20V \over 1000Ω+5Ω} =19.9mA$$
The voltage drop across the load resistance can be equal to $V_L = 1000Ω \times 19.9mA = 19.9V$, which shows that internal resistance of a voltage source effects the load voltage drop to some value depends upon the load resistance. Increasing the value of load resistance will make a sense of ignoring the internal resistance and relying on the load resistance only. But reducing the load resistance, will have a significant effect on the series current in the circuit, thus effect the load terminal voltage.
To overcome this effect, an extra voltage source value is added to the voltage source. Hence the effect of internal resistance is traded off due to that extra added voltage value to the source voltage.
For instance, In the above example, if the source voltage is made 20.1V, and Ohms law is applied again in the same way. The terminal voltage across the load resistance will now be equal to the total of 20 volts, which is our required voltage to be across the load resistance.
$$ I_L = {20.1V \over 1005Ω} = 20mA$$
$$ V_L = 20mA \times 1000Ω = 20V$$
#### What is no load and full load conditions

**Fig.no.3: ** No Load vs Full Load Conditions
#### How to find the internal resistance of a voltage source?

The internal resistance can be find through a test under no-load and full load conditions. Under No-Load conditions, the supply on output terminals gives full supply voltage with zero current due to open circuit shown in Fig.no.2(a). As we close the circuit with some output load resistance, the current level increases and the output terminal voltage drops similarly. The plot is shown in Fig.no.2(b), where the x-axis shows current and y-axis gives output terminal voltage of the supply.
**Fig.no.2: ** Plotting VL versus IL for the supply.
The slope of the line is defined by the internal resistance of the supply. That is,

which for the plot in Fig.no.2(b) results in
$$ \begin{array} {rcl}R_{int} &=& {\Delta V_L \over \Delta I_L}\\
& = &{20.1 V- 18.2 V \over 275.2mA - 0A}\\
&=&6Ω \end{array}$$
For supplies of any kind, the plot of particular importance is the output
voltage versus current drawn from the supply. Note that the maximum value is achieved under no-load (NL) conditions. Full-load (FL) conditions are defined by the maximum current the supply can provide on a
continuous basis.

(a) Ideal voltage source

(b) Battery with internal resistance

(a)

(b)

(a)

(b)

$$\bbox[5px,border:1px solid red] {\color{blue}{ R_{int} = {\Delta V_L \over \Delta I_L}}}$$ | Eq.(1) |