Internal Resistance of a Voltage Source. What is it?

May 122026

Internal Resistance of a Voltage Source

Voltage sources, whether batteries, generators, or others, are not ideal (perfect). A real voltage source is composed of an ideal voltage source in series with a resistance called internal resistance. This resistance does not really exist so that we can see it. It is a resistance deduced by the behavior of the real voltage sources.

See the diagrams of the ideal voltage source and the actual voltage source below.

VI = Internal resistance voltage.
VL = Load resistance voltage.
RI = Internal resistance.
RL = Load resistance.

Taking the following values: I = 4 amperes, RI = 3 ohms, and RL = 5 ohms.

Voltage drop on the internal resistance: VI = I x RI = 4 amps x 3 ohms = 12 volts
Voltage drop on the load resistor: VL = I x RL = 4 amps x 5 ohms = 20 volts

The total voltage drop will be VI + VL = 12 V + 20 V = 32 volts (equal to the voltage of the ideal source) (Kirchhoff voltage law).

It can be clearly seen that only 20 of the 32 volts are applied to the load (RL); the remaining voltage is lost in the internal resistance. Frequently, this voltage (VL = 20 volts) is called terminal voltage because it is measured at the terminals of the voltage source.

How is internal resistance obtained?

To obtain the internal resistance in a voltage source, these steps must be followed:

The voltage at the terminals of an uncharged voltage source is measured. The measured voltage will be VNL (voltage without load)
A load is connected to the voltage source, and the voltage is measured. The measured voltage will be VL (voltage with load)
The current of the source with a load is measured. The measured current will be I

Once these values are available, the following equation is applied: RI = (VNL – VL) / I (1)

For example:

If VNL = 12 volts, VL = 11.8 volts, and I = 10 amperes, then the internal resistance is RI = (12V – 11.8V) / 10A = 0.02 ohms.

From the last equation (1), we can get VL = VNL – (I x RI), so…

If the electric current I grows to 25 amps (meaning that the load resistor has a smaller value), the load voltage is VL = 20 volts – (25 amps x 0.02 ohms) = 11.5 volts.

We can conclude that the more electric current the load (RL) demands, the lower the terminal voltage due to the greater voltage drop in the internal resistance (RI).

Power transfer between a voltage source and a load is at its most efficient when the resistance of the load matches the internal resistance of the voltage source.

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